Maharashtra Board Class 11 Physics Chapter 5 Gravitation Solution

Maharashtra Board Class 11 Physics Solution Chapter 5 – Gravitation

Balbharati Maharashtra Board Class 11 Physics Solution Chapter 5: Gravitation. Marathi or English Medium Students of Class 11 get here Gravitation full Exercise Solution.

Std

Maharashtra Class 11
Subject

Physics

Chapter

5
Chapter Name

Gravitation


1.) Choose the correct option.

i) The value of acceleration due to gravity is maximum at

(A) the equator of the Earth .

(B) thecentre of the Earth.

(C) the pole of the Earth.

(D) slightly above the surface of the Earth.

Answer- C

The value of acceleration due to gravity is maximum at pole of earth and minimum at equator of earth.

ii) The weight of a particle at the centre of the Earth is

(A) infinite.

(B) zero.

(C) same as that at other places.

(D) greater than at the poles.

Answer- B

AT the centre of the Earth the value of acceleration due to gravity is zero hence the weight of a particle is also zero.

iii) The gravitational potential due to the Earth is minimum at

(A) the centre of the Earth.

(B) the surface of the Earth.

(C) a points inside the Earth but not at  its centre.

(D) infinite distance.

Answer- A

iv) The binding energy of a satellite revolving around planet in a circular orbit is 3×109 J. Its kinetic energy is

(A) 6×109J

(B) -3 ×109J

(C) -6 ×109J

(D) 3 ×109J

Answer- D

We know that binding energy and kinetic energy are equal .

Answer the following questions.

i) State Kepler’s law equal of area.

Answer-

The line that joins a planet and the Sun sweeps equal areas in equal intervals of time.

ii) State Kepler’s law of period.

Answer-

Kepler’s law of period is stated as The square of the time period of revolution of a planet around the Sun is proportional to the cube of the semi major axis of the ellipse traced by the planet.

iii) What are the dimensions of the universal gravitational constant?

Answer- M-1 L3 T-2

iv) Define binding energy of a satellite.

Answer-

The minimum energy required by a satellite to escape from Earth’s gravitational field is the binding energy of the satellite or it is minimum energy required to be provided to satellite to free it from Earth’s gravitational field.

v) What do you mean by geostationary satellite?

Answer-

Geostationary satellite is the type of satellite that revolves around the earth in equatorial plane with same time period as that of earth hence appears stationary from earth hence called geostationary satellite. As it go in synchronization with earth hence also called geosynchronous satellite.

vi) State Newton’s law of gravitation.

Answer-

Newton’s law of gravitation states that every particle of matter 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.

vii) Define escape velocity of a satellite.

Answer-

Minimum velocity required to throw an object from earth so that it escapes earth’s gravitational field is called escape velocity.

viii) What is the variation in acceleration due to gravity with altitude?

Answer-

We know the formula for the variation in acceleration due to gravity with altitude as

x) As we go from one planet to another planet, how will the mass and weight of a body change?

Answer-

As we go from one planet to another planet weight changes due to change in acceleration due to gravity of planet but mass of the body remains same as mass is just content of body.

xi) What is periodic time of a geostationary satellite?

Answer-

Periodic time of a geostationary satellite is the time required by satellite to complete one round of earth and it is same as time period of earth that is 24 hours.

xii) State Newton’s law of gravitation and express it in vector form.

Answer-

Newton’s law of gravitation states that every particle of matter 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.

If m1 and m2 are masses of two objects separated by distance r then in vector form,

Force on m2 due to m1 can be expressed in vector form as,

xiv) Why is a minimum two stage rocket necessary for launching of a satellite?

Answer-

For launching of satellite first satellite need to travel vertically upward up to desired height and after this it is provided with horizontal velocity equal o critical velocity so as to orbit it around the earth in given orbit. Hence two stages are required to launch the satellite in which when first stage of rocket is ignited it takes satellite to desired height then first stage is detached with the help of remote control.

After this satellite rotated by 90 degree and second stage is ignited which provide necessary horizontal velocity so as to orbit a satellite around the earth.

xv) State the conditions for various possible orbits of a satellite depending upon the tangential speed of projection.

Answer-

Various possible orbits of a satellite depending upon the tangential speed of projection.

Case (I) vh<vc

If tangential velocity of projection vh is less than the critical velocity, the orbit of satellite is an ellipse with point of projection farthest from the Earth and Earth at one of the foci.

Case (II) vh =vc

If the horizontal velocity is equal to the critical velocity, the satellite moves in a circular orbit round the Earth.

Case (III) VC<vh<ve

If horizontal velocity is greater than the critical velocity and less than the escape velocity at that height, the satellite again moves in an elliptical orbit round the Earth with the point of projection as perigee.

Case (IV) vh = ve

If horizontal speed of projection is equal to the escape speed at that height, the satellite travels along parabolic path and never returns to the point of projection. Its speed will be zero at infinity.

Case (V) vh > ve

If horizontal velocity is greater than the escape velocity, the satellite escapes from gravitational influence of Earth in a hyperbolic path.

3.) Answer the following questions in detail.

i) Derive an expression for critical velocity of a satellite.

Answer-

Consider a satellite of mass m revolving round the Earth at height h above its surface. Let m be the mass of the Earth and R be its radius. If the satellite is moving in a circular orbit of radius (R+h) = r, its speed must be the critical velocity vc.

The necessary centripetal force required for circular motion of satellite is provided by gravitational force exerted by the Earth on the satellite. Therefore

Centripetal force = Gravitational force

ii) State any four applications of a communication satellite.

