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Introduction to motion of celestial bodies

Kepler’s laws

Kepler’s law of orbits

Kepler’s law of areas

Central force

Kepler’s law of periods


Centripetal force

Universal law of gravitation

The gravitational force due to a hollow spherical shell

Finding the gravitational constant

Acceleration due to gravity of the earth

Acceleration due to gravity above the surface of earth

Acceleration due to gravity below the surface of earth

Gravitational potential energy

Escape velocity

Why is there no air on moon?

Earth’s satellites

Energy of an orbiting satellite

Geostationary satellites

Polar satellites




Introduction to Motion of Celestial Bodies

Ptolemy proposed a geocentric model of motion of celestial bodies, about 2000 years ago that all celestial objects, stars, the sun and the planets, revolved around the earth.

Similar theories were also proposed by Indian philosophers.

Aryabhatta (5th century A.D.) had proposed a ‘heliocentric’ model in which the Sun was the center around which the planets revolved

A Polish monk named Nicolas Copernicus (1473-1543) proposed a model in which the planets moved in circles around a fixed central sun, however the church discredited his model.

Tycho Brahe (1546-1601) from Denmark spent his entire lifetime recording observations of the planets with the naked eye.

His compiled data were analysed later by his assistant Johannes Kepler (1571-1640) and he propsed three laws that are known as Kepler’s laws.


  1. Kepler’s Law of Orbits: All planets move in elliptical orbits with the Sun situated at one of the foci of the ellipse.

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    Locus of a point, which moves in such a way that the sum of its distances from two fixed points always remains constant. The two fixed points are called foci (plural for focus).

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  1. Kepler’s Law of areas: The line that joins any planet to the sun sweeps equal areas in equal intervals of time.

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  1. Kepler’s Law of periods: The square of the time period of revolution of a planet is proportional to the cube of the semi-major axis of the ellipse traced out by the planet.

      T2 ∝ a3


Gravitation is the force of attraction between any two objects in the universe.

Italian physicist Galileo (1564-1642) recognised the fact that all bodies, irrespective of their masses, are accelerated towards the earth with a constant acceleration.

Galileo carried out experiments with bodies rolling down inclined planes and arrived at a value of the acceleration due to gravity.

Centripetal Force

The force directed towards the centre of the circle is called centripetal force.

Universal Law of Gravitation

Every body in the universe attracts every other body with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

The force is along the line joining the centres of two objects.



And F ∝ 1r2

Therefore, we can write,

F= G m1×m2r2

Where G is the universal gravitational constant, m1 and m2 are the masses of two bodies and r is the distance between them.

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The SI unit of G is N m2 kg-2.

The value of G is 6.673 × 10-11 N m2 kg-2.

The gravitational force due to a hollow spherical shell

Finding the Gravitational Constant

The value of gravitational constant G was first found by English scientist Henry Cavendish in 1798.

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The bar AB has two small lead spheres attached at its ends. The bar is suspended from a rigid support by a fine wire. Two large lead spheres are brought close to the small ones on opposite sides as shown. The big spheres attract the nearby small ones by equal and opposite force as shown. There is no net force on the bar but only a torque which is equal to F times the length of the bar, where F is the force of attraction between a big sphere and its neighbouring small sphere.

The bar AB rotates a little due to the torque. The angle of rotation can be measured experimentally.

Due to this torque, the suspended wire gets twisted till such time as the restoring torque of the wire equals the gravitational torque. If θ is the angle of twist of the suspended wire, the restoring torque is proportional to θ, equal to τθ, where τ is the restoring couple per unit angle of twist.

If d is the separation between the centers of the big and its neighbouring small ball, M and m their masses, the gravitational force between the big sphere and its neighouring small ball is,

F= G M×md2

And torque

G M×md2 L=τθ 

 G=τ θ d2M m L

All variables except G are known, hence, we can calculate the value of G.

Value of G = 6.67×10-11 N m2/kg2

Why do we say that Cavendish weighed the earth?

