Download Gravitation

P.1
AL Physics
Gravitation
1. Introduction of inverse square law
A story is sometimes told to the effect that Newton was led to his law of Universal
Gravitation by the thought that the centripetal force helding the moon in its orbit might
be the same gravitational force as pulls an apple to the ground but diminished in strength.
Fe = mg
....... (1)
Fm = mv2 / rm
.....(2)
Fe = force on the apple near earth’s surface
Fm = force on the apple in the moon’s orbit
g = 9.81 N/kg ( on the earth’s surface )
m = mass of the apple
v = speed of the moon = 1020 m/s
rm = radius of the moon’s orbit
= 3.84 x 108 m
re = radius of the earth = 6.38 x 106 m
T = period of the moon’s orbit = 27.3 days
From (3), (4) we know that the gravitational force acting on a body by the earth is
inversely proportional to its distance from the earth’s centre.
2. Law of Universal Gravitation
a)
It states that each particle in the universe attracts evergy other particle with a force
which is directly proportional to the product of their masses and inversely
proportional to the square of their distance apart.
F
b)
m1m2 / r2
G = universal gravitational constant
= 6.7 x 10 -11 Nm2kg-2
To extend the application of this law from particles to bodies of real size and real
shape. Newton showed that " The gravitational attraction between two spherically
symmetrical bodies is proportional to the product of their masses and inversely
proportional to the square of the distance between their centries. "
F = Gm1m2/r2
Example: Find the resultant gravitational force acting on m.
r1 = 6.7 x 104 m
r = 1.34 x 105 m
2
m1 = 2 x 1010 kg
m = 4 x 109 kg
2
m = 6 x 10 12 kg
G = 6.7 x 10 -11 Nm2kg-2
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3.Equivalence of gravitational mass and inertial mass
Consider a particle A on the earth's surface.
F = GMm’/r2 ( Gravitational law )
M = gravitational mass of the earth
r = radius of the earth
m’ = gravitational mass of A
m = inertial mass of A
( Newton’s law )
F = mg
m & m' appear to be equal experimentally.
Note :
i) In general cases,
F = GMm/r2
ii) If m takes a circular motion about M, then the gravitational force will contribute the
required centripetal force.
mg = GMm/r2 = mv2 /r
= mrw2
( use this equation when v = constant )
( use this equation when w = constant )
( g is not equal to 9.8 N/kg since m does not
lie on the earth (M) surface. )
4. Satellite moves round a planet
a) Determine the mass of the sun
( we have used the fact that
gravitational mass = inertial mass )
ve = speed of the earth
Te= period of earth’s orbit = 1 year
= 3 x 107 s
re = 1.5 x 1011 m
= radius of earth’s orbit
G = 6.7 x 10 -11 Nm2kg-2
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b) Determine the mass of the earth
vm = speed of the moon
Tm = period of earth’s orbit
= 1 month = 2.4 x 106 s
rm = 4 x 108 m
= radius of moon’s orbit
G = 6.7 x 10 -11 Nm2kg-2
c) Determine the period of the moon's orbit around earth
g = 10 ms-2 on earth’s surface
Re = radius of earth = 6.4 x 106 m
rm = distance of moon from earth’s centre
= 3.8 x 108 m
Consider the circular motion of the moon,
Consider a particle m’ on earth’s surface,
5. Gravitational field strength (g)
-
In the motion under gravity, g is treated as the acceleration due to gravity.
And g = 9.8ms-2 approximately.
-
Since in such case, the body is near earth's surface and hence g may be
considered as a constant with value 9.8 ms-2
-
In gravitational field, g is the gravitational dield strength. It is equal to the
gravitational force acting on a unit mass.
G = F/m
= 9.8 N/kg
( earth’s gravitational field strength on earth’s
surface )
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6. Relation between g and G
Consider a body of mass m on the earth's surface. Assuming the earth behaves as if its
whole mass M were concentrated at its centre O.
mg = GMm/r2
g = GM/r2
7. Variation of g with altitude
Consider g' the acceleration due to gravity at a distance 'a' from the centre of the earth. r
is the radius of the earth and g is the acceleration due to gravity on earth's surface.
i) a > r ( above earth's surface )
g’ = gravitational strength at a distance ‘a’ from
earth’s centre
g = gravitational strength on earth’s surface
ii) a = r ( on the earth's surface )
iii) a < r ( below earth's surface )
- Assumption : the earth is a sphere with uniform density.
