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Gravitation
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Gravitation
Newton’s law of gravitation
Every particle in the universe attracts every other particle with a force that is proportional to
the product of their masses and inversely proportional to the square of their distance apart.
This force is always attractive.
where G = 6.67 x 10-11 N m2 kg-2 is the gravitational constant
Note that gravity is a property associated with mass.
Examples:
1. Two particles of masses 3 kg and 15 kg are separated by a distance of 2.5 m. Which of
these particles exerts the greater gravitational force on the other. Calculate the
magnitude of the gravitational force each particle experiences.
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Note that both particles should
accelerate towards each other.
However, at this instant we are only
concerned with the force acting on
each particle.
As you can see gravitational forces are very weak forces.
2. Two particles exert a gravitational force of 3 x 10-5 N on each other. If they are 0.05 m
apart and the mass of one of them is 40 kg, what is the ratio of their masses?
Gravitational Fields
We have seen that two particles will each experience a force of attraction due to the other.
Another way of analyzing this phenomenon is to say that each particle finds itself in the
gravitational field created by the other. The force that one particle experiences is the result
of its interaction with the field produced by the other. A gravitational field is a region in
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space where a body experiences a force by virtue of its mass.
Field surrounding
point mass
Note that the field is stronger where the lines are
more packed together.
The gravitational field is represented by field lines. The direction of the lines indicate the
direction in which a gravitational force would act on a mass placed in the field.
The field surrounding a particle is said to be a radial field. Why this is so is obvious from the
figure above.
Orbits
The planets travel in circular orbits (actually they are elliptical) about the sun. Moons and
other satellites travel in circular orbits around planets. In all cases the force causing the
centripetal acceleration is the force of gravity.
The net force acting on m is
From Newton’s third law we know that the net force acting on a body is equal to the product
of the body’s mass and its acceleration.
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Note that the period and frequency of rotation do not depend on the mass of the rotating
body. It depends only on M and r. The same could be said for the linear speed of the orbiting
body.
Examples:
1. Find the period of revolution of a satellite moving in a circular orbit round the earth at
a height of 3.6 x 106 m above the earth’s surface. Assume the earth is a uniform sphere
of radius 6.4 x 106 m, the earth’s mass is 6 x 1024 kg and G is 6.7 x 10-11 N m2 kg. ans:
2.75 hours
2. A satellite of mass 66 kg is in orbit round the earth at a distance of 5.7 R above its
surface, where R is the value of the mean radius of the earth. If the gravitational field
strength at the earth’s surface is 9.8 N kg-1, calculate the centripetal force acting on
the satellite.
Assuming the earth’s mean radius to be 6400 km, calculate the period of the satellite in
orbit in hours. ans: 14.4 N, 24.5 h
Parking Orbits
A parking orbit is one in which the satellite revolves around the planet with a period equal to
that of the planet’s rotation. In such a case the satellite will always be positioned over the
same point on the planet. Sometimes such orbits are called gyro-synchronous orbits, or geostationary orbits.
As we have seen, the period of an orbit around a given planet depends only on its radius.
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Therefore all geo-stationary satellites orbit at a fixed distance from the equator.
The height h above the surface is (42 340 - 6400) = 35 940 km.
The speed of the satellite in orbit is easily calculated as
Example
A satellite is to be put into orbit 500 km above the earth’s surface. If its vertical velocity
after launching is 2000 m s-1 at this height, calculate the magnitude and direction of the
impulse required to put the satellite directly into orbit, if its mass is 50 kg. Assume g = 10 N
kg-1, radius of earth = 6400 km. ans: 4.0 x 105 N s
note:- On seeing that the orbit is only 500 km above the earth’s surface we can conclude
"right off the bat" that the orbit is not geo-stationary.
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