chapter 4 gravitation physics beyond 2000 gravity newton ry/newtongrav.html
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Chapter 4 Gravitation
Physics Beyond 2000
Gravity
• Newton
• http://csep10.phys.utk.edu/astr161/lect/history/newtongrav.html
• http://www.britannica.com/bcom/eb/article/9/0,5716,109169+2+106265,00.html
• http://www.nelsonitp.com/physics/guide/pages/gravity/g1.html
Gravity
• The moon is performing circular motion round the earth.
• The centripetal force comes from the gravity.
Fc
earthmoon
v
Gravity
• Newton found that the gravity on the moon is the same force making an apple fall.
W
Ground
Newton’s Law of Gravitation
• Objects attract each other with gravitational force.• In the diagram,
m1 and m2 are the masses of the objects and r is the distance between them.
m1 m2
F F
r
Newton’s Law of Gravitation
• Every particle of matter attracts every other particle with a force whose magnitude is
m1 m2
F F
r
221.
r
mmGF G is a universal constant
G = 6.67 10-11 m3kg-1s-2
Note that the law applies to particles only.
Example 1
• Find how small the gravitation is.
Shell Theorem
• Extends the formula
to spherical objects like a ball, the earth, the sun and all planets.
221.
r
mmGF
Theorem 1a. Outside a uniform spherical shell.• The shell attracts the external particle as if a
ll the shell’s mass were concentrated at its centre.
221.
r
mmGF
F F
r
m1m2O
Theorem 1b. Outside a uniform sphere.
• The sphere attracts the external particle as if all the sphere’s mass were concentrated at its centre.
221.
r
mmGF
F F
r
m1m2O
Example 2 Outside a uniform sphere.
• The earth is almost a uniform sphere.
221.
r
mmGF
F
r
m1 m2O
earth
F
Theorem 2a. Inside a uniform spherical shell.
• The net gravitational force is zero on an object inside a uniform shell.
m2
m1
The two forceson m2 cancel.
Theorem 2b. Inside a uniform sphere.
221.
r
mmGF
m2m1
r
F
where m1 is the mass of the corewith r the distance from the centre to the mass m2
Example 3
• Inside a uniform sphere.
m2m1
r
F
Gravitational Field
• A gravitational field is a region in which any mass will experience a gravitational force.
• A uniform gravitational field is a field in which the gravitational force in independent of the position.
• http://saturn.vcu.edu/~rgowdy/mod/g33/s.htm
Field strength, g
• The gravitational field strength, g, is the gravitational force per unit mass on a test mass.
test massF
m
m
Fg
F is the gravitational force
m is the mass of the test mass
g is a vector, in the same direction of F.SI unit of g is Nkg-1.
Field strength, g
• The gravitational field strength, g, is the gravitational force per unit mass on a test mass.
test massF
m
m
Fg
F is the gravitational force
m is the mass of the test mass
SI unit of g isNkg-1.
Field strength, g, outside an isolated sphere of mass M
• The gravitational field strength, g, outside an isolated sphere of mass M is
2r
GMg
Prove it by placing a test mass m at a point X with distance r from the centre of the isolated sphere M.
M r
field strengthat X
XO
Example 4
• The field strength of the earth at the position of the moon.
Field strength, g
• Unit of g is Nkg-1.
• g is also a measure of the acceleration of the test mass.
• g is also the acceleration due to gravity, unit is ms-2.
Field strength, g
• Field strength, g.
• Unit Nkg-1.
• A measure of the strength of the gravitational field.
• Acceleration due to gravity, g.
• Unit ms-2.• A description of
the motion of a test mass in free fall.
Field lines
• We can represent the field strength by drawing field lines.
• The field lines for a planet are radially inward.
Radial fieldplanet
Field lines
• We can represent the field strength by drawing field lines.
• The field lines for a uniform field are parallel.
Uniform field
earth’s surface
Field lines
• The density of the field lines indicates the relative field strength.
g1= 10 Nkg-1 g2= 5 Nkg-1
Field lines
• The arrow and the tangent to the field lines indicates the direction of the force acting on the test mass.
test mass
direction of the force
The earth’s gravitational field
• Mass of the earth Me 5.98 1024 kg
• Radius of the earth Re 6.37 106 m
Re
O
Gravity on the earth’s surface, go
• The gravitational field go near the earth’s surface is uniform and
2e
eo R
GMg
The value of go 9.8 Nkg-1
Example 5
• The gravity on the earth’s surface, go.
Apparent Weight
• Use a spring-balance to measure the weight of a body.
• Depending on the case, the measured weight R (the apparent weight) is not equal to the gravitational force mgo.
