# Physics Important Questions: Gravitation

 Maharashtra State Board > Science >  Gravitation > You are Here

### Very Short Answers  (1 Mark)

Q1. State dimensions of universal gravitational constant ‘G’.

• The dimensions of universal gravitational constant ‘G’ are [M-1 L3 T-2]

Q2. What do you mean by a satellite?

• Any object that revolves around a given planet in circular orbit under the influence of the planet’s gravitational force is called as a satellite. e.g. the moon is a natural satellite of the earth.

Q3. Define the critical velocity of a satellite.

• 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 around the earth is called critical velocity or orbital velocity.

Q4. Define periodic time of a satellite.

• The time taken by the satellite to complete one revolution around the planet is known as the period of revolution of the satellite.

Q5. State Kepler’s law of equal areas.

• The radius vector drawn from the Sun to the planet sweeps out equal areas in equal time. i.e. aerial velocity of the radius vector is constant. Thus the Aerial velocity of the satellite is always constant. i.e.  A/t = Constant.

Q6. Define binding energy of a satellite.

• The binding energy of a satellite can be defined as the minimum amount of energy required to be supplied to it in order to free the satellite from the gravitational influence of the planet (i.e. in order to take satellite from the orbit to a point at infinity).

Q7. State dimensions of gravitational potential.

• The dimensions of gravitational potential are [M0 L2 T-2]

Q8. Define the escape velocity of a satellite.

• The escape velocity of a body which is at rest on the earth’s surface Is defined as that minimum velocity with which It should be projected from the surface of the earth so that it escapes from the earth’s gravitational influence.

Q9. What do you mean by geostationary satellite?

• A geostationary satellite or a communication satellite is an artificial satellite which revolves around the earth in circular orbit in the equatorial plane such that, a) its direction of motion is the same as the direction of rotation of the ‘earth about its axis and b) its period is the same as the period of rotation of the earth, i.e. 24 hours.
• When observed from the earth’s surface, this satellite appears stationary. Therefore, it is called a geostationary satellite. Its motion Is synchronous with the rotational motion of the earth, it Is called a geosynchronous satellite.

Q10. State Newton’s law of gravitation.

• Newton’s law of gravitation states that “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.”

### Shot Answers – I (2 Marks)

Q1. State Newton’s law of gravitation. Express it in vector form.

• Newton’s law of gravitation states that “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.”
• Newton’s law of gravitation in vector form

Q2. State MKS and CGS units of the universal gravitational constant and obtain its dimensions.

• The SI unit of constant of gravitation is N m2 kg-2 and c.g.s. unit is dyne cm2 g-2 .

By Newton’s law of gravitation

Hence the dimensions of universal gravitation constant are [M-1 L3 T-2]

Q3. Obtain the relation between the universal gravitational constant and the gravitational acceleration on the surface of the earth.

• Let us consider a body of mass ‘m’ be at rest on the surface of the earth on which the acceleration due to gravity is ‘g’.  If ‘M’ and ‘R’ are mass and radius of earth (planet) respectively

Now the force of attraction on the body is equal to its weight ‘mg’

GM = R2g

This is the relation between the universal gravitational constant and the gravitational acceleration on the surface of the earth

Q4. How is an artificial satellite launched into a circular orbit around the earth?

• To launch a satellite in an orbit around the earth multistage 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. If the velocity is proper it starts revolving in stable circular orbit.
• To understand launching of satellite let us consider a simple case of the use of a two-stage rocket.  The satellite is kept at the tip of the two-stage 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. 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 the second stage also gets detached. Satellite starts orbiting around the earth.

Q5. State conditions for various possible orbits of a satellite.

• Depending upon Magnitude of Horizontal Velocity following four cases can arise for Motion of Satellite
• Case – 1 (vh < vc): 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 (vh = vc): 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 (Ve > Vh > Vc): 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 (vh = vc): 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 (Vh > Ve): 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.

Q6. Obtain an expression for the critical velocity of a satellite

• The constant horizontal velocity given to the satellite so as to put it into stable circular orbit around the earth is called as critical velocity and is denoted by Vc. It is also known as orbital speed or proper speed.
• Let us consider a satellite of mass “m” orbiting at height “h” from the surface of earth around the earth with critical velocity Vc 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

• 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. The expression does not contain the term, ‘m’. Hence we can conclude that the critical velocity of a satellite is independent of the mass of the satellite.

Q7. Show that the square of the period of revolution of a satellite is directly proportional to the cube of the radius of its orbit.

• Let us consider a satellite of mass “m” orbiting at height “h” from the surface of earth around the earth with critical velocity Vc 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
• 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.

Squaring both sides of the above equation, we get

For the given planet the quantity in the bracket is constant hence we can conclude that

T2  ∝  r3

• Thus the square of the period of a satellite is directly proportional to the cube of the radius of its orbit. (Kepler’s third law of planetary motion)

Q8. Assuming the expression for critical velocity, obtain an expression for the period of a satellite in a circular orbit.

