matter and interaction chapter 03 solutions

7/25/2019 Matter and Interaction Chapter 03 Solutions

1/60

1

3.X.1 According to the Principle of Reciprocity,

Fon star by planet

= Fon planet by star

= 4e25N

3.X.2According to Newtons gravitational force law, the magnitude of the gravitational force on a planet by a star is directlyproportional to the product of the mass of the star and the mass of the planet. Thus, increasing the mass of the planet by afactor of 2 will increase the gravitational force on the planet by a factor of 2. Thus,

Fon planet by star

= (2)(4e25 N)= 8e5 N

3-X-3According to Newtons law of gravitation, the magnitude of the gravitational force on a planet by a star is inverselyproportional to the distance between them squared. Tripling the distance will decrease the magnitude of the force by a factorof32. In other words, the gravitational force is smaller by a factor of(1/3)2 = 1/9.

Fgrav, on planet by star

= 19

(8e20 N)

8.89e19 N

3.X.4 According to Newtons law of gravitation, the gravitational force on M bym is directly proportional to the productof their masses and inversely proportional to the distance between them squared.

Tripling the mass of m will increase the force by a factor of 3.

Increasing the distance by a factor of 4 will decrease the force by 42, making the force smaller by (1/4)2 = 1/16.

Thus the objects will attract with a force.

F =

3

16

F

3.X.5 According to Newtons law of gravitation

Fgrav

= G m1m2r2,1

2=

6.6742 10

11 N m2

kg2

(3 kg)(5 kg)

(2 m)2

= 2.5e-10 N


2/60

2

3.X.6According to Newtons law of gravitation, the magnitude of the gravitational force on an object by Earth if the objeis on Earths surface is

F

grav

= Gm

Earthm

object

R2

Solve for the mass of Earth.

mEarth

=

Fgrav

R2Gm

object

= (9.8 N)(6.4e6 m)2

(6.6742 1011 N m2

kg2 )(1 kg)

= 6.0e24 kg

3.X.7

(a) Since we are asked for the force on Ganymede by Europa, the vectorrpoints from Europa (agent) to Ganymede (objecwhich is < 1, 0, 0>.

(b) r= < 1, 0, 0>=.

(c) It is the same as r, or < 1, 0, 0>.

In calculating the gravitational force on Europa by Ganymede:

(a) Since we are asked for the force on Europa by Ganymede, the vectorrpoints from Ganymede (agent) to Europa (objecwhich is < 1, 0, 0>.

(b) r= < 1, 0, 0>=.

(c) It is the same as r, or < 1, 0, 0>.

3.X.8

(a)

rME

= rM

rE

=

2.8 108, 0, 2.8 108

m 0, 0, 0 m

= 2.8 108, 0, 2.8 108 m(b)

rME

=

rME

x

2+

r

MEy

2+

r

MEz

2

=

(2.8 108 m)

2+ (0 m)

2+ (-2.8e8 m)

2

= 4.0e8m


3/60

3

(c)

rME

=r

MErME

=

2.8 108, 0, 2.8 108

m

4.0e8m=

(d)

FM,E

= Fg rav o n M by E

= Gm

Mm

ErM-E

2 rME=

6.6742 10

11 N m2

kg2

(7e22kg)(6e24kg)

(4.0e8m)2 ()

= 7.0e28N < 0.7, 0, +0.7>

= -4.9e28, 0, 4.9e28 N

Note that the force points from Moon toward Earth, exactly as expected for the force on Moon by Earth.

3.X.9

(a) The magnitude of the gravitational field at a distance|r| from a spherical body of mass M is

|g| = GM

|r|2

At the surface of Moon, |r|= R, the radius of Moon.

|g| =

6.6742 1011

N

m2

kg2

7e22kg(1.75e6m)2

= 1.5 N/kg

(b) The magnitude of the gravitational force on an object of massm by Moon is

Fgrav

= m |g|= (70 kg)(1.5 N/kg)

= 105 N

(c) On Earth

Fgrav

= m |g|= (70 kg)(9.8 N/kg)

= 686 N


4/60

4

3.X.10 The magnitude of the gravitational field at a distance|r| from a spherical body of mass M is

|g| = GM

|r|2

At the surface of Mars, |r|= R, the radius of Mars.

|g| =

6.6742 10

11 N m2

kg2

6.4e23kg

(3.4e6m)2

= 3.7 N/kg

3.X.11 The gravitational force on a person by Earth near its surface is

Fgrav, on person by Earth

= mperson

gEarth

(60 kg)(9.8 N/kg)

590 N

According to the Principle of Reciprocity,

Fgrav, on Earth by person

= Fgrav, on person by EarthFgrav, on Earth by person = Fgrav, on person by Earth F

grav, on Earth by person

590 N

3.X.12 The net force on the comet is toward Sun. Sketch the net force and its perpendicular and parallel components location B, as shown in Figure1.

Sun

A

B

pi

p

Fgrav

F

F

Figure 1: The net force on the comet at position B.


5/60

5

p

pi

pf

Figure 2: The final momentum after one time step, starting at position B.

The parallel component of the net force on the comet is opposite the momentum of the comet. As a result, the magnitudeof the momentum will decrease and the comet will slow down. This can be seen by sketching the momentum, change inmomentum, and final momentum after one step. It helps to draw the vectors on their own diagram, as shown in Figure2.

You can see in Figure 2 that the final momentum after one time step is smaller (i.e. shorter in length) than the initialmomentum at position B. This means that the comet slowed down, which is consistent with our earlier conclusion fromexamining the direction of the parallel component of the net force.

The perpendicular component of the net force is mostly upward in the +y direction (if +y is defined toward the top of the

page). As a result the net force changes the direction of the momentum, making the comet curve upward toward the topof the page along this part of the ellipse.

3.X.13 The magnitude of the electric force of one charged particle on another charged particle is given by Coulombs law.Substitute known values for charge and distance to calculate the electric force.

Felec = 1

4o

|q1q2|

|r|

2

= 1

4o

|qelectron

qproton

|

|r|2

=

9 10

9 N m2

C2

|(1.602 10

19

C)(1.602 1019

C)|

1e-10m2

= 2.3 108 N

Coulombs law is consistent with the Principle of Reciprocity. The magnitude of the force on the electron by the proton isequal to the magnitude of the force on the proton by the electron.

3.X.14

(a) The magnitude of the electric force of one charged particle on another charged particle is given by Coulombs law.Substitute known values for charge and distance to calculate the electric force. The charge of the gold nucleus is 79e =(79)(1.602 10

19

C).


6/60

6

Felec by gold nucleus on electron

= 14

o

|q1q2|

|r|2

= 1

4o

|qgold nucleus

qelectron

|

|r|2

=

9 10

9 N m2

C2 |(79)(1.602 1019 C)(1.602 1019 C)|

3e-9m2

= 2.0 109 N

(b) Coulombs law is consistent with the Principle of Reciprocity. The magnitude of the force on the electron by the gonucleus is equal to the magnitude of the force on the gold nucleus by the electron. Thus,

Felec by electron on gold nucleus

= Felec by gold nucleus on electron

= 2.0 109 N

3.X.15

Si: nprotons

= 14 nneutrons

= 28 14 = 14

Sn: nprotons

= 50 nneutrons

= 119 50 = 69

Au: nprotons

= 79 nneutrons

= 197 79 = 118

Th: nprotons

= 90 nneutrons

= 232 90 = 142

Examine the difference between the number of neutrons and number of protons for a given atom. For Si, it is 0. For Th,is142 90 = 52. It definitely increases with the number of protons in the atom.

3.X.16 Name the four stars as A, B, C, and D. The net force on star A is:

Fnet, A

= Fon A by B

+Fon A by C

+Fon A by D

All three of the above forces must be calculated directly from Newtons law of gravitation based on the positions of B, and D relative to A.

The net force on star B is

Fnet, B

= Fon B by A

+Fon B by C

+Fon B by D

But the force Fon B by A

=Fon A by B

can be obtained easily from what was already calculated by simply reversing the vect

ofFon A by B

. Thus, only two forces must be calculated directly from Newtons law of gravitation based on the positions of

and D relative to B. The other force is obtained using the Principle of Reciprocity.


7/60

7

The net force on star C is

Fnet, C

= Fon C by A

+Fon C by B

+Fon C by D

The first two forces were already computed and can be obtained by the Principle of Reciprocity. Only the last force must be

computed from Newtons law of gravitation.For star D, all forces on star D were already computed and can be obtained by the Principle of Reciprocity.

Thus, the total number of forces to be calculated fully from Newtons law of gravitation are 3 + 2 + 1 = 6, and the other 6forces are obtained by simply reversing the vectors of those forces already computed (i.e. from the Principle of Reciprocity).

3.X.17

(a) Use the non-relativistic approximation for the momentum of the blocks. The momentum of the two-block system is

psys

= p1

+ p12

= m1 v1 +m2 v2= (1 kg)5, 2, 0 m/s + (3 kg)3, 4, 0 m/s

= 5, 2, 0 kg m/s + 9, 12, 0 kg m/s

= 4, 14, 0 kg m/s

(b) According to the momentum principle, the momentum of an isolated system doesnt change.

psys

= 0

pf

= pi

pf = 4, 14, 0 kg

m/s

Therefore, the momentum of the system at any later time is 4, 14, 0 kg m/s.

3.X.18 The center of mass is a weighted average of the positions of the objects, with the mass of each object as theweighting factor.

rcm

=m

1r

1+m

2r

2

m1+m2

= (10 kg)3, 0, 0 m + (2 kg)8, 2, 0 m10 kg + 2 kg

= 30, 0, 0 kg m + 16, 4, 0 kg m

12 kg

= 46, 4, 0 kg m

12 kg

= 3.8, 0.33, 0 m


8/60

8

Indeed, the center of mass position3.8, 0.33, 0 m is fairly close to the position of the 10 kg ball which is at 4, 0, 0 m.