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8. Variation of g with latitude
- Assumption : the earth is a sphere with uniform density.
- The variation of g over earth's surface is due to the effect of earth's rotation.
i) at the equator
- there must be a centripetal force mrw2 to maintain its rotation.
w = constant
v changes with
radius of the circle
w = angular velocity of the earth
go = observed acc. due to gravity
g = actual acc. due to the gravity
r = radius of the earth
ii) at the pole
- there is no need to have a centripetal force.
Hence go is larger at the pole relatively.
iii) at latitude
- there must be a centripetal force mxw2 to maintain its rotation.
9. Gravitational field
a) Introduction
- Gravitational field is a region where any object which has mass will experience a force
( i.e. gravitational force ).
- The gravitational field can be produced by a large mass. ( e.g. a planet )
- The lines of force of the gravitational field produced by spherical planet M always
points towards its centre.
- The direction of the attractive gravitational force F experienced by another mass m
inside the gravitational field is along the tangent of the line of force.
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b) Gravitational field strength (g)
- The gravitational field strength (g) is defined as the gravitational force acting on unit
mass.
g=F/m
( Unit : N/kg )
10. Gravitational potential energy U
- the gravitational P.E. of a mass m at a certain point inside a gravitational field
( produced by M ) is equal to the work done in bringing it from infinite distance to that
point.
U(r) = gravitational P.E. of mass m at a
distance r from M
M = mass of the object ehcih produce the
gravitatyional field
Note:
U()=0
i.e. the potential energy at infinity is chosen to be zero.
U (  ) = maximum P.E. = 0
U ( r ) = - GMm/r
Example
i) Find the potential energy difference of a mass m when it is moved from A to B.
rB = 2 x 105 m
rA = 1 x 105 m
M = 2 x 1016 kg
m = 1 x 103 kg
G = 6.7 x 10 -11 Nm2kg-2
( Ans: increase 6700 J )
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ii) Given that the mass m is a rocket and no combustion of fuel occurs when it moves
from A to B, find
a) the speed of the rocket at position B, and
b) its maximum distance away from the planet M if its speed at A is 4 m/s
iii) If the engine of the rocket supplies 3000J to it when it moves from A to B, find the
speed of the rocket at position B if its speed at A is 10 m/s.
11. Gravitational potential ( V )
- The gravitational potential at a point inside a gravitational field ( produced by M ) is the
work done per unit mass in bringing a small body from infinite to that point.
V( r ) = gravitational potential at a distance r from M
U( r ) = gravitational potential energy of a mass m at
a distance r from M
Note:
1. The P.E. in the above case are considered under the condition that : " P.E. = 0 at
infinite distance. "
2. The equation P.E. = mgh ( always used for motion under gravity ) is considered under
the condition that : " P.E. = 0 at earth's surface and g = constant "
Example
1. Potential energy difference near earth's surface ( i.e. R >> h )
If we choose P.E. = 0 at earth surface ( i.e. position B ), the P.E. at A will be mgh.
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2. A mass of 3 kg is lifted from the surface of the earth by 1.6m. What is
a) the change of its P.E.,
b) the change of its potential ?
( Take g = 9.8 N/kg on the earth's surface. )
12. Equipotential lines/ surfaces
- Since V( r ) = - GM /r
The potential V(r) is the same at the spherical
surface which has a radius r from M.
- Equipotential lines/ surfaces are regions in which all the points have the same potential
- Equipotential lines are always perpendicular to the (field) lines of force.
Explanation:
Since A & B have the same potential
=> There is no work done when an object is moved from A to B.
=> Force is perpendicular to the displacement
=> Lines of force is perpendicular equipotential lines
Example The figure shows a region near the surface of the earth.
M = 6 x 1024 kg
G = 6.7 x 10-11 Nm2kg-2
a) Take gravitational P.E. = 0 at infinity, find
i) gravitational potential at A
ii) gravitational potential at B
b) What is the change of gravitational P.E. of a space capsule of mass 1500 kg when it
moves from A to B?
c) If later, it returns through B at a speed of 4 km/s, how fast will it be moving at C ?