Answer-

  1. For television transmission.
  2. Telephone and radio wave transmission.
  3. For GPS purpose
  4. For military use.

iv) Draw a labeled diagram to show different trajectories of a satellite depending upon the tangential projection speed.

Where,

Vc = critical velocity

Ve = escape velocity

Vh = horizontal speed of projection.

v) Derive an expression for binding energy of a body at rest on the Earth’s surface.

Answer-

The minimum energy required by a satellite to escape from Earth’s gravitational influence is the binding energy of the satellite.

Consider a satellite of mass m at rest on the Earth’s surface  its velocity is zero

vi) Why do astronauts in an orbiting satellite have a feeling of weightlessness?

Answer-

According to Newton’s second law of motion, F = ma, where F is the net force acting on an object having acceleration a. Though the weight of a body is the gravitational force acting upon it, we experience or feel our weight only due to the normal reaction force N exerted by the floor.

In the case of a revolving satellite, the satellite is performing a circular motion. The acceleration for this motion is centripetal, which is provided by the gravitational acceleration g at the location of the satellite. In this case, a = g, or the satelliteis in the state of free fall. Obviously, the apparent weight will be zero, giving the feeling of total weightlessness.

vii) Draw a graph showing the variation of gravitational acceleration due to the depth and altitude from the Earth’s surface.

Answer-

x) Define the binding energy of a satellite. Obtain an expression for binding energy of a satellite revolving around the Earth at certain attitude.

Answer-

The minimum energy required by a satellite to escape from Earth’s gravitational influence is the binding energy of the satellite.

Consider a satellite of mass m revolving at height h above the surface of the Earth in a circular orbit. It possesses potential energy as well as kinetic energy. Let m be the mass of the Earth, R be the Radius of  the Earth, vc be critical velocity of satellite, r = (R+h) be the radius of the orbit then

xi) Obtain the formula for acceleration due to gravity at the depth‘d’ below the Earth’s surface.

Answer-

When an object is on the surface of the Earth it experiences the gravitational force as if the entire mass of the Earth is concentrated at its centre.

The acceleration due to gravity is given as,

Consider a body at the depth d below the surface of the Earth.y. The net force at the depth d is only due to the material inside the inner sphere of radius R – d. Acceleration due to gravity because of this sphere is given by

This is the formula for acceleration due to gravity at the depth ‘d’ below the Earth’s surface  .

xii) State Kepler’s three laws of planetary motion.

Answer-

a) Law of orbit- All planets move in elliptical orbits around the Sun with the Sun at one of the foci of the ellipse.

b) Law of areas- The line that joins a planet and the Sun sweeps equal areas in equal intervals of time.

c) Law of periods-The Square of the time period of revolution of a planet around the Sun is proportional to the cube of the semi major axis of the ellipse traced by the planet.

xiii) State the formula for acceleration due to gravity at depth ‘d’ and altitude ‘h’ Hence show that their ratio is equal to (R-d)/(R-2h) by assuming that the altitude is very small as compared to the radius of the Earth.

Answer-

The expression of acceleration due to gravity at height h above the Earth’s surface is given by,

Hence proved.

xiv) What is critical velocity? Obtain an expression for critical velocity of an orbiting satellite. On what factors does it depend?

Critical velocity- minimum horizontal velocity provided to satellite so that it can revolve in a circular orbit round the earth called as critical velocity.

Consider a satellite of mass m revolving round the Earth at height h above its surface. Let m be the mass of the Earth and R be its radius. If the satellite is moving in a circular orbit of radius (R+h) = r, its speed must be the critical velocity vc.

The necessary centripetal force required for circular motion of satellite is provided by gravitational force exerted by the Earth on the satellite. Therefore

Centripetal force = Gravitational force

Hence critical velocity depends upon following factors

  1. Mass of earth.
  2. Height at which satellite revolving.
  3. Gravitational acceleration at given height.

xv) Define escape speed. Derive an expression for the escape speed of an object from the surface of the each.

Answer-

Escape speed- minimum vertical velocity require to escape from earth’s gravitational field is called escape velocity.

Let us consider the kinetic and potential energies of an object thrown vertically upwards with escape velocity ve, when it is at the surface of the Earth and when it reaches infinite distance.

On the surface of the Earth,

Kinetic energy of satellite = 1/2 mve2……..(1)

The gravitational potential at a distance R from the centre of the Earth is –GM/R. but

Potential energy of satellite = Gravitational potential × mass of satellite

P.E = m × ( –GM/R)

P.E = (-GMm)/R……..(2)

The total energy of satellite is given as

T.E. = K.E. + P.E.

From equation 1 and 2 we get,

T.E. = 1/2 mve2 – GMm/R…………..(1)

The kinetic energy of the object will go on decreasing with time and It will become zero when it reaches infinity and also at infinity potential energy will become zero as gravitational potential will become zero. Thus at infinite distance from the Earth T.E. = 0.

By conservation of energy,

After calculating we will get ve =11.2 km/s.

xvi) Describe how an artificial satellite using two stage rockets is launched in an orbit around the Earth.

Answer-

For launching of satellite first satellite need to travel vertically upward up to desired height and after this it is provided with horizontal velocity equal o critical velocity so as to orbit it around the earth in given orbit. Hence two stages are required to launch the satellite in which when first stage of rocket is ignited it takes satellite to desired height then first stage is detached with the help of remote control.  After this satellite rotated by 90 degree with the help of remote control.

After rotating satellite by 90 degree second stage comes into picture. Second stage is ignited here which provide necessary horizontal velocity so as to orbit a satellite around the earth. As soon as satellite attains the velocity equal to critical velocity then this second stage is also detached from satellite and satellite will start revolving around the earth.

Updated: April 26, 2022 — 1:06 am

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