• Let us consider a satellite of mass “m” orbiting at height “h” from the surface of earth around the earth with critical velocity Vc 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
• The critical velocity or orbital velocity of the satellite is given by

Now the period of satellite ‘T’ is given by

This is the expression for the time period of a satellite orbiting around the earth in a circular orbit.

Q9. State any four applications of a communication satellite.

• The communication satellites are used for sending microwave and TV signals from one place to another.
• They are used for weather forecasting.
• 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

Q10. State Kepler’s law of orbit and law of period.

• First Law (Keppler’s Law of Elliptical Orbits): Every planet moves around the sun in a closed elliptical orbit with the sun at one of its foci.
• Third Law (Keppler’s Law of Period): The square of the period of a satellite is directly proportional to the cube of the semimajor axis of its elliptical orbit.

T2  ∝  a3

Q11. Obtain an expression for the binding energy of a satellite revolving around the earth at a certain altitude.

• The binding energy of a satellite can be defined as the minimum amount of energy required to be supplied to it in order to free the satellite from the gravitational influence of the planet (i.e. in order to take satellite from the orbit to a point at infinity).
• Consider a satellite revolving around the earth in a circular orbit. Necessary centripetal force to keep the satellite orbiting in a stable circular orbit is provided by the force of gravitational attraction between the earth and the satellite.
• 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.

Let, M  = the mass of the earth

R   = the radius of the earth

h   = the height of the satellite above  the surface of the earth

vc  =  the critical velocity of the satellite

m  = the mass of the satellite.

r   =  the radius of a circular orbit of the satellite = (R + h)

• 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 planet-satellite 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.

Q12. Derive an expression for the binding energy of a body at rest on the earth’s surface.

• Consider a satellite of mass, ‘m’ which is at rest on the earth’s surface.  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 planet-satellite system.  This energy is known as the 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 stationary on earth’s surface.

Q13. Derive an expression for the escape speed of a body from the surface of the earth. Hence show that it is independent of the mass of a satellite.

• The escape velocity of a body which is at rest on the earth’s surface Is defined as that minimum velocity with which It should be projected from the surface of the earth so that it escapes from the earth’s gravitational influence.
• 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 Ve 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.

• The expression does not contain the term m, the mass of satellite. Hence escape velocity is independent of the mass of the satellite.

Q14. If R is the radius of the earth and9 is the mean density of the earth, then show that the escape speed of a body from the surface of the earth is

• Proof:

The escape velocity of a satellite on the surface of the planet is given by

Where G = Universal gravitational constant

M = the mass of the Planet

R = the radius of the Planet

Let d    = density of the material of the planet

Now, Mass of Planet   =  Volume of Planet x Density

Proved as required.

Q15. Explain why an astronaut in an orbiting satellite has a feeling of weightlessness.

• 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.
• 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. When we are moving down in accelerating lift, we can have the feeling of partial weightlessness.

Q16. Show that acceleration due to gravity at height ‘h’ is

• Proof:
• LetIf M be the mass of the earth, R be its radius, g be acceleration due to gravity on the surface of the earth and gh be acceleration due to gravity at height ‘h’ from the surface of the earth.

Proved as required.

Q17. What is a communication satellite? State any two uses of a communication satellite.

• A communication satellite is an artificial satellite which revolves around the earth in circular orbit in the equatorial plane such that, a) its direction of motion is the same as the direction of rotation of the ‘earth about its axis and b) its period is the same as the period of rotation of the earth, i.e. 24 hours.
• Uses of Communication Satellite:
• The communication satellites are used for sending microwave and TV signals from one place to another.
• They are used for weather forecasting.

Q18. Show that the escape velocity of a satellite from the surface of the earth is √  times the critical velocity for a satellite revolving very close to the earth’s surface.

For an orbiting satellite, critical velocity is given by

where G = Universal gravitational constant

M = the mass of the earth

R = the radius of the earth

h = height of the satellite above the earth’s surface.

• For satellite orbiting very close to the earth, h can be neglected as h < < R. Hence it can be neglected. Therefore the critical velocity of the satellite orbiting very close to earth’s surface.

The escape velocity of a satellite oh the surface of the earth is given by

Dividing equation (1) by (2)

• Thus the escape velocity of a body from the surface of the planet (earth) is √2  times the critical velocity of the body when it is orbiting close to the planet’s (earth’s) surface.

Q19. If ρ is the mean density of the planet and R is the radius of the planet, then  Show that the period of revolution of an artificial satellite is equal to

• Proof:

The time period of a satellite orbiting around the earth is given by

If satellite orbiting very close to the earth  (i.e. h < < R) then h can be neglected. Then R + h = R

Proved as required.

Q20. State expressions for acceleration due to gravity at depth, ‘d’ and altitude at height ‘h’ from the earth’s surface. Draw a graph showing the variation of the gravitational acceleration with depth and altitude.

