Physics  Chemistry  Biology  Mathematics 
Science > Physics > Gravitation > You are Here 
1.1 Statement of Newton’s Law of Gravitation:
 Every particle of matter in the universe attracts every other particle of matter with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.
1.2 Explanation of Newton’s Law of Gravitation:
 Let ‘m_{1}’ and ‘m_{2}’ be the masses of two particles separated by a distance r as shown. According to Newton’s Law of gravitation, these particles will attract each other by a force ‘F’ such that
 Where ‘G’ is constant of proportionality and known as Universal gravitation constant.
2.1 S.I. Unit of Universal Gravitational Constant (G):
By Newton’s law of gravitation
2.2 Dimensions of Universal Gravitational Constant (G):
By Newton’s law of gravitation
Hence the dimensions of universal gravitation constant are [M^{1} L^{3} T^{2}]
3.1 Keppler’s Laws of Planetary Motion:
 Keppler’s First Law (Law of Orbit): Every planet revolves in an elliptical orbit around the sun, with the sun situated at one of the focii of the ellipse.
 Keppler’s Second Law (Law of Equal Areas): The radius vector drawn from the sun to any planet sweeps equal areas in equal intervals of time OR the areal velocity of the radius vector is constant.
 Keppler’s Third Law (Law of Period or Harmonic Law): The square of the period of revolution of the planet around the sun is directly proportional to the cube of the semimajor axis of the elliptical orbit.
4.1 Launching or Projection of Satellite:
 To launch a satellite in an orbit around the earth minimum twostage rocket is used. The launching involves two steps. In the first step the satellite is taken to the desired height and then in the second step, it is projected horizontally with calculated speed in a definite direction.
 The satellite is kept at the tip of the twostage rocket. Initially, the first stage of the rocket is ignited on the ground so that rocket is raised to the desired height, the first stage is detached and falls back on the earth. Then the rocket is rotated by remote control to point it in the horizontal direction. Then the second stage is ignited so rocket gets push in the horizontal direction and acquires certain horizontal velocity (Vh). When the fuel is completely burnt second stage also gets detached. Satellite starts orbiting around the earth.
4.2 The Nature of All Possible Orbits of Satellite:
 Depending upon Magnitude of Horizontal Velocity following four cases can arise for Motion of Satellite
 Case – 1 (v_{h} < v_{c}): If the horizontal velocity imparted to the satellite is less than critical velocity Vc, then the satellite moves in long elliptical orbit with the centre of the Earth as the further focus. If the point of projection is apogee and in the orbit, the satellite comes closer to the earth with its perigee point lying at 180o. If enters earth’s atmosphere while coming towards perigee, it will lose energy and spiral down on the earth. Thus it will not complete the orbit. If it does not enter the atmosphere, it will continue to move in elliptical orbit.
 Case – 2 (v_{h} = v_{c}): If the horizontal velocity imparted to the satellite is exactly equal to the critical velocity Vc then satellite moves in stable circular orbit with earth as centre as shown in the diagram.
 Case – 3 (V_{e} > V_{h} > V_{c}): If the horizontal velocity Vh is greater than critical velocity Vc but less than escape velocity Ve then satellite orbits in
the elliptical path around the earth with the centre of the earth as one of the foci (nearer focus) of the elliptical orbit as shown.  Case – 4 (v_{h} = v_{c}): If the horizontal velocity imparted to the satellite is exactly equal to escape velocity of satellite Ve then satellite moves in a parabolic trajectory.
 Case – 5 (V_{h} > V_{e}): If the horizontal velocity is greater than or equal to escape velocity Ve then satellite overcomes the gravitational
attraction and escapes into infinite space along a hyperbolic trajectory.
5.1 Relation Between the Universal Gravitation Constant and Gravitational Acceleration on the Surface of the Earth:
Let m = mass of body resting on the surface of the earth
M = Mass of the earth
R = Radius of the earth
g = acceleration due to gravity on the surface of the earth.
∴ GM = R^{2}g
This is the required relation between G and g.
6.1 Need for TwoStage Rocket for Projection of Satellite:
 The launching of satellite involves two steps. In the first step the satellite is taken to the desired height and then in the second step, it is projected horizontally with calculated speed in a definite direction.
 To launch a satellite in an orbit around the earth minimum twostage rocket is used.
 The satellite is kept at the tip of the twostage rocket. Initially, the first stage of the rocket is ignited on the ground so that rocket is raised to the desired height, the first stage is detached and falls back on the earth. Then the rocket is rotated by remote control to point it in the horizontal direction. Then the second stage is ignited so rocket gets push in the horizontal direction and acquires certain horizontal velocity (Vh). When the fuel is completely burnt second stage also gets detached. Satellite starts orbiting around the earth.