3.X.19 The center of mass velocity can be found from the momentum of the system.

psys

= p1

+ p12

Also,

psys

= Mtotal

vcm

Equate the above expressions for the momentum of the system and solve for the center of mass velocity.

Mtotal

vcm

= p1

+ p12

= m1

v1

+m2

v2

= (2 kg)3, 4, 0 m/s + (5 kg)2, 6, 0 m/s

= 6, 8, 0 kg m/s + 10, 30, 0 kg m/s

= 4, 38, 0 kg m/s

vcm

=p

sys

Mtotal

= 4, 38, 0 kg m/s

2 kg + 5 kg

= 0.57, 5.4, 0 m/s

3.X.20

(a) System is ball A.The change in the momentum of the system is the change in the momentum of ball A.

psys

= pA,f

pA,i

psys

= mA

vA,i

psys

= mA

v

psys

= Mv

According to Conservation of Momentum, the change in momentum of the surroundings is

psurr

= psys

= (Mv)

= Mv


9/60

9

(b) System is ball B.The change in the momentum of the system is the change in the momentum of ball B.

psys

= pB,f

pB,i

psys

= mB

vB,f

< 0, 0, 0>

psys

= Mv

According to Conservation of Momentum, the change in momentum of the surroundings is

psurr

= psys

= (Mv)

= Mv

(c) System is both balls. The total momentum of the system is the sum of the momentum of both balls.

psys,i

= pA,i

+ pB,i

= mA

vA,i

+< 0, 0, 0>

= Mv

psys,f

= pA,f

+ pB,f

= +mB

vB,f

= Mv

The change in the momentum of the system is

psys

= psys,f

psys,i

= Mv Mv

= 0

The momentum of the system did not change. It was constant. Therefore, the change in momentum of the surroundingsis also zero, according to Conservation of Momentum.

psurr

= psys

= 0

3.X.21


10/60

10

(a) Which object in the surroundings interacts with you as you are falling? Earth.

(b) Define the system as you and Earth. There is nothing in the surroundings that exert a force on the system, exceperhaps Sun and Moon. However, these can be neglected. Since the net external force on the system is zero, tmomentum of the system is constant. This comes from the momentum principle.

p =

Fnet tp

sys,f p

sys,i= 0

psys,f

= psys,i

The total momentum of the system does not change. It is constant.

The momentum of the system of you and Earth also equals zero since the net external force on the system is zero. Thuas your momentum increases in the -y direction as you are falling, Earths momentum increases in the +y direction ait rises to meet you. Your momentum and Earths momentum have the same magnitude, but opposite direction. the total momentum is zero and remains zero throughout the motion.

3.X.22

(a) The total initial momentum of the system is the sum of the momenta of the two balls. Name them A and B.

psys,i

= pA,i

+ pB,i

= mA

vA,i

+mB

vB,i

= 0.3 kg4, 3, 2 m/s + 0.5 kg2, 1, 4 m/s

= 1.2, 0.9, 0.6 kg m/s + 1, 0.5, 2 kg m/s

= 2.2, 0.4, 2.6 kg m/s

(b) The system is near Earth, so the gravitational force on the system is

Fgrav by Earth on sys

= mg

=

=

= N

(c) Use the momentum principle

pf

= pi

+Fnet

t

= 2.2, 0.4, 2.6 kg m/s + ( N)(0.1 s)

= 2.2, 0.4, 2.6 kg m/s + kg m/s

= 2.2, 1.2, 2.6 kg m/s


11/60

11

Note that the x and z momenta of the system are constant (i.e. they did not change) since the net force on the system isonly in the y-direction.

3.X.23 According to the momentum principle,

Fnet

= p

t

= mv

cm

t= ma

cm

acm

=F

net

m=

300, 500, 200 N

100 kg

= 3, 5, 2 m/s2

3.X.24The total initial momentum of the system is the sum of the momenta of the two lumps of clay. Name them A and B.

psys,i = pA,i + pB,i= m

Av

A,i+m

Bv

B,i

= 0.02 kg(5, 2, 3 m/s) + 0.02 kg(3, 0, 2 m/s)

= 0.1, 0.04, 0.06 kg m/s + 0.06, 0, 0.04 kg m/s

= 0.16, 0.04, 0.1 kg m/s

According to the momentum principle, during the small time interval of the impact, the momentum of the system is approx-imately.

pf

= pi

+

0F

nett

pf

= 0.16, 0.04, 0.1 kg m/s

Its velocity is


12/60

12

psys,f

= Mtotal

vcm,f

Mtotal

vcm

= 0.16, 0.04, 0.1 kg m/s

vcm

= 0.16, 0.04, 0.1 kg m/s

Mtotal

vcm = 0.16, 0.04, 0.1 kg

m/s0.2 kg + 0.2 kg

vcm

= 0.4, 0.1, 2.5 m/s

3.X.25 The total initial momentum of the system is the sum of the momenta of the two rocks. Name them A and B.

psys,i

= pA,i

+ pB,i

= 10, 20, 5 kg m/s + 8, 6, 12 kg m/s

= 2, 14, 7 kg m/s

According to the momentum principle, since the net external force on the system is zero, the momentum of the system constant. So the momentum after the collision is

pf

= pi

+

0F

nett

= 2, 14, 7 kg m/s

3.X.26

(a) Neutron decay is governed by the weak interaction.

(b) Proton-neutron attraction is governed by the strong interaction.

(c) Earth-Moon attraction is governed by the gravitational interaction.

(d) Proton repulsion is governed by the electromagnetic interaction.

3.X.27 The magnitude of the gravitational interaction between two particles is proportional to the product of the particle

masses. Changing either mass by a factor n will change the product, and thus the magnitude of the gravitational interactioby the same factor n. In this example, if the satellites mass triples (changes by a factor of three), then the magnitude the gravitational interaction also triples and becomes 9e23N. The magnitude of the gravitational interaction between twparticles is inversely proportional to distance squared. Changing this distance by a factor n will change the magnitude by tsame factor, but inversely. If distance increases, then magnitude decreases. If distance decreases, then magnitude increaseIn this example, the distance increases by a factor of three, so the magnitude would decrease by a factor of nine and becom13

e23N.


13/60

13

3.X.28The magnitude of the gravitational interaction between two particles is inversely proportionalto the distance betweenthe particles. If the distance increases by a factor n, the magnitude decreases by a factor n2. If the distance decreases by afactorn, the magnitude increases by a factor n2. In this example, increasing the distance by a factor of three will decreasethe magnitude by a factor of nine.

3.X.29The magnitude of the gravitational interaction between two particles is inversely proportionalto the distance betweenthe particles. If the distance increases by a factor n, the magnitude decreases by a factor n2. If the distance decreases by afactorn, the magnitude increases by a factor n2. In this example, decreasing the distance by a factor of four will increasethe magnitude by a factor of sixteen. (Note that increasing by a factor of 1

4is equivalent to decreasing by a factor of4.)

3.X.30 Here are the relative position unit vectors at each point:

rA

= 1, 0, 0

rB

= 0, 1, 0

rC

= 1, 0, 0

rD

= 0, 1, 0

Here are the force unit vectors at each point:

FA

= rA

=1, 0, 0

FB

= rB

=0, 1, 0

FC

= rC

=1, 0, 0

FD

= rB

=0, 1, 0

Manipulating directions like this may seem strange at first, but it is perfectly valid.

3.X.31

(a) This is a straightforward numerical application. Proceed as follows:

Fgrav on Mer by Sun

= G mMer mSunrMer,Sun

2=

6.6742 10

11 N m2

kg2

(3.3e23kg) (2e30kg)

(4.8e10 m)2

2e22N

(b) By reciprocity, we must have

Fgrav on Mer by Sun

=Fgrav on Sun by Mer

2e22N.

Think about the problem this way. Between Mercury and Sun, there can be only one gravitational interaction, so therecan be only one magnitude associated with that interaction.

3.X.32 According to Newtons law of gravitation, the magnitude of the gravitational force on an object by Jupiter if theobject is on Jupiters surface is


14/60

14

Fgrav

= G mJupiter mobjectR2

Solve for the mass of Jupiter.

mJupiter

=

Fgrav

R2Gm

object

= (24.9 N)(71.5e6 m)2

(6.6742 1011 N m2

kg2 )(1 kg)

= 1.9e27 kg

3.X.33

(a) The relative position vector of the star, relative to the planet, is

rfrom planet to the star

= rstar

rplanet

=

2 1011, 3 1011, 0

m

5 1011, 2 1011, 0

m

=

7 1011, 5 1011, 0

m

(b) The distance is


= (7 1011 m)2 + (5 1011 m)2 + (0 m)2= 8.6e11m

(c) The direction of the relative position vector of the star, relative to the planet, is


=r

from planet to the starrfrom planet to the star

=

7 1011, 5 1011, 0

m

8.6e11m=

(d) Newtons law of gravitation is

Fgrav on planet by star

= G mplanet mstar|r|

2

=

6.6742 10

11 N m2

kg2

(4e24kg)(5e30kg)

(8.6e11m)2

= 1.8e21N


15/60

15

(e) Because of the Principle of Reciprocity, the planet and star exert equal magnitude forces on one another. Thus,


= Fgrav on star by planet

= 1.8e21N

(f) The (vector) force exerted on the planet by the star is

Fon the planet by the star

=F

on the planet by the star

rfrom planet to star

= (1.8e21N) < 0.81, 0.58, 0>

=

1.5 1021, 1.0 1021, 0

N

(g) The (vector) force exerted on the star by the planet is

Fon the star by the planet

=F

on the star by the planet

(rfrom planet to star

)

= (1.8e21N)(< 0.81, 0.58, 0>)

=

1.5 10

21

, 1.0 10

21

, 0

N

3.X.34 First, sketch a rough picture so that your answer can be compared to the picture. A sample is shown in Figure 3.

rfrom star to planet


Figure 3: A star and planet.

Apply Newtons law of gravitation.