Acceleration g is readily measurable. RE is a known quantity. The measurement of G by Cavendish’s experiment (or otherwise), enables us to estimate ME. This is the reason why there is a popular statement “Cavendish weighed the earth”.

Acceleration Due to Gravity of the Earth

The acceleration experienced by the mass m, which is denoted by ‘g’ on the surface of earth is given by

g =Fm= G MERE2

Where RE = radius of earth and ME = mass of earth.

Acceleration due to gravity above the surface of earth

Consider a point mass m at a height h above the surface of the earth. If the radius of the earth is RE, its distance from the centre of the earth is (RE + h).

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If F(h) denotes the the magnitude of the force on the point mass m at a height h from earth’s surface, we get,

Fh= G ME×m(RE+h)2 

  gh=G ME(RE+h)2= GMERE21 1+hRE2

gh=g 1 1+hRE2

gh=g 1+hRE-2

For hRE << 1, using binomial expansion, we have,

gh= g 1-2hRE

Acceleration due to gravity below the surface of earth

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Consider a point mass m at a depth d below the surface of the earth.

The distance from the center of the earth is (RE – d).

The earth can be thought of as being composed of a smaller sphere of radius (RE – d) and a spherical shell of thickness d.

The force on m due to the outer shell of thickness d is zero

For smaller sphere of radius (RE – d), the point mass is outside it and hence, the force due to this smaller sphere is just as if the entire mass of the smaller sphere is concentrated at the centre.

If MS is the mass of the smaller sphere, since mass of a sphere is proportional to be cube of its radius,


 MS = MERE  d3RE3

Thus the force on the point mass is

F(d)= G MS×m(Re-d)2 

  gd=G MS(Re-d)2

gd=GMERE  dRE3 

gd=g RE  d RE


Gravitational Potential Energy

If the position of the particle changes on account of forces acting on it, the change in its potential energy is equal to the amount of work done on the body by the force.

Forces for which the work done is independent of the path are the conservative forces. Gravitational force is conservative.

The force of gravity is practically a constant equal to mg, directed towards the center of the earth.

If we consider a point at a height h1 from the surface of the earth and another point vertically above it at a height h2 from the surface, the work done in lifting the particle of mass m from the first to the second position

If we associate a potential energy W(h) at a point at a height h above the surface such that


Setting h = 0 in the last equation, we get

h = 0 means points on the surface of the earth. Thus, Wo is the potential energy on the surface of the earth.

Now let us consider points at arbitrary distance from the surface of the earth. The gravitational force is no longer constant.

Gravitational force outside the earth at distance r from the centre of earth is,

F(r)= G ME×mr2

where ME = mass of earth, m = mass of the particle and r its distance from the center of the earth.

If we now calculate the work done in lifting a particle from r = r1 to r = r2 (r2 > r1) along a vertical path, we get,

W12 =r1r2G MEmr2dr

We can thus write, for r > RE

Wr= - GMEmr+ W1

Hence, W12 = W(r2) – W(r1)

Setting r = infinity in the last equation, we get

W (∞) = W1.

Thus, W1 is the potential energy at infinity.

Only the difference of potential energy between two points has a definite meaning. Actual value at any point has no meaning.

We set W1 = 0, so that the potential energy at a point is just the amount of work done in displacing the particle from infinity to that point.

Escape Velocity

Escape velocity is the minimum speed needed for an object to escape from the gravitational influence of a massive body.

Suppose the object reached infinity with speed Vf. The energy of an object is the sum of potential and kinetic energy. If W1 denotes that gravitational potential energy of the object at infinity, the total energy of the projectile at infinity is

E() = W1 + mVf22

If the object was thrown initially with a speed Vi from a point at a distance (h + RE) from the center of the earth (RE = radius of the earth), its energy initially was

Eh + RE= -GMEmh + RE+W1 + mVf22

By principle of conservation of energy, we have

W1 + mVf22= -GMEmh + RE+W1 + mVi22

-GMEmh + RE + mVi22=  mVf22

The R.H.S. is a positive ( > 0 ) hence the L.H.S must also be positive ( > 0 ), i.e.