7.1 Derivation of Expression of Period of Satellite:
Let m = mass of the satellite
M = Mass of the earth
R = Radius of the earth
h = Height of satellite above the surface of the earth
r = R + h = Radius of orbit of the satellite
G = Universal gravitational constant
 The necessary centripetal force for the circular motion of satellite is provided by the gravitational attraction between the satellite and the earth.
Now, Centripetal force = Gravitational force
Now the period of satellite ‘T’ is given by
This is the expression for the time period of a satellite orbiting around the earth at height h from the surface of the earth.
8.1 To prove that the square of the Period of revolution of a satellite is directly proportional to the cube of the orbital radius:
Let m = mass of the satellite
M = Mass of the earth
R = Radius of the earth
h = Height of satellite above the surface of the earth
r = R + h = Radius of orbit of the satellite
G = Universal gravitational constant
 The necessary centripetal force for the circular motion of satellite is provided by the gravitational attraction between the satellite and the earth.
Now, Centripetal force = Gravitational force
Now the period of satellite ‘T’ is given by
Squaring both sides of above equation, we get
For the given planet the quantity in the bracket is constant hence we can conclude that
T^{2} ∝ r^{3}
 Thus the square of the Period of revolution of a satellite is directly proportional to the cube of the orbital radius:
9.1 Definition of Critical Velocity:
 The minimum horizontal velocity of projection that must be given to a satellite at a certain height, so that it can revolve in a circular orbit round the earth is called critical velocity or orbital velocity.
9.2 Derivation of the Expression of Critical Velocity of a Satellite:
Let m = mass of the satellite
M = Mass of the earth
R = Radius of the earth
h = Height of satellite above the surface of the earth
r = R + h = Radius of orbit of the satellite
G = Universal gravitational constant
 The necessary centripetal force for the circular motion of satellite is provided by the gravitational attraction between the satellite and the earth.
Now, Centripetal force = Gravitational force
 This is the expression for the critical velocity of a satellite orbiting around the earth at height h from the surface of the earth.
10.1 Definition of Binding Energy:
 The minimum amount of energy required for a satellite to escape from earth’s gravitational influence is called the binding energy of the satellite.
10.2 Derivation of the Expression for Binding Energy of a Satellite Revolving in a cIrcular Orbit Around the Earth:
 Let us consider a satellite of mass “m” orbiting at height “h” from the surface of earth around the earth with critical velocity V_{c} as shown in the diagram. Let M and R be the mass and radius of earth respectively. The radius ’r’ of the orbit is r = R + h
 When the satellite is orbiting around the earth it possesses two types of mechanical energies. The kinetic energy due to its orbital motion and the potential energy due to its position in the gravitational field of the earth.
 Kinetic Energy of satellite:
 As the gravitational force is providing the necessary centripetal force required for circular motion,
Now, Centripetal force = Gravitational force
Where G is Universal gravitational constant.
The kinetic energy of a satellite orbiting around the earth is given by
 Potential Energy:
 Now, the satellite is in the gravitational field of the earth. The gravitational potential at a point on the surface of the earth is given by
The potential energy of a body (satellite) Is given by
P.E. = The gravitational potential x the mass of a satellite
 Total Energy of Satellite:
The total mechanical energy of the satellite In orbit is given by
T.E. = K.E. + P.E.
 The negative sign indicates that the satellite is bound to the earth by attractive forces and cannot leave it on its own. To move the satellite to infinity .we have to supply energy from outside the planetsatellite system. This energy Is known as binding energy of a satellite.
 Binding Energy of Satellite:
B.E. = (Energy of a Satellite at Infinity) – (The energy of the satellite in the orbit)
 This is an expression for Binding Energy of a satellite orbiting around the earth In stable circular orbit. Numerically, Binding Energy Is equal to the total energy of a satellite In the orbit.
11.1 Derivation of the Expression for Binding Energy of a Satellite at Rest on the Earth’s Surface:
 Consider a satellite of mass ‘m’ which is at rest on the earth’s surface. Let M be the mass an R be the radius of the earth.
 As the satellite is at rest, it will not possess any kinetic energy. i.e. K.E. = 0.
 Now, the satellite is in the gravitational field of the earth. The gravitational potential at a point on the surface of the earth is given by
The potential energy of a body (satellite) Is given by
P.E. = The gravitational potential x the mass of a satellite
The total mechanical energy of the satellite on the surface of the earth is given by
T.E. = K.E. + P.E.
 The negative sign indicates that the satellite is bound to the earth by attractive forces and cannot leave it on its own. To move the satellite to infinity we have to supply energy from outside the planetsatellite system. This energy is known as binding energy of a satellite.
 Binding Energy:
B.E. = (Energy of a Satellite at Infinity) – (The energy of the satellite in the orbit)
 This is an expression for the binding energy of a satellite at rest on earth’s surface.
