Class 11th Physics - Gravitation Case Study Questions and Answers 2022 - 2023
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Gravitation Case Study Questions With Answer Key
11th Standard CBSE
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Reg.No. :
Physics
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Galileo was the first to recognise the fact that irrespective of their masses, bodies fall towards the earth with a constant acceleration. Later on gravity and its laws were given by Newton in his universal law of gravitation and motion of planets around the sun was explained by Kepler in his laws of planetary motion. Newton's universal law of gravitation and third law of motion have a similarity that two bodies exert equal and opposite force on each other.
(i) What does the word 'Gravity' mean?
(ii) Define acceleration due to gravity
(iii) State Newton's law of gravitation.
(iv) What would be the gravitational force on a point mass body situated insde the hollow spherical shell of uniform density?
(v) State Kepler's law of period for planetary motion.
(vi) Give relationship between angular momentum and areal velocity of a planet around sun.
(vii) At what distance force of gravitation between two bodies would be zero? -
There are three identical point mass bodies each of mass m located at the vertices of an equilateral triangle with side r. They are exerting gravitational force of attraction on each other, which can be given by newton's law of gravitation. Each mass body produces its gravitational field in the surrounding region. The magnitude of gravitational field at a point due to a point mass body is the measure of gravitational intensity at that point. The gravitational potential at a point in a gravitational field is the amount of work done in bringing a unit mass body from infinity to the given point without acceleration.
(i) What is the magnitude of the gravitational force on one body due to other two bodies in the given arrangement.
(ii) What is the work done in taking one body far away from the other two bodies?
(iii) Determine the gravitational potential at the centroid point O.
(iv) When a body falls towards earth, earth also moves towards the body. Why is earth's motion not noticed? -
A rocket is fired vertically upwards with speed v = 5 kms-1 from the surface of earth. It goes up to a height h before returning to earth. At height h a body is thrown from the rocket with speed vo in such a way so that the body becomes satellite of earth. Let the mass of the earth, M = 6 x 1024kg, mean radius of the earth, R = 6.4 x 10 6 m, G = 6.67 x 10- 11 Nm2 kg-2, g = 9.8 ms-2.
(i) Determine the value of height h above the surface of earth from which body is thrown out from the, rocket.
(ii) Determine the orbital velocity of the satellite.
(iii) Calculate the time period of revolution of satellite around the earth.
(iv) If this satellite is to be taken at double of the present height from the surface of the earth, then find the new time period of revolution of satellite.
(v) What is the value of gravitational field on the surface of earth?
(vi) Is the potential energy of a system of bodies positive or negative? Give reason.
(vii) What is the maximum value of gravitational potential energy and where? -
Geostationary satellite is a particular type of satellite used in communication. A number of communication satellites are launched which remain in fixed position at a specified height above the equator. They are called synchronous satellites. These appear fixed at a position above a certain place on the earth, it must corotate with the earth so that its orbital period around the earth is exactly equal to the rotational period of the earth about its axis of rotation.
(i) What is the height of geostationary satellite above the surface of the earth?
(ii) What is the time period and direction of revolution of a geostationary satellite respectively?
(iii) What is the orbital-speed of geostationary satellite?
(iv) Which electromagnetic wave is used in satellite communication?
(v) Can a Pendulum vibrate in an artificial satellite?
(vi) From where does a satellite revolving around planet get the required centripetal Force? -
A body moving in an orbit around the earth is called earth satellite. The First artificial satellite was put into earth's orbit in 1956. Artificial satellites are put into orbit at an altitude of a few hundred kilometers. The satellite is carried in a rocket which is launched from the earth with a velocity greater than the escape velocity. The escape velocity is the velocity with which a body must be projected in order that is may escape the gravitational pull of the earth. When the rocket has achieved the desired height, the satellite is released horizontally by imparting to it a very high speed so that it remains moving in a nearly circular orbit around the earth. This velocity is called the orbital velocity which is about 8 kms-1 for a satellite at a few hundred kilometeres above the earth.
(i) What are the factors on which escape velocity of a rocket fired from the earth depends upon?
(ii) What provides the necessary centripetal force to keep a satellite in a circular orbit around the earth?
(iii) An artificial satellite is orbiting the earth at an altitude of 500 km. A bomb is released from the satellite. How will the bomb move after its release?
(iv) How escape velocity is related to orbital velocity of satellite?