R
mgo
Apparent Weight
• The reading on the spring-balance is affected by the following factors:
1. The density of the earth crust is not uniform.
2. The earth is not a perfect sphere.
3. The earth is rotating.
Apparent Weight
1.The density of the earth crust is not uniform.
• Places have different density underneath. Thus the gravitational force is not uniform.
Apparent Weight
2. The earth is not a perfect sphere.
Points at the poles are closer to the centre than points on the equators.
rpole < requator
gpole > gequator N-pole
S-pole
Equator
Apparent Weight
3. The earth is rotating. Except at the pole, all
points on earth are performing circular motion with the same angular velocity . However the radii of the circles may be different.
X
Y
Apparent Weight
3. The earth is rotating. Consider a mass m is at p
oint X with latitude .
The radius of the circle is r = Re.cos .
X
rRe
m
O
Y
Apparent Weight
3. The earth is rotating. The net force on the mass
m must be equal to the centripetal force.
X
rRe
m
2mrFF cnet cos... 2
eRm
O
YFc
Note that Fnet points to Y.
Apparent Weight R
3. The earth is rotating. The net force on the mass m
must be equal to the centripetal force.
So the apparent weight (normal reaction) R does not cancel the gravitational force mgo.
X
r m
O
YFc
R
mgo
co FgmR
Apparent Weight R
3. The earth is rotating. The apparent weight R is
not equal to the gravitational force mgo in magnitude.
X
r m
O
YFc
R
mgo
co FgmR
Apparent weight R on the equator
2eo mRRmg 2eo mRmgR
emgThe apparent field strengthon the equator is
2eoe Rgg
mgo R
Apparent weight R at the poles
0 Rmgo
omgR pmg
The apparent field strengthat the poles is
op gg
mgo
R
Example 6
• Compare the apparent weights.
Apparent weight at latitude
X
r m
O
YFc
R
mgo
2mrFF cnet cos... 2
eRm
co FgmR
Note that the apparent weight Ris not exactly along the line throughthe centre of the earth.
Variation of g with height and depth
• Outside the earth at height h.
h = height of the mass m from the earth’s surface
r m
hRe
Meg
O
Variation of g with height and depth
• Outside the earth at height h.
r m
hRe
Meg
O
2
22
2
).(
).(
r
Rg
r
R
R
GMr
GMg
eo
e
e
e
e
where go is the field strength on the earth’s surface.
2
1
rg
Variation of g with height and depth
• Outside the earth at height h.
r m
hRe
Meg
O
where go is the field strength on the earth’ssurface.
2
2
)1.(
).(
eo
e
eo
R
hg
hR
Rgg
Variation of g with height and depth
• Outside the earth at height h close to the earth’s surface. h<<Re.
r m
hRe
Meg
O
where go is the field strength on the earth’ssurface.
)2
1.(e
o R
hgg
ee R
h
R
h 21)1( 2
Variation of g with height and depth
• Below the earth’s surface.
Re
rO d
Me
g
Only the core withcolour gives thegravitational force.
r = Re-d
Variation of g with height and depth
• Below the earth’s surface.
Re
rO d
Me
g
Find the mass Mr of
3)(e
er R
rMM
r = Re-d
Variation of g with height and depth
• Below the earth’s surface.
Re
rO d
Me
g 32
2
)(e
e
r
R
r
r
GMr
GMg
r = Re-d
Variation of g with height and depth
• Below the earth’s surface.
Re
rO d
Me
g )1(
)(2
eo
ee
e
R
dg
R
r
R
GMg
r = Re-dg r
Variation of g with height and depth
earth
ggo
0r distance from the centreof the earthRe
1. r < Re , g r.
2. r > Re , 2
1
rg
Gravitational potential energy Up
• Object inside a gravitational field has gravitational potential energy.• When object falls towards the earth, it gains kinetic energy and
loses gravitational potential energy.
This objectpossesses Up
earth
Zero potential energy
• By convention, the gravitational potential energy of the object is zero when its separation x from the centre of the earth is .
Up = 0earth
x O
Negative potential energy
• For separation less than r, the gravitational potential energy of the object is less than zero. So it is negative.
Up < 0earth
O r
Gravitational potential energy Up
• Definition 1
• It is the negative of the work done by the gravitational force FG as the object moves from infinity to that point.
earth
O r
FG
dx
Gravitational potential energy Up
• Definition 1
earth
O r
FG
dx
WU p
Gravitational potential energy Up
• Definition 2
• It is the negative of the work done by the external force F to bring the object from that point to infinity.
earth
O r
F
dx
Me
m
Gravitational potential energy Up
• Definition 2
earth
O r
F
dx
r
P dxFU .
Me
m
Gravitational potential energy Up
2x
mGMFFrom e
r
mGMU eP
earth
O r
Me
m
Example 7
• Conservation of kinetic and gravitational potential energy.