The relative position vector of the planet, relative to the star, is


= rplanet

rstar

=

6 1011, 3 1011, 0

m

5 1011, 3 1011, 0

m

=

11 1011, 6 1011, 0

m


16/60

16

The distance between the star and planet is


= (11 1011 m)2 + (6 1011 m)2 + (0 m)2= 1.25e12m

The unit vector is


=r

from star to planetrfrom star to planet

=

11 1011, 6 1011, 0

m

12.5e11m=

Newtons law of gravitation is



2

=

6.6742 10

11 N m2

kg2

(4e24kg)(6e30kg)

(1.25e12m)2

= 1.0e21N

The (vector) force exerted on the planet by the star is


=F

on planet by star

(rfrom star to planet

)

= (1.0e21N)(< 0.88, 0.58, 0>)

=

0.88 1021, 0.58 1021, 0

N

=

8.8 1020, 5.8 1020, 0

N

Note that this points to the right and downward which is in agreement with the force vector drawn in Figure 3.

3.P.35This problem is an exercise in estimation and approximation. We first have to assume that we can treat the two bookas particles so we can apply Newtons expression for gravitational interaction without any modifications. Two books standinright next to each other on a bookshelf have centers separated by, to the nearest order of magnitude, 1e-2m. What abothe books masses? Well, if the books are old fashioned physics books, they probably have masses on the order of1e0kwhich is just a fancy way of saying a few kilogram. We can now estimate the magnitude of the books mutual interaction

Fbook on book

GMbook Mbookr

book,book

2 G

M2bookr

book,book

2 6.6742 1011 N m2

kg2

(1 kg)2

(1e-2m)2

6.6742e-7N


17/60

17

Note that its the order of magnitude thats important here. Multiples and factors of two or three will not change the orderof magnitude. Now lets estimate the magnitude of a books gravitation interaction with Earth. Obviously we need Earthsmass,6 10

24

kg, and well approximate the book-Earth distance to be simply Earths radius,6.4 106

m.

Fbook on Earth G

Mbook

MEarthrbook,Earth 2

6.6742 10

11 N m2

kg2 (1 kg)(6

1024

kg)

(6.4 10

6

m)2 9.78 N

Now we can compare the two magnitudes by taking their ratio.

Fbook on Earth

Fbook on book

9.78 N

6.6742e-7N1.47e7

So the book-Earth interaction is approximately ten million times stronger than the book-book interaction.

3.P.36

(a) If Mathilde were not present, the NEAR spacecraft would travel through space with constant momentum. However,Mathilde will interact gravitationally with NEAR, deflecting the spacecrafts path slightly toward the asteroid as thespacecraft approaches and passes the asteroid. The interaction will probably have more effect on NEARs directionthan its speed.

Mathilda

NEAR

104

m/s 1200 km

d = 2400 km

nearlycontstantvelocity

Figure 4: Approximate deflection of NEAR as it interacts with Mathilde.

(b) NEARs change in momentum resulting from this encounter can be approximated by estimating the net force on NEAR

due to Mathilde, FNEAR,Mathilde

, along with an estimate of the duration, t, of the interaction. The magnitude and the

direction of the force on NEAR both vary with time, so we have no choice but to use rather crude estimates. Letstake t represent the duration of NEARs travel across the 2400 km distance in the diagram. Lets take the force tohave an approximate magnitude corresponding to a distance of1200 km. Since were really only interested in NEARsdeviation from an otherwise straight trajectory, lets also treat the force as having only ay component, which is anothervery crude approximation. But wait! We also need an estimate of Mathildes mass. We can get this from its assumedmass density and rough physical dimensions.


18/60

18

MMathilde

(7e4m)(5e4m)(5e4m)(3e3kg/m3) 5.25e17kg

Now lets apply the momentum principle.

pNEAR,y

FNEAR,Mathilde,y

t

G MNEAR MMathilder

NEAR,Mathilde

2 dv

NEAR

6.6742 10

11 N m2

kg2

(805 kg)(5.25e17kg)

(1.2e6m)2

(2.4e6m)

(1e4m/s)

5 kg m/s

Be certain you understand where the negative sign came from!

(c) In one day, NEAR will travel approximately (1e4m/s)(24 60 60 s) 9e8m. As NEAR gets farther and farthaway from Mathilde, the y component of its momentum (and its velocity) is nearly constant. Therefore, in one dNEARs deviation from its initial trajectory will be approximately

yNEAR

p

NEAR,y

MNEAR

(1 day)

5 kg m/s

805 kg

(24 60 60 s) 500 m

(d) Astronomers observed a deviation significantly less than was predicted. Gravitational attraction is proportional mass, and mass is proportional to density, so a smaller deviation indicated a smaller mass. NEARs mass was constanso Mathildes mass must have been less than the predicted value.

3.X.37 At the surface of a planet, the magnitude of the gravitational field g is

gat surface

= G mplanetR2

Earth and Moon have different masses and different radii. Therefore, they have different gravitational field strengths at thesurfaces. Incidentally, the gravitational field of Moon at its surface is about 1/6 of Earths gravitational field at the surfaof Earth.

3.X.38At an altitude y, the distance from Earths surface is R + y. At this height, the gravitational field is 99% of the fieat Earths surface.

gat surface

= G mEarthR2

gat altitude y

= G mEarth(R+y)2


19/60

19

gat altitude y

= 0.99 gat altitude surface

G

mEarth

(R+y)2 = 0.99G

mEarth

R2

GmEarth(R+y)2

= 0.99GmEarth

R2

1(R+y)2

= 0.99 1R2

(R+y)2 = (1/0.99)R2

R2 +y2 + 2Ry = (1/0.99)R2

y2 + 2Ry+R2 (1/0.99)R2 = 0

y2 + 2Ry+R2(1 (1/0.99)) = 0

Substitute Earths radius and solve using the quadratic formula or the solvefunction on your calculator.

y2 + 2(6.4e6m)y+ (6.4e6m)2(1 (1/0.99)) = 0y2 + 2(6.4e6m)y+ 4.137e11m2 = 0

y = 3.2e4m

Check the answer by calculating g.

gat altitude y

= G mEarth(R+y)2

= 6.6742 1011 N m2

kg2 6e24kg

(6.4e6m + 32000)2

= 9.7 N/kg

and 9.79.8

= 99% as expected.

3.X.39 Assumptions:

Earth is treated as a particle (not a bad assumption given Earths nearly spherically symmetric mass distribution)

humans are treated as particles (a rather crude assumption, but not bad over large distances such as Earths radius)

neglect the distance between a humans center of mass and Earths surface, making the the distance from a humans

center of mass to Earths center of mass approximately Earths radius

humans mass is approximately60 kg(lots of variation here, but the important part is an order of magnitude of10 kg)

Fon human by Earth

G mhuman mEarthrhuman,Earth

2

6.6742 1011 N m2

kg2

6 1024 kg (60 kg)6.4 106 m

2 600 N


20/60

20

Now lets do the same calculation for two humans with equal masses, separated by 3 m. Note that Earth doesnt participain this interaction to the answer must not depend on Earths mass or radius.

Fon human by human

G mhuman mhumanr

human,human

2

6.6742 10

11 N m2

kg2

(60 kg)(60 kg)

(3 m)2

3e-8N

Finally, lets compare these two quantities by looking at their ratio.

Fon human by Earth

Fon human by human

600 N

3e-8N

6e2N

3e-8N2e10

The human-Earth interaction is approximately twenty billion times stronger than the human-human interaction.

3.P.40Lets assume this experiment takes place near Earths surface so we can approximate gravitational interactions betwe

Earth and object as

F

object,Earth m

objectg. Let t be the very short duration of contact, and assume that t is so sho

that the downward change in the balls momentum due to its interaction with Earth during contact is much smaller than tupward change in the balls momentum due to its interaction with the scale. The ball initially has a downward momentup

ball,i,y=p just before hitting the scale. Shortly after interacting with the scale, the ball has an upward momentump

ball,f

So the ball undergoes a change in momentum ofpball

=pball,f

pball,i

pball,f,y

pball,i,y

which, remembering that (a) we

assuming the ball rebounds to its initial height and (b) pball,i,y

is a negative number, is just 2pball,i

or just 2p. (No

thatp is a negative number!) This change in momentum is, remember, almost exclusively due to the balls interaction withe scale and is directed upward. Therefore, the average force on the ball due to its interaction with the scale must also upward.

LetT be the long time between the instants when the ball is at its maximum height (its really the period of the balls osclatory motion), neglecting any energy dissipation. The momentum gives us the magnitude of the average force immediate

asF

avg

|p

ball|

T 2p

T (remember p is a negative number!).

In each fall, the magnitude of the balls momentum increases from 0 to p (remember p is a negative number!) due to igravitational interaction with Earth. This happens within a duration ofT /2. Applying the momentum principle to the baldownward motion allows us to express T in terms of known quantities.

pball,y

p 0F

ball,Earth

t p m

ballg

T

2

Solving for T gives T 2pm

ballg

, and now we can get a final expression for the magnitude of the average force:

Favg 2p

T

2p 2pm

ballg

mball

g

This surprising result happens to be exactly the same magnitude of the ball-scale interaction if the ball were at rest on tscale! This unusual result will play a role later in our study of gases.

Alternatively, we can get this surprising result another way. Let Tbe the duration of the balls fall from rest back to thsame place (again at rest). The net change in the balls momentum is zero. Let F

avgbe the average force on the ball due

the scale. Apply the momentum principle.


21/60

21

pball

= Favg

T+ Fball,Earth

T

0 F

avg

T mball

gT

F

avg

mball

g

3.X.41 There is only one gravitational interaction between Earth and tennis ball, so there can be only one associatedmagnitude. Therefore, statement C is correct and the others are incorrect.

NOTE:This question is not carefully worded. Force is a vector, and one vector cant be greater than or less than another.Their magnitudes, however, can be so compared. The questions intent is to compare the magnitudes of the Earth-tennisball interaction and the tennis ball-Earth interaction.