12.1 Definition of Escape Velocity:
 The minimum velocity with which a body should be projected from the surface of the earth so that it escapes from the earth’s gravitational field, is called escape velocity of the body.
12.2 Derivation of the Expression for Escape Velocity of a Satellite:
 Consider a body of mass ‘m’ which is at rest on the surface of the earth. Let M be the mass of the earth and R be the radius of the earth. Then the binding energy of the body on the surface of the earth is given by
Where G is Universal gravitational constant.
Let V_{e} be the escape velocity, then the kinetic energy given to the body is given by
 It means Satellite should be given this much kinetic energy so that it can go out of earth’s gravitational influence.
K.E. = B.E.
This is an expression for escape velocity of a satellite on the surface of the earth.
13.1 Feeling of Weightlessness Experienced by an Astronaut in Satellite:
 By the definition, the weight of a body is equal to the gravitational force with which the body is attracted towards the centre of the earth. When the astronaut is on the surface of the earth, his weight acts vertically downwards. At the same time, the earth’s surface exerts an equal and opposite force of reaction on the astronaut. Due to this force of reaction, the astronaut feels his weight.
 When the astronaut is in an orbiting satellite, a gravitational force still acts upon him. However, in this case, both the astronaut as well as the satellite are now attracted towards the earth and have the same centripetal acceleration due to gravity at that place. It is the condition of free fall of the body.
 As both astronaut and the surface of the satellite are attracted towards the earth centre with the same acceleration, and hence the astronaut can’t produce any action on the floor of the satellite. So the floor does not give any reaction on the astronaut. Hence the astronaut has a feeling of weightlessness.
14.1 Variation of g with a) altitude b) depth and c) latitude
 Let m = mass of body resting on the surface of the earth, M = Mass of the earth, R = Radius of the earth, g = acceleration due to gravity on the surface of the earth.
 Variation in Acceleration Due to Gravity Due to Altitude:
g_{h} = acceleration due to gravity at height ‘h’
Expanding binomially and neglecting terms of the higher power of (h/R) we get
 This is an expression for the acceleration due to gravity at small height ‘h’ from the surface of the earth. The acceleration due to gravity decreases as we move away from the surface of the earth.

Variation in acceleration due to gravity Due to Depth:
The acceleration due to gravity on the surface of the earth is given by
Let ‘ρ’ be the density of the material of the earth.
Now, mass = volume x density
Substituting in the equation for g we get
 Now, let the body be taken to the depth, ’d’ below the surface of the earth. Then acceleration due to gravity g_{d }at the depth, ’d’ below the surface of the earth is given by
Dividing equation (3) by (2) we get
 This is an expression for the acceleration due to gravity at the depth, ’d’ below the surface of the earth. The acceleration due to gravity decreases as we move down into the earth.

Variation in acceleration due to gravity Due to Latitude of the Place:
 The latitude of a point is the angle Φ between the equatorial plane and the line joining that point to the centre of the earth.
 Let us consider a body of mass ‘m’ at a point P with latitude ‘Φ’ as shown on the surface of the earth. Let ‘g_{Φ}’ be the acceleration due to gravity at point P.
OP = Radius of earth = R
O’P = Distance of point P from the axis of earth
 Due to rotational motion of the earth about its axis, the body at P experiences a centrifugal force which is given by mrω^{2 }which acts radially outward. The component of centrifugal force along the radius of the earth is mrω^{2 }cosΦ.
 Now the body is acted upon by two forces its weight mg acting towards the centre of the earth and the component mrω^{2 }cosΦ acting radially outward. The difference between the two forces gives the weight of that body at that point.
mg_{Φ} = mg – mrω^{2}cosΦ ………….. (1)
Now cos Φ = O’P / OP = r/R
∴ r = R cos Φ
Substituting in equation (1)
mg_{Φ} = mg – m(R cos Φ)ω^{2}cos Φ
∴ g_{Φ} = g – R ω^{2}cos^{2} Φ
 This is an expression for acceleration due to gravity at a point P on the surface of the earth having latitude Φ.
15. 1 Geostationary Satellite:
 An artificial satellite revolving in a circular orbit around the earth in the equatorial plane, in the same sense of rotation of the earth and having the same period of revolution as the period of rotation of the earth i.e 24 hours, is called geostationary satellite.
15.2 Uses of Satellites:
 They are used for sending microwave and TV signals from one place to another.
 They are used for weather forecasting and to give warnings of cyclones to coastal villages.
 They are used for detecting water resource locations and areas rich in ores.
 They are used for spying In enemy countries i.e. It can be used for military purposes
 They are used for global positioning system (GPS).
 They are used for astronomical observations and to study of solar and cosmic radiations.
Science > Physics > Gravitation > You are Here 
Physics  Chemistry  Biology  Mathematics 