Example 8
• - Work done
= gravitational potential energy
Example 9
• Two particles are each in the other’s gravitational field.
• Thus each particle possesses gravitational energy.
System of three particles• Each particle is in another two particles’
gravitational fields.
• Each particle possesses gravitational potential energy due to the other two particles.
m
M1 M2
r1r2
Up of
2
2
1
1
r
GmM
r
GmM
System of three particlesUp of
Up of
m
M1 M2
r1r2
r3
2
2
3
12
r
GmM
r
MGM
1
1
3
12
r
GmM
r
MGM
Example 10
• Up of the moon due to the earth’s gravitational field.
r
earth
moon
What is the Up of the earth due to the moon’s gravitational field?
Escape speed ve
• Escape speed ve is the minimum projection speed required for any object to escape from the surface of a planet without return.
ve
Escape speed ve
• Escape speed ve is the minimum projection speed required for any object to escape from the surface of a planet without return.
ve
Escape speed ve
• On the surface of the planet, the body possesses both kinetic energy Uk and gravitational potential energy Up.
veR
mM
Uk
2
2
1emv
UP
R
GMm
Escape speed ve
• If the body is able to escape away, it means the body still possesses kinetic energy at infinity.
• Note that the gravitational energy of the body at infinity is zero.
veR
mM
Escape speed ve
veR
mM
• If there is not any loss of energy on projection,
the total energy of the body at lift-off
= the total energy of the body at infinity
Escape speed ve
veR
mM
)(2
1 2
r
GMmmve = kinetic energy at infinity
≧0
Escape speed ve
veR
mM
RgR
GMv o2
2where go is the gravitational acceleration on the surfaceof the earth.
Escape speed ve
veR
mM
RgR
GMv oe 2
2
So the escape speed from earth is
Escape speed ve
veR
mM
Example: Find ve
Gravitational potential V
• Definition: The gravitational potential at a point is the
gravitational potential energy per unit test mass.
m
UV
where U is the gravitational potential energy of a mass m at the point
Gravitational potential V
• Definition: The gravitational potential at a point is the
gravitational potential energy per unit test mass.
m
UV unit of V is J kg-1
Gravitational potential V
• Example 12 – to find the change in gravitational potential energy.
• ΔU = U – Uo
• If ΔU >0, there is a gain in U.
• If ΔU <0, there is a loss in U.
Equipotentials
• Equipotentials are lines or surfaces on which all points have the same potential.
• The equipotentials are always perpendicular to the field lines.
Equipotentials• The equipotentials around the earth are ima
ginary spherical shells centered at the earth’s centre.
Equipotentials• The field is radial.
Equipotentials• The equipotentials near the earth’s surface a
re parallel and evenly spaced surface.
• The field is uniform.
surface
Equipotentials
• Example 13 – Earth’s equipotential.
Potential V and field strength g
dr
dVg
rr
MGV
2r
GMg
Potential V and field strength g
rr
MGV
2r
GMg
If we consider the magnitude of g only,
dr
dVg
Earth-moon system
• http://tycho.usno.navy.mil/vphase.html• The potential is the sum of the potentials due to the earth and the moon.
earth
moonP
D
r
D-r
Me
Mm
Earth-moon system
earth
moonP
D
r
D-r
Me
Mm
rD
GM
r
GMV meP
Earth-moon system
earth
moon
r
V
0
Earth-moon system
earth
moon
r
V
0
dr
dVg
Earth-moon system
earth
moon
r
V
0
dr
dVg
g
Earth-moon system
earth
moon
r
V
0g=0
g = 0 at a point X between the earthand the moon. X is a neutral point.
X
Earth-moon system
earth
moon
r
V
0
g>0
g points to the centre of the earth if it is positive.
X
Earth-moon system
earth
moon
r
V
0g<0
g points to the centre of the moon if it is negative.
X
Earth-moon systemGiven: Me = 5.98 × 1024 kg Mm = 7.35 × 1022 kg D = 3.84 × 108 m G = 6.67 × 10-11 Nm2kg-2
Find: the position X at which g = 0.
earth
moon
X
x
Hint: 0dr
dVg
Earth-moon systemGiven: Me = 5.98 × 1024 kg Mm = 7.35 × 1022 kg D = 3.84 × 108 m G = 6.67 × 10-11 Nm2kg-2
Find: the position X at which g = 0.
earth
moon
X
x
Answer: x = 3.46 × 108 m
Earth-moon system
• Example 14 – potential difference near the earth’s surface.