(v) A satellite is orbiting around the earth with a speed v. To make the satellite escape, what is the minimum percentage increase in its speed?
(vi) A satellite is launched into a circular orbit of radius R around the earth. A second satellite is launched into an orbit of radius 1.01 R. By what percentage the period of second satellite is larger than that of first satellite?
(vii) Does the speed of a satellite remain constant in a particular orbit? -
Considering the earth as an isolated mass, a force is experienced by a body at any distance from it. This force is directed towards the centre of the earth and has a magnitude mg, where m is the mass of the body and g is the acceleration due to gravity. The value of the acceleration due to gravity decreases with increase in the height above the surface of the earth and with increase in the depth below the surface of the earth. Even on the surface of the earth the value of g varies from place to place and decreases with decrease in the latitude of the place.
(i) Assuming the earth to be a sphere of uniform mass density draw the graph representing the variation of 'g' with distance r from the centre of earth.
(ii) A body weighs 63 N on the surface of the earth. How much will it weigh at a height equal to half the radius of the earth?
(iii) Assuming the earth to be a sphere of uniform mass density, what is the weight of a body when it is taken to the end of tunnel 32 km below the surface. (Radius = 6400 km).
(iv) If gp is the acceleration due to gravity at the poles and ge that at the equator, then which one is greater?
(v) If a tunnel is dug along a diameter of the earth and body is dropped from one end of the tunnel then how wiu the body move in the tunnel?
Case Study
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Answers
Gravitation Case Study Questions With Answer Key Answer Keys
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(i) Gravity is the force of attraction exerted by earth towards its centre on a body lying on or near the surface of earth. Weight of body is equal to gravitational pull.
(ii) It is the force of gravity acting on unit mass of the body placed on or near the surface of earth.
(iii) Every body in this universe attracts every other body with a force which is directly proportional to the product of their masses and is inversly proportional to the square of the distance between them
\(F=G \frac{m_{1} m_{2}}{r^{2}}\)
(iv) Zero, it is so because gravitational forces on the point mass due to various small regions ofthe spherical shell will be acting in various directions which cancel out each other.
(v) It states that the square of the time period of revolution of a planet around the sun is directly proportional to the cube of semi major axis of its elliptical orbit. i.e \(\tau^{2} \alpha R^{3}\)
(vi) \(\vec{L}=2 m \times \frac{d \overrightarrow{\mathrm{A}}}{d t}\)
(vii) Force of gravitation between two bodies would be zero if the separation between them becomes infinity. -
(i) Gravitational force on body at C due to body at A is
\(F_{1}=\frac{G m m}{r^{2}} \text { along } \mathrm{CA}\)
Similarly due to body at B \(F_{2}=\frac{G m \times m}{r^{2}}\) along CB Angle between \(F_{1} \text { and } \vec{F}_{2} \) is 60°.
Total force F \(=\sqrt{F_{1}^{2}+F_{2}^{2}+2 F_{1} F_{2} \cos 60^{\circ}}\)
\(=\frac{\sqrt{3}}{r^{2}} G m^{2}\)
(ii) Gravitational Potential energy of the system is
\(U=\frac{-3 G m^{2}}{r}\)
when one body is 'taken away, then the gravitational potenetial energy of the system is \(U_{2}=\frac{-G m^{2}}{r}\)
Work done is taking the body at C far away from other bodies.
\(W=U_{2}-U_{1}=\frac{-G m^{2}}{r}+\frac{3 G m^{2}}{r}=\frac{2 G m^{2}}{r}\)
(iii) Here \(\mathrm{AO}=\mathrm{BO}=\mathrm{CO}=\frac{r}{\sqrt{3}}\)
Gravitational potential at 0 due to masses at A, B and C is
\(=\left(\frac{-G M}{\frac{r}{\sqrt{3}}}\right)+\left(\frac{-G M}{\frac{r}{\sqrt{3}}}\right)+\left(\frac{-G M}{\frac{r}{\sqrt{3}}}\right)=\frac{-3 \sqrt{3} G M}{r}\)
(iv) Earth motion is not noticed because acceleration produced in earth is negligibly small, due to large mass of the earth. -
(i) According to law of Conservation of total mechanical energy
\(\frac{1}{2} m v^{2}-\frac{G m m}{R}=\frac{-G m m}{R+h}\)
\(\Rightarrow \ h=\frac{R v^{2}}{2 g R-v^{2}}\)
\(=\frac{6.4 \times 10^{6} \times\left(5 \times 10^{3}\right)^{2}}{\left(2 \times 9.8 \times 6.4 \times 10^{6}\right)-\left(5 \times 10^{3}\right)^{2}}\)
= 1.6 x 106 m
(ii) The orbital velocity of satellite is
\(v_{0}=\sqrt{\frac{G M}{R+h}}=\left[\frac{6.67 \times 10^{-11} \times 6 \times 10^{24}}{6.4 \times 10^{6}+1.6 \times 10^{6}}\right]^{\frac{1}{2}}\)
= 7.1 x 103 ms-1
(iii) Tie period of revolution of satellite \(T=\frac{2 \pi(\mathrm{R}+h)}{v_{0}}\)
\(T=\frac{2 \times 3.14 \times\left(6.4 \times 10^{6}+1.6 \times 10^{6}\right)}{7.1 \times 10^{3}}\)
= 7076 = 7100 sec.