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13. Relation between gravitational potential ( V ) and gravitational field strength
(g)
Note : Actually, when consider both the direction and magnitude, we have:
V increases in this direction
Example The figure shows part of the gravitational field near earth's surface. On this
scale the field strength g is approximately uniform. ( g = 9.8 N/kg )
a) What is the distance apart of equipotential surface ?
b) What is the gravitational force on a mass of 2 kg placed at ( i ) A and ( ii ) B
c) What is the gravitational P.E. of a mass of 2kg at (i) A and (ii) B
d) How much work must be done to move the mass from B to A ?
e) Suppose a stone of mass 2 kg has been thrown so that it passes through A and B ( on a
curved path ) : its speed when pass through A is 3 m/s. What is its speed when it (i) is at
B and (ii) hits the ground
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14. Velocity of Escape (v)
- A body of mass m leaves the earth with speed v and just escapes from its gravitational
field. (i.e. can move from re to infinity )
re = radius of the earth
M = mass of the earth
m = mass of the body e.g. spacecraft
Consider the mass m at the earth’s surface,
( g = gravitational field strength on
earth’s surface )
( v = escaping speed )
If there is continuous supply of energy, rocket can leaves the earth even if
Example:
Given that g = 9.8 N/kg on earth’s surface,
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re = radius of earth = 6.38 x10 m
G = 6.7 x 10-11 Nm2kg-2
M = mass of the earth = 6 x 1024 kg
i)
ii)
Find the escaping speed of a rocket which lies on the earth’s surface.
If the speed of the rocket has a value doubled the escaping speed, find its speed
when it is far away from the earth.
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15. Potential energy and Kinetic energy of satellite
- The centripetal force of the satellite is produced by the gravitational force.
P.E. of satellite = - GMm/ r
K.E. of satellite = 1/2 mv2.
Question : Is there anything wrong with the total energy which has a negative value ?
If a stellite moves from a lower orbit to a higher orbit,
e.g. from r1 to r2 with r1 < r2
Since
hence, the total energy is increased. Extra energy is required in this case which should be
supplied by the combustion of fuel.
Example:
Plot the graph of satellite energy (P.E. : K.E. & Et ) against r ( distance
from planet's centre ).
16. Calculation on radius of Parking orbit around earth
a) Parking orbit
If the period of a satellite in revolving the earth is equal to the period of earth's rotation.
The satellite is said to be moving in a parking orbit. Such satellite is called synchronous
satellite and remains in the same position above the earth, apparently satationary. They
make continuous, world-wide communications ( e.g. TV programmes ) possible.
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b) Calculation
Consider a satellite moves around the earth.
g = 9.8 N/kg on earth’s surface,
re = radius of earth = 6.4 x106 m
r = distance of satellite from earth’s centre
T = period for the satellite to have one complete revolution = 24 hours
17. Weightless
- An astronaut orbiting the earth in a space vehicle with its rocket motors off is said to be
weightless
- 'Weightless' is only the sensation of unsupported. It is because the astronaut with
constant speed along the orbit implies he has a centripetal acceleration and this value is
the same as that of the spaceship which is equal to g ( at that height ).
- Therefore, the walls of the space vehicle exert no force on him. Hence, he feels
weightless.
M = mass of the spaceship
m = mass of the astronaut
N = interaction between the astronaut
and the space ship
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18. Kepler’s laws ( about solar system )
1. Each planet moves in an ellipse which has the sun at one focus.
2. The line joining the sun to the moving planet sweeps out equal areas in equal time.
3. The squares of the times of revolution of the planets about the sun are proportional to
the cubes of their mean orbit distance r from it.
i.e. r3/T2 = constant
Derivation of Kepler’s third law
r = mean orbit radius
M = mass of the sun
m = mass of the planet
v = speed of the planet
Verification on Kepler’s second law
In time interval t,
Area swept out A =
rate of area swept out,
For this central force problem, there is no external torque.
Thus angular momentum mrw2 is a constant.
Hence, rate of the sweeping area is also a constant.
Kepler’s first law
- The elliptical orbit is due to the effect of other planets moving round the sun.