-GMEmh + RE + mVi22>0  

  mVimin22>GMEmh + RE 

  Vimin2=2GMEh + RE

  Vimin=2GMEh + RE

If h = 0, that is the object is thrown from the surface of the earth, then

Vimin=2GME RE

Escape velocity from Earth’s surface is about 11.186 km/s.

Why is there no air on moon?

The escape speed for the moon is 2.3 km/s, about five times smaller than that on earth. This is the reason that moon has no atmosphere. Gas molecules if formed on the surface of the moon having velocities larger than this will escape the gravitational pull of the moon.

Earth’s Satellites

Earth satellites are objects which revolve around the earth.

Their motion is very similar to the motion of planets around the Sun and hence Kepler’s laws of planetary motion are equally applicable to them.

Their orbits around the earth are circular or elliptic. Moon is the only natural satellite of the earth with a near circular orbit with a time period of approximately 27.3 days which is also roughly equal to the rotational period of the moon about its own axis.

Let us consider a satellite in a circular orbit of a distance (RE + h) from the centre of the earth, where RE = radius of the earth. If m is the mass of the satellite and V its speed, the centripetal force required for this orbit is,

FC=  mV2(RE+h)

This centripetal force is provided by the gravitational force,

Fg= G ME×m(RE+h)2

Equating the two forces,

 mV2(RE+h)=  GME(RE+h)

V=  GME(RE+h)

So, time period of revolution,

T = 2πRE+hV

T= 2π(RE+h)3/2GME

Squaring both sides,

T 2=  4π2(RE+h)3GME=k(RE+h)3

T 2 r3

This is the Kepler’s law of periods.

If h = 0, that is the satellite is very near the earth, we get,


To = 2π RE3GME= 2π REg

If we substitute the numerical values g = 9.8 m s-2 and RE = 6400 km, we get,

To = 2π 6.4 ×1069.8=5079.62 s  85 minutes   

Energy of an orbiting satellite

The kinetic energy of the satellite in a circular orbit with speed v is,

KE =12mv2=12  GMEm(RE+h)  

Since gravitational potential energy at infinity is zero, the potential energy at distance (RE + h) from the center of the earth is,

PE= - GMEm(RE+h)

Total energy therefore is,


=12  GMEmRE+h- GMEmRE+h

=-12  GMEm(RE+h)

Q: Why is the total energy of a satellite negative?

If the total energy is positive or zero, the object escapes to infinity. Satellites are always at finite distance from the earth and hence their energies cannot be positive or zero.

Geostationary satellites

Satellites in a circular orbits around the earth in the equatorial plane with T = 24 hours are called geostationery Satellites. Since the earth rotates with the same period, the satellite would appear fixed from any point on earth.


T = 2π(RE+h)3/2GME

RE+h= T2 G ME4π2 1/3

For T = 24 hours, h works out to be 35800 km, which is much larger than RE.

Polar satellites

These are low altitude (h 500 to 800 km) satellites going around the poles of the earth in a north-south direction whereas the earth rotates around its axis in an east-west direction.

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Since its time period is around 100 minutes it crosses any altitude many times a day.

A camera fixed on it can view only small strips of the earth in one orbit. Adjacent strips are viewed in the next orbit, so that in effect the whole earth can be viewed strip by strip during the entire day.

These satellites can view polar and equatorial regions at close distances with good resolution.

Information gathered from such satellites is useful for remote sensing, meterology as well as for environmental studies of the earth.


Weight of an object is the force with which the earth attracts it.

When bodies are falling with the same acceleration as that the acceleration due to gravity, they are said to be in free fall and it will feel weightless. This phenomenon is called the phenomenon of weightlessness.

Examples of free fall