Gravitational Force

Gravitational Force

Newton’s Law of Gravitation:

G = 6.67 x 10-11 Nm2/kg2

= gravitational constant

rr

mGmFG

ˆ2

21F

-F

r

Gravitational Force

A uniform spherical shell of matter attracts a particle that is outside the shell as if all the shell’s mass were concentrated at its center

A uniform shell of matter exerts no net gravitational force on a particle located inside it.

rr

mGmFG

ˆ2

21r

FG = 0

Orbits

The centripetal force of an object orbiting around a planet or star (sun) is provided by the gravitational force:

2

2

2

2

211

r

Gm

r

v

r

mGmam

FF

c

Gc

r

m1

m2

Geosynchronous Orbit

Geosynchronous orbit:

orbit period (T) equals 1 day T

rv

2

2

2

32

2

2

22

2

21

2

1

2

21

2

1

4

4

2

TGm

r

r

Gm

T

r

r

mGm

r

T

r

m

r

mGm

r

vm

Can figure out r, m or T if you know the other two variables.

Gravitational Potential Energy

Gravitational potential energy is defined as zero (0) at infinite distance (r∞).

Total energy: ET = K + UG

r

mGmUG

21

Kepler’s 1st and 2nd Laws

1st: All planets move in elliptical orbits, with the sun at one focus.

2nd: A line that connects a planet to the sun sweeps out equal areas in equal times

Angular momentum is conserved!

Kepler’s 3rd Law

The square of the period of any planet is proportional to the cube of the semimajor axis of its orbit.

For circular objects this follows directly from

Fc = FG

32

2 4r

GMT

Orbit Problem

In March 1999 the Mars Global Surveyor (GS) entered its final orbit about Mars. Assume a circular orbit with a period of 7080s and orbital speed of 3400m/s. The mass of the GS is 930kg and the radius of Mars is 3.43 x 106 m.

a) Calculate the radius of the GS orbit.

Orbit Problem

a) Calculate the radius:

m

ssm

vTr

T

rv

61083.3

2

)7080)(/3400(

2

2

Orbit Problem

b) Calculate the mass of Mars.

T = 7080 s

v = 3400 m/s

mGS = 930 kg

rM = 3.43 x 106 m

rorbit = 3.83 x 106 m

Orbit Problem

b) Calculate the mass of Mars:

Set Fc = FG:

kg

msm

G

rvm

r

Gmv

r

mGm

r

vm

kg

mN

M

M

MGSGS

23

11

62

2

2

2

2

1064.6

1067.6

)1083.3()/3400(

2

2

Orbit Problem

c) Calculate the total mechanical energy of the GS in this orbit.

T = 7080 s v = 3400 m/s mGS = 930 kg mM = 6.64 x 1023 kg rM = 3.43 x 106 m rorbit = 3.83 x 106 m

Orbit Problem

c) Calculate the total mechanical energy of the GS in this orbit.

J

m

kgkgsmkg

r

mGmvm

UKE

kg

mN

MGSGS

GT

9

6

2311

2

21

2

21

1038.5

1083.3

)1064.6)(930)(1067.6()/3400)(930(

2

2

Orbit Problem

d) If the GS was to be placed in a lower circular orbit (closer to the surface of Mars), would the new orbital period of the GS be greater than or less than the given period? Justify your answer.

Answer: Less than. T2 is proportional to r3 so if r decreases

then T also decreases

Orbit Problem

e) In fact, the GS orbit was slightly elliptical with its closest approach to Mars at 3.71 x 105 m above the surface and its furthest distance at 4.36 x 105 m above the surface.

If the speed of the GS at closest

approach is 3.40 x 103 m/s, calculate the speed at the furthest point of the orbit.

Orbit Problem

e) Calculate the speed at the furthest point.

Angular momentum is conserved:

mmm

smmm

r

vrv

vrvr

r

vrm

r

vrm

II

f

ccf

ffcc

f

f

fGS

c

ccGS

ffcc

3

65

365

22

1034.3)1043.31036.4(

)/1040.3)(1043.31071.3(