3.P.42

Before the loop, you should

Define constants such as G

Specify the initial (vector) momentum of each object

Specify an appropriate value for the time step

Specify the mass of each object

Specify the initial (vector) position of each object

Inside the loop, you should

Calculate the (vector) forces acting on the objects

Update the (vector) momentum of each object

Update the (vector) position of each object

3.P.43

(a) Sketch a picture of the situation in 2-D, like the one shown in Figure5.

Apply Newtons law of gravitation. The relative position vector of the spaceship, relative to the asteroid, is

rfrom asteroid to spaceship

= rspaceship

rasteroid

=

3 105, 7 105, 4 105

m

9 105, 3 105, 12 105

m

=

6 105

, 10 105

, 8 105

m

The distance between the asteroid and spaceship is


= (6 105 m)2 + (10 1011 m)2 + (16 105 m)2= 1.41e6m


22/60

22

Fgrav on spaceship by asteroid

rspaceship rel to asteroid

Figure 5: The spaceship and asteroid. z-components of the positions are not shown.

The unit vector is


=r

from asteroid to spaceshiprfrom asteroid to spaceship

=

6 105, 10 105, 8 105

m

1.41e6m=

Newtons law of gravitation is


= G mspaceship masteroid|r|

2

=

6.6742 10

11 N m2

kg2

(1.4e3kg)(7e15kg)

(1.41e6m)2

= 3.29e-4N

The (vector) force exerted on the spaceship by the asteroid is


=F

on spaceship by asteroid

(rfrom asteroid to spaceship

)

= (3.29e-4N)(< 0.424, 0.707, 0.566>)

=

1.39 104, 2.32 104, 1.86 104

N

In the x-y plane, the force points to the right and downward which is in agreement with the force vector drawn Figure5.


23/60

23

(b) Apply the momentum principle

pf

= pi

+Fnet

t

pf

pi

= Fnett

pf

pi

= (

1.39 104, 2.32 104, 1.86 104

N)(538 s 532 s)

=

8.4 104

, 1.40 103

, 1.12 103

kg m/s

3.P.44

(a) Sketch a picture of the situation showing relevant vectors.

m1 m2

Figure 6: The positions of block 1 and block 2.

Find the relative position vector of block 2, relative to block 1.

r2 relative to 1

= r2

r1

= 18, 11, 0 m 7, 11, 0 m

= 11, 0, 0 m

The magnitude of the vector is easy in this case since its a one-dimensional vector.

r2 relative to 1

= 11 m

The gravitational force is given by Newtons law of gravitation.

Fgrav on 2 by 1

= G m1 m2|r|

2

=

6.6742 10

11 N m2

kg2

(40 kg)(1000 kg)

(11 m)2

= 2.2e-9N


24/60

24

(b) Use the momentum principle. Define the system as block 2.

p2,t

= p2,i

+Fnet

t

p2,f

= +(

2.2 109, 0, 0

N)(4.7 s 4.6 s)

p2,f

= 2.2 1010, 0, 0 kg m/s

(c) Define the system to be both blocks. Att = 4.6 s the momentum of the system is

psys,i

= p1,i

+ p2,i

= 0

The net external force on the system is zero, assuming its an isolated system. Thus,

psys,f

= psys,i

+

0F

nett

psys,f

= 0

The momentum of the system is the sum of the momenta of the blocks.

psys,f

= p1,f

+ p2,f

0 = p1,f

+ p2,f

p1,f

= p2,t

=

2.2 1010, 0, 0

kg m/s

=

2.2 1010, 0, 0

kg m/s

Block 2s momentum increases as it moves left. Block 1s momentum increases as it moves right. Their momenta aequal in magnitude, but opposite in direction so that the sum of their momenta is zero.

(d) p

1,f

=p

2,f

m

1v

1,f = m

2v

2,f

They have the same magnitude momentum, but their masses are different. The smaller mass has the greater speeSo, block 1 is moving faster at t = 4.7 s.

3.P.45


25/60

25

Star

Planet

Figure 7: The star and planet with relevant vectors

(a) Sketch a picture of the star and planet.

Calculate the net force on the planet, which is the gravitational force by the star.

r = rplanet

rstar

=

3 1012, 4 1012, 0

m

5 1012, 2 1012, 0

m

=

2 1012, 2 1012, 0

m

|r| =

(2 1012 )

2+ (2 1012 )

2+ (0 )

2m

= 2.83e12m

r = r

|r|

= 0.707, 0.707, 0



2

= (6.67 1011)(3e24kg)(7e30kg)

(2.83e12m)2

= 1.75e20N


26/60

26


=F

grav

(r)= (1.75e20N)(0.707, 0.707, 0)

=

1.24 1020, 1.24 1020, 0

N

Apply the momentum principle to calculate the momentum of the planet after 1e6s. Define the system to be t

planet.

pf

= pi

+Fnet

t

= mvi

+Fnet

t

= (3e24kg)(

0.3 104, 1.5 104, 0

m/s) + (

1.24 1020, 1.24 1020, 0

N)(1e6s)

=

9.12 1027, 4.49 1028, 0

kg m/s

Solve for the final velocity of the planet.

vf

=p

f

m

=

9.12 1027, 4.49 1028, 0

kg m/s

3e24kg

=

3.04 103, 1.496 104, 0

m/s

(b) The position of the planet is found by using the definition of average velocity.

rf

= ri

+ vavg

t

Use the approximationvavg

vf.

rf

=

3.004 1012, 4 1012, 0

m + (

3.04 103, 1.496 104, 0

m/s)(1e6s)

=

3.004 1012, 4.01 1012, 0

m

(c) The final velocity and final position are both approximate since the net force on the planet is not constant and sin

the final velocity is used in place of the average velocity. These approximations improve in the limit as t 0. Thuthe approximations would be worse for larger tlike 1e9 s.

3.P.46

(a) Sketch a picture of the two stars att = 0.

Apply Newtons law of gravitation to calculate the force on star 1.


27/60

27

Star 1

Star 2

Figure 8: The stars and relevant vectors

r = rstar 1

rstar 2

=

2.00 1012, 5.00 1012, 4.00 1012

m

2.03 1012, 4.94 1012, 3.94 1012

m

= 3 1010, 6 1010, 5 1010 m

|r| =

(3 1010 )

2+ (6 1010 )

2+ (5 1010 )

2m

= 8.367e10m

r = r

|r|

= 0.3586, 0.7171, 0.5976

Fgrav on 1 by 2

= G

mstar 1

mstar 2

|r|2

= (6.67 1011)(4e30kg)(3e30kg)

(8.367e10m)2

= 1.143e29N

Fgrav on star 1 by star 2

=F

grav

(r)= (1.143e29N)(0.3586, 0.7171, 0.5976)

=

4.100 1028, 8.200 1028, 6.833 1028

N

Apply the momentum principle to calculate the momentum of star 1 after 1e5 s. Define the system to be star 1.

pf

= pi

+Fnet

t

= mvstar 1, i

+Fnet

t

=

2.841 1035, 2.482 1035, 3.268 1035

kg m/s


28/60

28

Solve for the final velocity of the planet.

(b) The momentum of the first star is approximate because the net force on it is not constant, but rather it depends othe distance between the stars. The approximation improves as t 0.

(c)

rf

= ri

+ vavg

t

Approximate the average velocity as vavg

vf.

vf

=p

f

mstar 1

=

7.103 104, 6.205 104, 8.171 104

m/s

rf

=

2.00 1012, 5.00 1012, 4.00 1012

m + (

7.103 104, 6.205 104, 4.00 1012

m/s)(1 105 s)

=

2.007 1012

, 4.994 1012

, 3.992 1012

m

The final position was computed to four significant figures because for a small time interval, its displacement is quismall. Sometimes rounding to three or fewer significant figures will make it seem like there is no displacement at all

(d) This result is approximate because the velocity is changing andvavg

=vf. However, for small time intervals,v

avg v

3.X.47Each positively charged particle will attract the negatively charged particle. We are to assume that the particles aat the vertices of an isosceles triangle. If this is true, then the vertical components of each positive-negative interaction wnullify each other, while the horizontal components of each positive-negative interaction will reinforce each other. The nforce on the negatively charged particle will therefore be directed away from the negatively charged particle and will poitoward the left. Geometrically, the net force will lie along the perpendicular bisector of the line connecting the two positivecharged particles. Note that no formula is necessary to answer this question. The solution is in the geometry.

3.X.48 If the positively charged particle were at the midpoint of the line joining the two negatively charged particles, thnet force on it would be zero (a symmetry argument). But the symmetry is broken, and the positively charged particlecloser to the negatively charged particle on the right hand side. Thus, the interaction between the positively charged particand the right hand negatively charged particle dominated the net force on the positively charged particle. Therefore, the nforce will be toward the right. Note that no formula is necessary to answer this question. The solution is in the geometry

3.X.49 The direction of the electric force on the electron by the proton is toward the top edge of the page, along the linfrom the electron to the proton. Be careful about calling that directionup if your book is lying open flat on your desk! Al

be careful about calling that direction toward the protonbecause there are infinitely many such directions! The direction the electric force on the proton by the electron is toward the bottom edge of the page, along the line from the proton to thelectron. There can be only one electric interaction between the proton and the electron, so there can be only one magnitudTherefore, the magnitude of the proton-electron interaction must equal the magnitude of the electron-proton interaction.

The direction of the electric force on electron A by electron B is toward the bottom edge of the page, along the line connectinthe two electrons. The direction of the electric force on electron B by electron A is toward the top edge of the page, alothe line connecting the two electrons.

Note that no formula is necessary to answer this question. The solution follows directly from reciprocity.


29/60

29

3.X.50 There can be only one electric interaction between the alpha particle and the proton, so there can be only oneassociated magnitude. Therefore, statement C is correct and the others are incorrect.

3.P.51This is a straightforward problem, assuming we treat electrons as particles.

Felec,electrons

14

o

Qelectron

Qelectronr

electron,electron

2 1

4o

Q2electronr

electron,electron

2F

grav,electrons

G melectron melectronrelectron,electron

2 Gm2

electronrelectron,electron

2Now take the ratio of these two magnitudes and youll see that the distance dependence divides out, and that the comparisondepends only on fundamental physical constants!