The acceleration due to gravity at height h from the surface of the earth is given by

The acceleration due to gravity at depth d from the surface of the earth is given by

The variation of gravitational acceleration with depth and altitude is as follows

Q21. Show that the escape velocity of a body of mass, ‘m’ from the surface of the earth is equal to √2FR/m  where F is the gravitational force and R is the radius of the earth.

• Let M be the mass of planet and R be its radius. Mass of satellite is ‘m’.

By Newton’s law of gravitation, the gravitational force is given by

22. Draw a diagram showing different stages of projection for an artificial satellite

#### Short Answers II (3 Marks)

Q1. Define critical speed of a satellite and obtain an expression for it. On what factors does it depend?

• The constant horizontal velocity given to the satellite so as to put it into stable circular orbit around the earth is called as critical velocity and is denoted by Vc. It is also known as orbital speed or proper speed.
• Let us consider a satellite of mass “m” orbiting at height “h” from the surface of earth around the earth with critical velocity Vc 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

• 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. The expression does not contain the term ‘m’. Hence we can conclude that the critical velocity of a satellite is independent of the mass of the satellite.

Q2. Define the period of revolution of a satellite. Derive an expression for the period of revolution of a satellite in a circular orbit.

• The time taken by the satellite to complete one revolution around the planet is known as the period of revolution of the satellite.
• Let us consider a satellite of mass “m” orbiting at height “h” from the surface of earth around the earth with critical velocity Vc 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
• 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 in a circular orbit.

Q3. State Kepler’s laws of planetary motion.

• First Law (Keppler’s Law of Elliptical Orbits): Every planet moves around the sun in a closed elliptical orbit with the sun at one of its foci.
• Second Law (Keppler’s Law of Equal Areas): The radius vector drawn from the Sun to the planet sweeps out equal areas in equal time. i.e. aerial velocity of the radius vector is constant. Thus the Aerial velocity of the satellite is always constant. i.e.  A/t = Constant.

• Third Law (Keppler’s Law of Period): The square of the period of a satellite is directly proportional to the cube of the semimajor axis of its elliptical orbit.

T2  ∝  a3

Q4. Define binding energy of a satellite. Obtain an expression for the binding energy of a satellite revolving around the earth at a certain altitude.

• The binding energy of a satellite can be defined as the minimum amount of energy required to be supplied to it in order to free the satellite from the gravitational influence of the planet (i.e. in order to take satellite from the orbit to a point at infinity).
• Consider a satellite revolving around the earth in a circular orbit. Necessary centripetal force to keep the satellite orbiting in a stable circular orbit is provided by the force of gravitational attraction between the earth and the satellite.
• 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.

Let, M  = the mass of the earth

R   = the radius of the earth

h   = the height of the satellite above  the surface of the earth

vc  =  the critical velocity of the satellite

m  = the mass of the satellite.

r   =  the radius of a circular orbit of the satellite = (R + h)

• 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 planet-satellite 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.

Q5. Derive an expression for the gravitational acceleration on the earth’s surface at a latitude ‘Φ‘.

• 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. Latitude of the equator is 0° and that of poles is 90°.
• Let us consider a body of mass ‘m’ at a point P with latitude ‘Φ’ as shown on the surface of the earth. Let ‘gΦ’ 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 mrω2. Let us resolve this centrifugal force into two rectangular components. Its component along the radius of the earth is mrωcosΦ.
• Now the body is acted upon by two forces its weight mg acting towards the centre of the earth and the component mrωcosΦ acting radially outward. The difference between the two forces gives the weight of that body at that point.

mgΦ = mg – mrω2cosΦ ………….. (1)

Now cos Φ = O’P / OP = r/R

∴  r  = R cos Φ

Substituting in equation (1)

mgΦ = mg – m(R cos Φ)ω2cos Φ

∴   gΦ = g – R ω2cos2 Φ

• This is an expression for acceleration due to gravity at a point P on the surface of the earth having latitude Φ.
• At the equator  Φ = 0°. Hence ‘g’ is minimum on the equator. For the poles Φ = 90°. Hence ‘g’ is maximum on the poles.

Q6. Derive an expression for acceleration due to gravity at a depth ‘d‘ below the earth’s surface.

• 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 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 gat 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. This expression shows acceleration due to gravity decreases as we move down into the earth. At the centre of the earth d = R, hence acceleration due to gravity at the centre of the earth is zero.

Q7. State expression for acceleration due to gravity at depth ‘d’ and altitude ‘h’. Hence show that their ratio is equal to(R – d)/(R- 2h), assuming that h<<<<R, where R is the radius of the earth

The acceleration due to gravity at height h from the surface of the earth is given by

The acceleration due to gravity at depth d from the surface of the earth is given by

Proved as required

 Maharashtra State Board > Science >  Gravitation > You are Here

### One Comment

1. ऋषिकेश देवकर

सर , please maths चे question Bank of all chapters update करा.