Orbital motion• The description of the motion of a planet
round the sun.
sun
Orbital motion• Kepler’s law:
1. The law of orbits.
All planets move in elliptical orbits, with the sun at one focus.
sun
Orbital motion• Kepler’s law:
2. The law of areas.
The area swept out in a given time by the line joining any planet to the sun is always the same.
sun
Orbital motion• Kepler’s law:
3. The law of periods.
The square of the period T of any planet about the sun is proportional to the cube of their mean distance r from the sun.
sun 32 rT
Orbital motion
• Basically, we only study the simple case of circular orbit.
r
Orbital motionA satellite of mass m performs circular motion round the earth with speed vc .The radius of the orbit is r.
r
satellite
earth
vc
Orbital motion
The centripetal force is provided by the gravitational force.
r
satellite
earth
vc
Fc
Orbital motion
Show that
r
satellite
earth
vc
Fc
r
GMv ec where Me is the
mass of the earth
Orbital motion
• Example 15 – find the speed of a satellite.
r
satellite
earth
vc
Proof of Kepler’s 3rd law in a circular orbit
32 rT
r1
satellite 1
earth
vc1
r2satellite 2
vc2
Proof of Kepler’s 3rd law in a circular orbit
r1
satellite 1
earth
vc1
r2satellite 2
vc2
Note that the proof is true for satellites round the same planet.
Kepler’s 3rd law
• Example 16 – apply Kepler’s 3rd law.
Satellites
• Natural satellites – e.g. moon.
• Artificial satellites –
e.g. communication satellites,
weather satellites.
http://weather.yahoo.com/graphics/satellite/US.html
http://www.smgaels.org/physics/97/MGRAHLFS.HTM
Geosynchronous satellites
• A geosynchronous satellite is above the earth’s equator.
• It rotates about the earth with the same angular speed as the earth and in the same direction.
• It seems stationary by observers on earth.
Geosynchronous satellites
ω
equator
axis
satellite
h
Re
Geosynchronous satellitesFind the radius of the orbit of a geosynchornoussatellite.
ω
equator
axis
satelliteh Re
rs
h + Re = rs
Geosynchronous satellites
rs = 4.23×107 m
ω
equator
axis
satelliteh Re
r
h + Re = rs
Geosynchronous satellites
h = 3.59×107 m
ω
equator
axis
satelliteh Re
r
h + Re = rs
Parking OrbitNote that there is only one such orbit.It is called a parking orbit.
ω
equator
axis
satelliteh Re
r
h + Re = rs
Satellites Near the Earth’s surface
• Assume that the orbit is circular with radius r Re , the radius of the earth.
• The gravitational field strength go is almost a constant (9.8 N kg-1).
• The gravitational force provides the required centripetal force.
Satellites Near the Earth’s surface
r Re satellite
earth
vr
Find vr
Energy and Satellite Motion
• Find v and the kinetic energy Uk of the satellite.
r
satellite
earth Me
v
m
Energy and Satellite Motion• The satellite in the orbit possesses both
kinetic energy and gravitational energy.
r
satellite
earth Me
v
m
Energy and Satellite Motion
r
satellite
earth Me
v
m
r
GMv e2
r
mGMU ek 2
Note that Uk > 0
Energy and Satellite Motion
• Find Up the gravitational potential of the satellite.
r
satellite
earth Me
v
m
Energy and Satellite Motion
r
satellite
earth Me
v
mr
mGMU ep
Note that Up < 0
Energy and Satellite Motion
r
satellite
earth Me
v
m
Find U, the total energy of the satellite.
Energy and Satellite Motion
r
satellite
earth Me
v
m
r
mGMUUU epk 2
Note that U < 0
Energy and Satellite Motion
r
satellite
earth Me
v
m
U : Up : Uk = -1 : -2 : 1
Falling to the earth
r
satellite
earth Me
v
m
The satellite may lose energy due to airresistance. The total energy becomes more negative and r becomes less.
r
mGMU e
2
Falling to the earth
r
satellite
earth Me
v
m
The satellite follows a spiral path towardsthe earth.
r
mGMU e
2
Falling to the earth
r
satellite
earth Me
v
m
As r decreases, the kinetic energy of the satellite increases and the satellite moves faster.
r
mGMU ek 2
Falling to the earth
Example 17 – Loss of energy
Weightlessness in spacecraft
mg
vv
The astronaut is weightless.
Weightlessness in spacecraft
• We fell our weight because there is normal reaction on us.
mg
Normal reaction
ground
Weightlessness in spacecraft
• If there is not any normal reaction on us, we feel weightless. e.g. free falling
mg
Weightlessness in spacecraft
v
mg The gravitational forcemg on the astronaut isthe required centripetalforce. He does not requireany normal reaction toact on him.
Weightlessness in spacecraft
v
mg
The astronaut isweightless.
http://www.nasm.edu/galleries/gal109/NEWHTF/HTF611A.HTM