Felec,electrons

Fgrav,electrons

=1

4o Q2

electron

Gm2electron

9 10

9 N m2

C2

1.602 10

19

C2

6.6742 1011 N m2

kg2

9.109 1031 kg

2 4e42

If the two electrons are initially at rest, they will experience both a gravitational attraction and an electrical repulsion.However, the electrical repulsion overwhelms the gravitational attraction by a factor of about 4e42! The net interaction willbe electrical repulsion.

3.P.52

(a)

r21 = r2 r1 =0.4, 0.4, 0 m 0.5, 0.2, 0 mr

21 0.9, 0.6, 0 m

(b)

r21

=

r21,x

2+

r21,y

2+

r21,z

2

((0.9))

2+ ((0.6))

2+ (0)

2

mr

21

1.08 m(c)

r21

=r

21r21 0.9, 0.6, 0 m

1.08 m

0.833, 0.556, 0

(d) Fgrav,21

= G m1 m2r21

2Fgrav,21

6.6742 1011 N m2kg2

(2e-3kg)2

(1.08 m)2

2.29e-16N


30/60

30

(e)

Fgrav,21

=F

grav,21

Fgrav,21

=F

grav,21

r21

F

grav,21 (2.29e-16N)(0.833, 0.556, 0) 1.91e-16, 1.27e-16 , 0 N

Make sure this result indicates an attractive interaction and agrees with your diagram.

(f) Felec,21

= 14

o

q1

q2

r21

2Felec,21

9 109 N m2C2

|(-2e-9C)(-4e-9C)|

(1.08 m)2

6.17e-8N

(g)

Felec,21

=F

elec,21

Felec,21

=F

elec,21

r21

F

elec,21 (6.17e-8N) (0.833, 0.556, 0) -5.14e-8, 3.43e-8, 0 N

Make sure this result indicates a repulsive interaction and agrees with your diagram.

(h) Felec,21

Fgrav,21

6.17e-8N

2.29e-16N2.69e8

(i) Both gravitational and electric interactions vary with distance squared, but the ratio of these two interactionsindependent of distance for a given pair of interacting particles. Therefore, the ratio of the magnitude of the electrforce to the magnitude of the gravitational force is still 2.69e8.

3.P.53

(a) Assuming a negligible change in the cars speed, the magnitude of the mosquitos momentum goes from0 to m |v| pmosquito

=m |v|.(b) There is only one interaction shared by car and mosquito, so there can be only one magnitude,

F.(c) Neglecting interactions with the road and air, the car and mosquito have changes in momentum that are equal

magnitude, but opposite in direction. Sop

car

= pmosquito

m |v|. Given the mosquitos small mass, this negligible to the magnitude of the cars initial momentum

pcar,i

=M|v|.(d) Although the car and mosquito experience the same force magnitude (different directions), the change in velocity

greater for the mosquito because the mosquito has a much smaller mass than the car.

pcar

= pmosquitop

car

= pmosquito

Mv

car

m vmosquito

vmosquito

Mm

vcar

Note that we used the Newtonian approximation for momentum.


31/60

31

3.X.54 Lets take system = both rocks + string. Were told the strings mass is negligible, which means the stringscontribution to the systems momentum is also negligible. Therefore, we only need account for the momenta of the tworocks. Lets also assume both rocks have velocities with magnitudes negligible compared to light speed, so we can use theNewtonian approximation for momentum.

psys

= prock 1

+ prock 2

+0pstring prock 1 + prock 2 m

rock 1v

rock 1+m

rock 2v

rock 2

(0.1 kg) 0, 5, 0 m/s + (0.25 kg) 7.5, 0, 0 m/s

0, 0.5, 0 kg m/s + 1.875, 0, 0 kg m/s

1.875, 0.5, 0 kg m/s

3.X.55

(a) The system for which the change in momentum is zero is that system on which the net force is zero. There is a nonzeronet force on the car. There is a nonzero net force on the bug. There is, however, no net force on the system consisting

of car + bug.

(b) There is only one interaction between the car and bug, and thus only one force relating these two entities. Force andimpulse (change in momentum) are directly proportional to one another, so there can be only oneimpulse relating thecar and the bug. Therefore, neither is greater than the other; they must be equal.

(c) If the magnitude of the bugs change in momentum is equal to that of the car, then their respective changes in velocitiesmust have different magnitudes. The magnitude of the change in velocity is inversely proportional to mass, so the bugmust experience the greater changein velocity than the car.

3.X.56 Choose system = both protons and apply the momentum principle to this system. Let the first proton mentionedbe 1 and the other proton be 2. Lets also assume velocities with magnitudes negligible to that of light, so we can use thenewtonian approximation for momentum.

psys,f

= psys,i

+

0F

nett

p1,f

+ p2,f

= p1,i

+

0p

2,i

p2,f

= p1,i

p1,f

= 3.4e-21, 0, 0 kg m/s 2.4e-21, 1.6e-21, 0 kg m/s

= 1.0e-21, -1.6e-21, 0 kg m/s

3.P.57

(a) Lets choose system = all clay and apply the momentum principle. Lets also use the Newtonian approximation formomentum.

psys,i

p1,i

+ p2,i

m1

v1,i

+m2

v2,i

psys,i

(0.03 kg)(3, 3, 3 m/s) + (0.03 kg) (3, 0, 3 m/s)

psys,i

0.09, 0.09, 0.09 kg m/s + 0.09, 0, 0.09 kg m/s 0, 0.09, 0.18 kg m/s


32/60

32

(b) Just after the collision, the clay exists as one lump whose mass is the sum of the individual masses. Apply the momentuprinciple to the system just after the collision.

psys,f

psys,f

(m1

+m2

)vf

But we cant evaluate this because we dont know vf. If, however, we assume there are no significant interactions o

our system, then the momentum principle predicts that psys,f

=psys,i

. Therefore,psys,f

= 0, 0.09, 0.18 kg m/s.

(c) Solve the above application of the momentum principle forvf.

psys,f

(m1

+m2

)vf

vf

p

sys,i

(m1

+m2

)

0, 0.09, 0.18 kg m/s

0.06 kg 0, 1.5, 3 m/s

3.P.58

(a) Lets choose system = car + truck + road and apply the momentum principle to the system just after the collisioassuming low speeds so we can use Newtons expression for momentum.

psys,i

mcar

vcar,i

+mtruck

vtruck,i

psys,f

psys,f

(mcar

+mtruck

)vf

The momentum principle predictspsys,i

=psys,f

so we can solve forvf.

psys,f

(mcar

+mcar

)vf

vf

psys,i(m

car+m

truck)

mcar vcar,i +mtruck vtruck,i

(mcar

+m2

)

vf

(2800 kg)40, 0, 0 m/s + (4700 kg)14, 0, 29 m/s

(7500 kg)

vf

1.12e5, 0, 0 kg m/s + 6.58e4, 0, 1.363e5 kg m/s

(7500 kg)

vf

6.16, 0, 18.2 kg m/s

(b) Since our choice of system included the road, the road cant exert an external force on the system and so has no effeon the systems momentum.

3.X.59

(a) You should choose the system on which the net force is zero. With net forces on the bullet and the block, the bechoice is the system consisting of bullet + block.

(b) If the net force on a system is zero, then the systems final and initial momenta must be equal. Therefore, statementis true.


33/60

33

3.X.60This is a nuclear decay problem, and so is governed by nuclear interactions. Choice of system is extremely importantin this problem. We cannot choose system = radium nucleus because the radium nucleus does not exist after the nuclearinteraction takes place. Similarly, we cannot choose system = alpha particle + radon nucleus because neither of these particlesexists prior to the nuclear interaction happening. The best choice of system is system = all particles. The systems initialstate is such that the total momentum is zero (only one stationary particle). Note that in the initial state, the systems totalenergy is notzero (stationary particles have energy). With no net force acting on our chosen system, the total momentum

must be zero in both the initial and final states. So if the newly created alpha particle moves in the +z direction in the finalstate, then the newly created radon nucleus must move in thez direction so as to conserve momentum.

3.X.61

(a) Choosing the system consisting of bowling ball + ping pong ball and applying the momentum principle to that systemtells us that the systems momentum is conserved. Therefore p

sys,z = 0.

(b) Choosing the system consisting of just the bowling ball and applying the momentum principle to that system tells usthat the net force on the bowling ball must be in the same direction as the bowling balls change in momentum. Theforce on the bowling ball by the ping pong ball is in the +z direction, and that must also be the direction of the bowlingballs change in momentum. Thus, for this system, p

sys,zhas a positive sign.

3.P.62

Assume the bullet travels in the+xdirection,vi

=300, 0, 0 m/s. Define the system to be the bullet and the block. Assumethat the net external force on the system is zero. Apply the momentum principle with the initial momentum being beforethe collision and the final momentum being after the collision.

psys,f

= psys,i

+

0F

nett

The momentum of the system is the sum of the momenta of the bullet and block. The final velocity of the bullet is the same

as the block.

pbullet,i

+ pblock,i

= pbullet,f

+ pblock,f

mbullet

vbullet,i

+mblock

0

vblock,i

= mbullet

vf

+mblock

vf

mbullet

vbullet,i

= (mbullet

+mblock

)vf

vf

=

m

bullet

mbullet

+mblock

v

bullet,i

vf

=

0.105 kg

0.105 kg + 2 kg

300, 0, 0 m/s

= 15, 0, 0 m/svf

= 15 m/s

3.P.63

(a)


34/60

34

psys,i

= pA,i

+ pB,i

= 20, 5, 0 kg m/s + 5, 6, 0 kg m/s

= 25, 1, 0 kg m/s

(b) The impulse on the system of the two objects is due to the net external force on the system. During the small timinterval of the collision, the net force on the system is approximately zero, so the impulse on the system is approximatezero.