(iv) Initial orbital radius,
R1 = 6.4 x 106 + 1.6 x 106 = 8 x 106 m
T1 = 7100 S
R2 = 8 x 10 6 + 1.6 x 106 = 9.6 x 10 6 m
T2 = ?
\(T_{2}=T_{1}\left(\frac{R_{2}}{R_{1}}\right)^{\frac{3}{2}}=7100\left(\frac{9.6 \times 10^{6}}{8 \times 10^{6}}\right)^{\frac{3}{2}}\)
= 9333 sec.
(v) Strength of gravitational field on the surface of earth \(=\frac{G M}{R^{2}}\) = g i.e acceleration due to gravity.
(vi) Since the gravitational force between the different bodies of a system are attractive in nature so the potential energy of the system is negative.
(vii) The value of gravitational potential energy increases as we move away from the earth and becomes maximum i.e., zero at infinity. -
(i) Height of a geostationary satellite above the surface of earth is nearly 36000 km.
(ii) Time period of revolution of geostationary satellite is 24 hours and direction of rotation is from west to east.
(iii) Orbital speed of geostationary satellite is about 3 kms-1 .
(iv) Microwaves
(v) No, this is because inside the satellite, there is no gravity i.e., g = 0. As t = \(2 \pi \sqrt{l / g}\) hence t = \(\infty\) for g = 0. Thus pendulum will not vibrate.
(vi) Satellite revolving around planet get the required centripetal Force from the gravitational attraction of planet exerted on satellite. -
(i) Escape velocity depends on the acceleration due to gravity only.
(ii) The gravitational attraction between the earth and satellite provides the necessary centripetal force to keep it in a circular orbit around the earth.
(iii) The bomb will orbit the earth along with the satellite.
(iv) Escape speed = \(\sqrt{2}\) x orbital speed. Here speed means magnitude of velocity.
(v) % increase in speed = \(\frac{\left(v_{e}-v_{0}\right)}{v_{0}} \times 100\)
\(=\left(\frac{v_{e}}{v_{0}}-1\right) \times 100\)
\(=(\sqrt{2}-1) \times 100=41.4 \%\)
(vi) According to Kepler's third law of planetary motion
T2 = KR3
\(2 \frac{\Delta T}{T}=3 \frac{\Delta R}{R} \Rightarrow \frac{\Delta T}{T}=\frac{3}{2}\left(\frac{1.01 R-R}{R}\right)\)
\(\Rightarrow \quad \frac{\Delta T}{T} \times 100=\frac{3}{2} \times 0.01 \times 100=1.5 \%\)
(vii) Yes, the speed of a satellite in a particular orbit remain constant. Since orbital velocity v = \(\sqrt{G m / r}\) shows that v is constant if r is constant, because Gm is constant for earth. If the orbital path of a satellite is circular, then its speed is constant. -
(i)
(ii) \(W=W_{0}\left(\frac{R}{R+h}\right)^{2}=63 \times\left(\frac{R}{R+\frac{R}{2}}\right)^{2}=28 N\)
(iii) \(g=g_{0}\left(1-\frac{d}{R}\right)=g_{0}\left(1-\frac{32}{6400}\right)=\frac{199}{200} g_{0}\)
Decrease in weight = \(m g_{0}-m g=m g_{0}\left(1-\frac{199}{200}\right)\)
\(=\frac{m g_{0}}{200}\)
(iv) ge > gp, since Rp < Re.
(v) Body will execute simple harmonic motion about the centre of the earth .
Case Study