(c)

psys,f

= pA,i

+

0F

nett

psys,f

= psys,i

= 25, 1, 0 kg m/s

(d)

psys,f

= pA,f

+ pB,f

pB,f

= psys,f

pA,f

= 25, 1, 0 kg m/s 18, 5, 0 kg m/s

= 7, 4, 0 kg m/s

3.P.64Define the system to be the two rocks. The net force on the system is zero, so

psys,f

= psys,i

+

0F

nett

psmall rock,i

+0p

large rock,i= p

small rock,f+ p

large rock,f

plarge rock,f

= psmall rock,i

psmall rock,f

= msmall rock

(vi

vf)

= 5 kg(0, 1500, 0 m/s 0, 1800, 0 m/s)

= 5 kg(0, 3300, 0 m/s)

= 0, 16500, 0 kg m/s

3.P.65

Define the system to be the two rocks, named A and B. The net external force on the system is zero. Apply the momentuprinciple.


35/60

35

psys,f

= psys,i

+

0F

nett

psys,f

= psys,i

pA,f

+ pB,f

= pA,i

+ pB,i

mA,f

vA,f

+mB,f

mB,f

= mA,i

vA,i

+mB,i

vB,i

vB,f

=m

A,iv

A,i+m

B,iv

B,i m

A,fv

A,f

mb,f

= (9 kg)(4100, 2600, 2800 m/s) + (6 kg)(450, 1800, 3500 m/s) (7 kg)(1300, 200, 1800 m/s)

8 kg

= 3138, 1750, 4200 m/s

3.P.66

Define the system as rock 1 and rock 2. Apply the momentum principle to the system. The net external force on the systemis zero, so

psys,f

= psys,i

+

0F

nett

psys,f

= psys,i

p1,f

+ p2,f

= p1,i

+ p2,i

m1

v1,f

+m2

v2,f

= m1

v1,f

+m2

v2,f

v2,f =

m1

v1,i

+m2

v2,i

m1

v1,f

m2

= (5 kg)(30, 45, 20 m/s) + (8 kg)(9, 5, 4 m/s) (5 kg)(10, 50, 5 m/s)

8 kg

= 16, 1.88, 5.38 m/s

3.P.67

Define the system as the two rocks. The net external force on the system is zero. Apply the momentum principle.

psys,f

= psys,i

+

0F

nett

psys,f

= psys,i

The momentum of the system is the sum of the momenta of the two rocks. The final velocity of each rock is the same sincethey stick together.


36/60

36

pA,f

+ pB,f

= pA,i

+ pB,i

mA

vf

+mB

vf

= mA

vA,i

+mB

vB,i

(mA

+mB

)vf

= mA

vA,i

+mB

vB,i

vf

=m

Av

A,i+m

Bv

B,i

mA +mB

= (15 kg)(10, 30, 0 m/s) + (32 kg)(15, 12, 0 m/s)

15 kg + 32 kg

= 13.4, 1.4, 0 m/s

3.P.68

First, sketch a picture of the situation showing the people and Earth before they jump and the people and Earth at thmoment the people leave the ground, as shown in Figure 9. Treat the people as a single particle.

Earth

People

Before people jump

Earth

People

Just after peopleleave the ground

Figure 9: Sketch of people and Earth before and just after they jump.

Define the system to be the people and Earth. The net external force on the system is zero. The system is initially at reso the initial momentum of the system is zero. Apply the momentum principle.

psys,f

= psys,i

+

0F

nett

psys,f

= psys,i

psys,f

= 0

The momentum of the system is the sum of the momenta of the people and Earth.


37/60

37

psys,f

= ppeople,f

+ pEarth,f

0 = ppeople,f

+ pEarth,f

pEarth,f

= ppeople,f

mEarth

vEarth,f

= mpeople

vpeople,f

vEarth,f

= mpeople vpeople,fm

EarthvEarth,f

=

mpeople

mEarth

vpeople,f

Assume that the mass of each person is 50 kg (110 lb). Assume that the average speed of each person as she or he leavesthe ground is about 2 m/s.

v

Earth,f =

(6 109)(50 kg)

6e24kg (2 m/s)

= 1e-13m/s

Earth would move a distance less than the diameter of a single hydrogen atom in second 1.

3.P.69

Assume the bullet travels in the +x direction, vi

=< v, 0, 0 >. Define the system to be the bullet and the block. Assumethat the net external force on the system is zero. Apply the momentum principle with the initial momentum being beforethe collision and the final momentum being after the collision.

psys,f

= psys,i

+

0

Fnet

t

The momentum of the system is the sum of the momenta of the bullet and block. The final velocity of the bullet is the sameas the block.

pbullet,i

+ pblock,i

= pbullet,f

+ pblock,f

mvbullet,i

+M

0

vblock,i

= mvf

+Mvf

mvbullet,i

= (m+M)vf

vf = m

m+M

vbullet,ivf

= mm+M

vbullet,i

v

f

= mm+M

v


38/60

38

Before Collision

M

m

After Collision

M

m

Figure 10: The meteor and satellite before and after the collision.

3.P.70

First, sketch a picture showing the objects before and after the collision (see Figure10).

Though the satellite is rotating, treat it as a particle. Define the system to be the meteor and satellite. Apply the momentuprinciple with net external force equal to zero.

psys,f

= psys,i

+

0F

nett

psys,f

= psys,i

pmeteor,f

+ psatellite,f

= pmeteor,i

+ psatellite,i

mvmeteor,f

+Mvsatellite,f

= mvmeteor,i

+Mvsatellite,i

m < v2x

, v2y

, 0> +M = m < v1x

, v1y

, 0> +M

Write the above equation in component form in the x and y directions.

x:

mv2x

+Mvfx

= mv1x

+Mv

Mvfx

= mv1x

+Mv mv2x

vfx =

mv1x

+Mv mv2x

M

vfx

=m(v

1x v

2x) +Mv

M

Note thatvfx

>v as a result of the collision. Thus, the satellite will be moving faster in the +x direction due to the collisio

Now for y,


39/60

39

y:

mv2y

+Mvfy

= mv1y

+ 0

Mvfy

= mv1y

mv2y

vfy

=mv

1y mv

2y

M

vfy =

m(v1y

v2y

)

M

Note thatvfy

is positive, thus after the collision, the satellites velocity has a slight upward component in the +y direction.

3.P.71

(a) First, sketch a picture showing the balls before and after the collision (see Fig. 11).

Before Collision

mB

mA

After Collision

mB

mA

Figure 11: The balls before and after the collision.

Define the system as the two balls, A and B, where A is the lighter ball. Apply the momentum principle. The netexternal force on the system is zero.

psys,f

= psys,i

+

0F

nett

psys,f

= psys,i

pA,f

+ pB,f

= vA,i

+ pB,i

mA vA,f +mB vB,f = mA vA,i +mB vB,i

vA,f

=m

Av

A,i+m

Bv

B,i m

Bv

B,f

mA

= (0.05 kg)(17, 0, 0 m/s) + 0 (0.1 kg)(3, 3, 0 m/s)

0.05 kg

= 11, 6, 0 m/s


40/60

40

(b) Impulse on ball A is

pA

= pA,f

pA,i

= mA

vA,f

mA

vA,i

= mA

(vA,f

vA,i

)

= (0.05 kg)(11, 6, 0 m/s 17, 0, 0 m/s)= 0.3, 0.3, 0 kg m/s

(c) Define the system as the lighter ball A. Apply the momentum principle.

Fnet on A

=p

A

t

= 0.03, 0.03, 0 kg m/s

0.03 s= 10, 10, 0 N

3.P.72

Sketch a picture of the system before and after the package is launched, as shown in Figure 12.

Before Launch

M

m

After Launch

M

m

Figure 12: The system before and after the package is launched.

Treat the objects as point particles. Define the system to be the space station and the package. Assume that m


41/60

41

pstation,f

= ppackage,f

Mvf

= mv

vf

= m

M

v

vf

= mM vcos , vsin , 0 m/s

Written in component form:

vfx

= m

M

v cos

and

vfy

= mM v sin

Basically, the space station recoils in the opposite direction with equal magnitude and opposite momentum as the package,as a result of launching the package.

3.P.73

(a) Write a VPython program to solve the problem. Here is an example.

1 from __future__ import d i v i s i o n2 from v i s u a l import 3

4 RE = 6. 4 e6 #r a d i u s o f E ar th 5

6 s p a c e c r a f t = s p h e r e ( p o s =(10RE, 0 , 0 ) , c o l o r =c o l o r . c ya n , r a d i u s = 0. 25RE)7 E a rt h = s p h e r e ( c o l o r =c o l o r . b l u e , r a d i u s =RE )8

9 m=1.5e4 #mass o f s p a c e c r a f t 10 ME = 6e 2411 G = 6 . 6 7 e 1112

13 v=vect or (0 ,0 ,0 ) #i n i t i a l v e l o c i ty o f s p ac e cr a ft 14 p=mv #i n i t i a l momentum o f s p a c e c r a f t 15

16 t=017 dt= 0. 13600 #t i me s t e p18

19 rmag=mag( sp a ce cr af t . pos ) ; #d i s t an c e o f s p a c e c r a f t from E ar th 20

21 t r a i l =c u r ve ( c o l o r = s p a c e c r a f t . c o l o r )22

23 while rmag>RE: #s t o p i f rmag < RE


42/60

42

24 r a t e ( 1 0 0 0 )25 r= s p a c e c r a f t . p o s26 rmag=mag( r )27 runit=r/rmag28

29 Fgrav=GmME/rmag2r u n i t30 Fnet=Fgrav31

32 p = p + F ne tdt33 v = p/m34 s p a c e c r a f t . p o s = s p a c e c r a f t . p os + vdt35

36 t r a i l . a pp en d ( p o s= s p a c e c r a f t . p o s )37

38 t = t + dt

Begin with an initial velocity of zero. The spacecraft accelerates to the right, crashing into Earth.

(b) To 3 significant figures, using a time step of 0.1 h (i.e. (0.1)(3600 s)), the minimum velocity isvi

=0, 1.06e3, 0 m/However, if you watch many orbits you will find that its not a closed orbit. This is due to the fact that the time st

is too large and therefore produces inaccurate results.

(c) When trying smaller time steps, t = 0.01 h (or rather (0.1)(3600 s)) gave better results that were consistent fnumerous orbits. However, the initial speed had to be increased to v

i=0, 1.07e3, 0 m/sto avoid crashing into Eart

(d) Straight lines could be seen near Earth, and in fact the orbit wasnt even closed fort= 0.2 h(which is (0.2 h)(3600 s/hAny larger time interval will show straight line segments and an open orbit, especially noticeable as the spacecraft aproaches Earth. The error is mostly due to the the approximation that v

avg v

f, when in fact the average velocity w

likely be greater than or less than (in magnitude) the final velocity. The other approximation is that net force is costant during the time step, when in fact it changes as the spacecrafts distance from Earth changes during the time ste

(e) It helps to print rmag inside the loop so that you can examine the distance of the spacecraft from Earth. For a perfectcircular orbit of radius 10(6.4e6m) = 6.4e7m, the distance of the spacecraft from Earth will not deviate from thvalue. You can adjust the rate statement to speed up or slow down the calculations. Note that printing will slow dowthe program. For a time step of 0.01 h, a speed of2.5e3m/s gives a nearly circular orbit that is off by only abo0.2% (1/640 0.002).

(f) If the launch speed is slightly less than or slightly greater than the speed for a circular orbit, the path of the spacecrais an ellipse.

(g) Using a time step of(0.01)(3600 s), the spacecraft escapes if its initial speed is greater than 3.5e3m/s.

(h) The spacecraft travels to the left, slowing down until its speed is zero, and then it speeds up toward Earth.

(i) The speed needed to leave Earth, moving to the left, is the same as in part (g).

3.P.74


43/60

43

(a) Modify the program in 3.P.74. You will need to look up the radius of Moon and its orbital radius (i.e. average center-to-center distance from Earth). Heres a sample program for part (a). The rate() statement has been removed inorder to speed up the animation. When drawing the sphere for Moon, make its radius arbitrarily large so that it canbe seen in the animation. Also, note that the net force on the spacecraft is the sum of the gravitational force by Earthand the gravitational force by Moon.

In this example, Earth is at the origin. Therefore, the distance of the spacecraft from Earth is simply the magnitudeof the spacecrafts position.


4 RE = 6. 4 e6 #r a d i u s o f E ar th 5 RM = 1. 73 7 e6 #r a d i u s o f Moon 6

7 s p a c e c r a f t = s p h e r e ( p o s =(10RE, 0 , 0 ) , c o l o r =c o l o r . c ya n , r a d i u s = 0. 25RE)8 Earth = spher e ( co l o r=co l o r . bl ue , rad i us=RE)9 Moon = s p h e r e ( p o s = ( 3 .8 4 e 8 , 0 , 0 ) , c o l o r =c o l o r . w h i te , r a d i u s = 0. 5RE)

10

11 m=1.5e4 #mass o f s p a c e c r a f t 12 ME = 6e 24 #m as s o f E a r th 13 MM = 7 e22 #mass of Moon14 G = 6 . 6 7 e 1115

16 v= vector (0 , 3. 264 e3 , 0 ) #i n i t i a l v e l o c i t y o f s p ac e cr a ft 17 p=mv #i n i t i a l momentum o f s p a c e c r a f t 18

19 t=020 dt= 0. 013600 #t i me s t e p21

22 rmag=mag( sp a ce cr af t . pos ) ; #d i s t an c e o f s p a c e c r a f t from E ar th 23 rrelmoonmag=mag( s pa ce cr a ft . posMoon. pos ) #d i s t a n c e o f s p a c e c r a f t fr om Moon 24

25 t r a i l =c u r ve ( c o l o r = s p a c e c r a f t . c o l o r )26

27 while rmag>RE and rre lm oo nm ag>RM: # s t o p i f r ma g < RE o r r re lm oo nm ag < RM 28 #c a l c u l a t e F gr av on s p a c e cr a f t by Ea rt h

29 r= s p a c e c r a f t . p o s30 rmag=mag( r )31 runit=r/rmag32 FgravE=GmME/rmag2r u n i t33

34 #c a l c u l a t e F gr av on s p a c e c r a f t b y Moon

35 rrel m oon= spa ce cr af t . pos Moon. pos36 rrelmoonmag=mag( rre lmoo n )37

rrel m oonuni t= rrel m oon/rrel m oonm ag38 FgravM=GmMM/ rr elm oo nma g 2 r r e l m o o n u n i t39

40 #c a l c u l a t e n et f o rc e

41 Fne t=FgravE + FgravM42

43 #u p d a te momentum a nd p o s i t i o n

44 p = p + F ne tdt45 v = p/m


44/60

44

46 s p a c e c r a f t . p o s = s p a c e c r a f t . p os + vdt47

48 t r a i l . a pp en d ( p o s= s p a c e c r a f t . p o s )49

50 t = t + dt

The minimum initial velocity for the spaceship to crash into Moon, to four significant figures, is 3.264e3m/s.

(b) Though not perfectly symmetric, a speed of3.259e3m/s will cause the spacecraft to loop around Moon and return Earth in a figure-8 like-pattern. Increasing or decreasing the speed by 1 m/s will allow you to see various orbits threturn the spacecraft to Earth.

(c) An orbit where the spacecraft reaches zero speed and returns along its same path is achieved with an initial speed 3.2534e3m/s. Decreasing the time step by 1/2 slightly improves the accuracy of the orbit. A reasonable time step use is about 10 seconds.

(d) Increasing the speed by only 0.1 m/s yields a very different orbit that does not repeat itself and is not stable. T3.2534e3m/s for example.

(e) When the spacecraft is approaching Earth or Moon, it speeds up because the parallel component of the net force othe spacecraft is in the same direction as the momentum of the spacecraft. As a result, the spacecrafts momentuincreases in magnitude and the spacecraft speeds up.

3.P.75Create three bodies of the same mass. It helps to use typical values of radius and mass for a planet or star. In texample below, masses and radii were chosen to be similar as Sun. The program below has initial conditions for a particularinteresting orbit in which identical mass stars are arranged in an equilateral triangle and the initial velocities are tangent a circle which circumscribes the triangle, as shown in the figure.

a a

a

Figure 13: Initial positions and velocities of three stars of the same mass.

The code for this example is shown below. The code used to calculate the initial velocities ensures that the velocity vectoare tangent to the circle.

1 from __future__ import d i v i s i o n2 from v i s u a l import


45/60

45

a=1e11 # co ns ta nt us e d i n i n i t i a l p o s i t i on s o f s t a r s R=7e8 # r ad iu s o f t he s t a r s G=6.67e11

x=0.288675a # u s e d t o f i n d t he r ad iu s o f t he i n i t i a l c i r c l e t h at c ir cu ms c r ib es t he s t a rs R c i r c l e = ( s q r t ( 0 . 7 5 ) 0.288675) a # ra di us o f t he i n i t i a l c i r c l e t h at c ir cu ms cr ib es t he

s t a r s

st ar1= sphere ( pos=(a/2,x , 0 ) , c o l o r =c o l o r . y e l lo w , r a d i u s = 10R) st ar2= sphere ( pos=(a/2, x , 0 ) , c o l o r =c o l o r . c ya n , r a d i u s = 10R) s t a r 3 =s p h e r e ( p o s = (0 , R c i r c l e , 0 ) , c o l o r =c o l o r . m ag en ta , r a d i u s = 10R)

st ar 1 .m=1.989e30 st ar 2 .m=sta r1 .m st ar 3 .m=sta r1 .m

si=3e4 #i n i t i a l s pe e d

#f i g ur e o u t i n i t i a l v e l o c i t y v e ct o rs s o t he y a re t an ge nt t o t h e c i r c l e

s t a r 1 . v= s i cr os s (norm ( sta r1 . pos) , vec tor (0 , 0 , 1 ) ) s t a r 2 . v= s i cr os s (norm ( sta r2 . pos) , vec tor (0 , 0 , 1 ) ) s t a r 3 . v= s i cr os s (norm ( sta r3 . pos) , vec tor (0 , 0 , 1 ) )

print "r_1(m) : " , st ar 1 . posprint "r_2(m) : " , st ar 2 . posprint "r_3(m) : " , st ar 3 . posprint "v_1(m/s ) : " , sta r1 . vprint "v_2(m/s ) : " , sta r2 . vprint "v_3(m/s ) : " , sta r3 . v

st ar1 . p= star 1 . m s t a r 1 . v

st ar2 . p= star 2 . m s t a r 2 . v st ar3 . p= star 3 . m s t a r 3 . v

t=0dt=1e2

t r a i l 1 = c u rv e ( c o l o r =s t a r 1 . c o l o r ) t r a i l 2 = c u rv e ( c o l o r =s t a r 2 . c o l o r ) t r a i l 3 = c u rv e ( c o l o r =s t a r 3 . c o l o r )

while 1 :

#c a l c u l a t e Fgrav on s t ar 1 by s t a r 2

r12=st ar 1 . pos sta r2 . pos r12mag=mag( r1 2 ) r12unit=r12/r12mag Fgrav12 = G s t a r 1 .m st ar 2 .m/r12mag2( r 1 2 u n i t )

#c a l c u l a t e Fgrav on s t ar 1 by s t a r 3

r13=st ar 1 . pos sta r3 . pos r13mag=mag( r1 3 )


46/60

46

56 r13unit=r13/r13mag57 Fgrav13 = G s t a r 1 .m st ar 3 .m/r13mag2( r 1 3 u n i t )58

59 #c a l c u l a t e Fnet on s t a r 1

60 Fnet1 = Fgrav12 + Fgrav1361

62 #c a l c u l a t e Fgrav on s t ar 2 by s t a r 1

63

Fgrav21 = Fgrav1264


66 r23=st ar 2 . pos sta r3 . pos67 r23mag=mag( r2 3 )68 r23unit=r23/r23mag69 Fgrav23 = G s t a r 2 .m st ar 3 .m/r23mag2( r 2 3 u n i t )70




75 Fgrav31 = Fgrav13

76


78 Fgrav32 = Fgrav2379



83 #u p da t e momentum an d p o s i t i o n o f s t a r 1

84 s t a r 1 . p = s t a r 1 . p + F ne t1dt85 s t a r 1 . v = s t a r 1 . p / s t a r 1 .m86 s t a r 1 . p os = s t a r 1 . p os + s t a r 1 . v dt87 t r a i l 1 . append( pos= star 1 . pos )88

89 #u p da t e momentum an d p o s i t i o n o f s t a r 2 90 s t a r 2 . p = s t a r 2 . p + F ne t2dt91 s t a r 2 . v = s t a r 2 . p / s t a r 2 .m92 s t a r 2 . p os = s t a r 2 . p os + s t a r 2 . v dt93 t r a i l 2 . append( pos= star 2 . pos )94

95 #u p da t e momentum an d p o s i t i o n o f s t a r 3

96 s t a r 3 . p = s t a r 3 . p + F ne t3dt97 s t a r 3 . v = s t a r 3 . p / s t a r 3 .m98 s t a r 3 . p os = s t a r 3 . p os + s t a r 3 . v dt99 t r a i l 3 . append( pos= star 3 . pos )

100

101 t=t+dt

Shown below are some initial conditions for various orbits. The initial conditions for the long-lasting orbit are for thridentical masses arranged in an equilateral triangle with velocities tangent to a circle that circumscribes the triangle.


47/60

47

long-lasting orbit: r1 : -5E+10,-2.88675E+10,0m

r2 : 5E+10,-2.88675E+10,0 m

r3 : 0, 5.7735E+10,0m

v1 : 15000, 25980.8, 0m/s

v2 : 15000, 25980.8, 0m/s

v3 : 30000, 0, 0m/s

For an orbit that produces a collision, many different initial conditions will suffice. Note that the animation will look awkwardbecause there is no if statement to stop the animation when a collision takes place. But rather, the stars may go througheach other, and due to a very large force when their centers become very close, be pulled back through one another, givingthe impression that they bounce off of each other. Heres a sample set of initial conditions that causes this to occur.

collision: r1: -5E+10,0, 0 m

r2: 5E+10,0, 0m

r3: 0, 1E+11,0mv1 : 20000, 0, 0m/s

v2 : 0, 20000, 0m/s

v3 : 20000, 20000, 0m/s

Again, there are many different initial conditions that will cause one of the stars to escape.

escape: r1 : -5E+10 ,0, 0m

r2 : 5E+10,0, 0 m

r3 : 0, 1E+11,0 mv1 : 0, 40000, 0 m/s

v2 : 20000, 0, 0 m/s

v3 : 20000, 0, 0 m/s

3.P.76

(a) Assume that Earths orbit is a circle. Average speed is

|v| = distance in one revolution

time interval for one revolution

=2R

orbit

T

= 2(1.5e11 m)

(365day)(24 h/day)(3600 s/h)

= 3.0e4m/s


48/60

48

(b) A period is 365 days. Its reasonable to maket= 0.001Twhich will result in 1000 steps to compute one orbit. Forcircle, each step will be less than 1 degree around the circle. So anything less than 12 hours (1/2 day) is a reasonabsmall time step. For example, use t= (0.0010)(365)(24)(3600) s.

(c) Here is a sample program. Though Sun is placed at the origin, the program is written so that if Sun is not at torigin, the calculation of the gravitational force on Earth is still correct. The variable au in the program refers to aastronomical unit which is defined as the average distance of Earth from Sun.


4 au=1.5e11 #d i s t a n c e o f E ar th fr om Sun 5 R=7e8 #r a d i u s o f Sun 6 G=6.67e 117

8 Sun=spher e ( pos = (0 , 0 , 0) , co l o r=co l o r . yel l ow , rad i us =10R)9 E a rt h=s p h e r e ( p o s =( au , 0 , 0 ) , c o l o r =c o l o r . b l u e , r a d i u s = 0. 5Sun . radi us )

10

11 Sun .m=2e3 012 Ea rth .m=6e2 413

14 T = ( 3 6 5243 6 0 0 ) #p er io d f o r a c i r c u l a r o r b i t 15 s i = 2p i au/T #i n i t i a l s pe e d f or a c i r c u l a r o r b i t 16

17 #i n i t i a l v e l o c i t y

18 Earth . v=vec to r (0 , si , 0)19 Eart h . p=Earth .mEarth . v20

21 t=022 dt=0.001T23

24 tr ai l E ar th=curve ( co l o r=Earth . co l o r )25

26 while 1 :27 r a t e ( 1 0 0 )28 #c a l c u l a t e F gr av on E ar th by Sun

29 r=Eart h . pos Sun. pos30 rmag=mag( r )31 runit=r/rmag32 Fgrav = GSun.mEa rth .m/rmag2( r u n i t )33

34 #c a l c u l a t e F ne t on E ar th

35 Fnet = Fgrav36

37 #u p d a t e momentum a nd p o s i t i o n o f E a r t h

38 Earth . p = Earth . p + Fnetdt39 Earth . v = Earth . p/Earth .m40 Earth . pos = Earth . pos + Earth . vdt41 tr ai l E ar th . append( pos=Earth . pos )42

43 t=t+dt

(d) The path of Earth in the simulation is indeed a circle. Add a print statement to the while block to print the position Earth. Also, change the while statement to while t


49/60

49

component of Earths position means that it hasnt quite reached its initial position yet. Thus, there is numerical errorin the simulation with this time step.

To print t after Earth completes one orbit, change the while statement to while 1: and use an if statement to checkthe clock reading and then the position of Earth. If the clock reading is greater than one period, then check to seewhen Earth crosses the y-axis. The sample program below illustrates how to do this. Note theif statement that iswithin the while loop which prints the data you need when the condition is met.


4 au=1.5e11 #d i s t a n c e o f E ar th f ro m S un 5 R=7e8 #r a d i u s o f Sun 6 G=6.67e 117


10

11 Sun .m=2e3 012 Ea rth .m=6e2 413



18 Earth . v=vect or (0 , si , 0)19 Eart h . p=Earth .mEarth . v20

21 t=022 dt=0.001T23



29 r=Ear th . pos Sun. pos30 rmag=mag( r )31 runit=r/rmag32 Fgrav = GSun.mEar th .m/rmag2( r u n i t )33


35 Fnet = Fgrav36


38 Earth . p = Earth . p + Fnetdt39

Earth . v = Earth . p/Earth . m40 Earth . pos = Earth . pos + Earth . vdt41 tr ai l E ar th . append( pos=Earth . pos)42

43 i f ( t>T) :44 print "T=" , T ," t=" , t , "r=" , Earth . pos45 i f( Earth . pos . y>0) :46 break

47


50/60

50

48 t=t+dt

The printed results are:

T= 31536000 t= 31567536.0 r=

T= 31536000 t= 31599072.0 r=

T= 31536000 t= 31630608.0 r=

T= 31536000 t= 31662144.0 r= T= 31536000 t= 31693680.0 r=

T= 31536000 t= 31725216.0 r=

T= 31536000 t= 31756752.0 r=

T= 31536000 t= 31788288.0 r=

In the simulation, Earth completes an orbit at t = 31788288.0 s, yet the actual period of Earths orbit is closer T= 31536000 s. This is fairly close, and decreasingt tot= 0.0001T doesnt make much of an improvement. Yincreasing it to t= 0.1Tgives a non-circular orbit with lots of straight lines. t = 0.01T is about the largest timinterval that can be used and still get a circular orbit with a period close to what is obtained with much smaller timintervals.

(e) Based on the previous part, begin witht= 0.001T. Take out the if statement(s) from the previous part. Heres

sample program.1 from __future__ import d i v i s i o n2 from v i s u a l import 3

4 au=1.5e11 #d i s t a n c e o f E ar th fr om Sun 5 R=7e8 #r a d i u s o f Sun 6 G=6.67e 117


10

11 Sun .m=2e3 012 Ea rth .m=6e2 413



18 Earth . v=1.2 v e c t o r ( 0 , s i , 0 )19 Eart h . p=Earth .mEarth . v20

21 t=022 dt=0.001T23



29 r=Eart h . pos Sun. pos30 rmag=mag( r )31 runit=r/rmag32 Fgrav = GSun.mEa rth .m/rmag2( r u n i t )33


51/60

51


35 Fnet = Fgrav36


38 Earth . p = Earth . p + Fnetdt39 Earth . v = Earth . p/Earth . m40 Earth . pos = Earth . pos + Earth . vdt41

tr ai l E ar th . append( pos=Earth . pos)42

43 t=t+dt

The result is an ellipse. The largest time step recommended ist= 0.01T. The more eccentric the ellipse, the fasterEarth travels when near Sun. In this case, smaller time steps such as t= 1 104Tmust be used.

If a large time step is used (i.e. too large for accuracy), you will see straight lines on the ellipse, especially when Earthtravels fast, near Sun.

A highly eccentric ellipse like this would cause enormous climate change. Im not sure it would kill life, but it wouldaffect which life survives. Life that evolves to handle such climate change would be the life that would survive.

(f) First, add print statements to print the gravitational force on Earth and its momentum. This helps you know how toscale the arrows that you will draw to represent the force and momentum vectors. The sample program below shows

how to draw and scale arrows to represent the force and momentum vectors for Earth. Run the program, and you willnotice that the force on Earth is always toward Sun, as expected. The momentum of Earth is always tangent to its pathand in the direction of motion (as it should be). Earths momentum is greatest when Earth is nearest Sun (perihelion)and is least when Earth is furthest from Sun (aphelion).

1 from __future__ import d i v i s i o n2 from v i s

matter and interaction chapter 03 solutions

Documents