QUESTIONS FROM TEXTBOOK ( Class 11 Physics Gravitation )

Question 8. 1. Answer the following:
(a) You can shield a charge from electrical forces by putting it inside a hollow conductor. Can you shield a body from the gravitational influence of nearby matter by putting it inside a hollow sphere or by some other means?
(b) An astronaut inside a small spaceship orbiting around the Earth cannot detect gravity. If the space station orbiting around the Earth has a large size, can he hope to detect gravity?
(c) If you compare the gravitational force on the Earth due to the Sun to that due to the Moon, you would find that the Sun’s pull is greater than the Moon’s pull. (You can check this yourself using the data available in the succeeding exercises). However, the tidal effect of the Moon’s pull is greater than the tidal effect of Sun. Why?
Answer: (a) No. Gravitational forces are independent of medium. A body cannot be shielded from the gravitational influence of nearby matter.
(b) Yes. If the size of the spaceship is extremely large, then the gravitational effect of the spaceship may become measurable. The variation in g can also be detected.
(c) Tidal effect depends inversely on the cube of the distance, unlike force which depends inversely on the square of the distance. Since the distance of moon from the ocean water is very small as compared to the distance of sun from the ocean water on earth. Therefore, the tidal effect of Moon’s pull is greater than the tidal effect of the sun.

Question 8. 2. Choose the correct alternative:
(a) Acceleration due to gravity increases/decreases with increasing altitude.
(b) Acceleration due to gravity increases/decreases with increasing depth (assume the Earth to be a sphere of uniform density).
(c) Acceleration due to gravity is independent of the mass of the Earth/mass of the body.
(d) The formula – GMm (1/r₂-1/r₁) is more/less accurate than the formula mg (r₂ – r₁) for the difference of potential energy between two points r2 and r1 distance away from the centre of the Earth.
Answer: (a) decreases
(b) decreases
(c) mass of the body
(d) more

Question 8. 6. Choose the correct alternative:
(a) If the zero of potential energy is at infinity, the total energy of an orbiting satellite is negative of its kinetic/potential energy.
(b) The energy required to launch an orbiting satellite out of Earth’s gravitational influence is more/less than the energy required to project a stationary object at the same height (as the satellite) out of Earth’s influence.
Answer:  (a) If the zero of potential energy is at infinity, the total energy of an orbiting satellite is negative of its kinetic energy.
(b) The energy required to launch an orbiting satellite out of Earth’s gravitational influence is less than the energy required to project a stationary object at the same height (as the satellite) out of Earth’s influence.

Question 8. 8. A comet orbits the Sun in a highly elliptical orbit. Does the comet have a constant (a) linear speed (b) angular speed (c) angular momentum (d) kinetic energy (e) potential energy (f) total energy throughout its orbit? Neglect any mass loss of the comet when it comes very close to the Sun.
Answer: (a) The linear speed of the comet is variable in accordance with Kepler7s second law. When comet is near the sun, its speed is higher. When the comet is far away from the sun, its speed is very less.
(b) Angular speed also varies slightly.
(c) Comet has constant angular momentum.
(d) Kinetic energy does not remain constant.
(e) Potential energy varies along the path.
(f) Total energy throughout the orbit remains constant.

Question 8. 9.Which of the following symptoms is likely to afflict an astronaut in space (a) swollen feet, (b) swollen face, (c) headache, (d) orientational problem.
Answer: (a) The blood flow in feet would be lesser in zero gravity. So, the astronaut will not get swollen feet.
(b) In the conditions of weightlessness, the face of the astronaut is expected to get more supply. Due to it, the astronaut may develop swollen face.
(c) Due to more blood supply to face, the astronaut may get headache.
(d) Space also has orientation. We also have the frames of reference in space. Hence, orientational problem will affect the astronaut in space.

Question 8. 11. For the above problem, the direction of the gravitational intensity at an arbitrary point P is indicated by the arrow (i) d, (ii), e, (iii) f (iv) g.
Answer: Using the explanation given in the solution of the previous problem, the direction of the gravitational field intensity at P will be along e. So, option (ii) is correct.

Question 8. 13. How will you ‘weigh the sun’, that is, estimate its mass? The mean orbital radius of the earth around the sun is 1.5 x 10⁸ km.
Answer: The mean orbital radius of the Earth around the Sun
R = 1.5 x 10⁸ km = 1.5 x 10¹¹ m
Time period, T = 365.25 x 24 x 60 x 60 s
Let the mass of the Sun be M and that of Earth be m.

Question 8. 17. A rocket is fired vertically with a speed of 5 km s^-1 from the earth’s surface. How far from the earth does the rocket go before returning to the earth? Mass of the earth = 6.0 x 10²⁴ kg; mean radius of the earth = 6.4 x 10⁶ m; G = 6.67 x 10^-11 N m² kg^-2.
Answer:  Initial kinetic energy of rocket = 1/2 mv² = 1/2 x m x (5000)2 = 1.25 x 10⁷ mJ
At distance r from centre of earth, kinetic energy becomes zero
.•. Change in kinetic energy = 1.25 x 10⁷ – 0 = 1.25 x 10⁷ m J
This energy changes into potential energy.
Initial potential energy at the surface of earth = GM_em/’r

Question 8. 20. Two stars each of one solar mass (=2 x 10³⁰ kg) are approaching each other for a head on collision.When they are at a distance 10⁹ km, their speeds are negligible. What is the speed with which they collide? The radius of each star is 10⁴ km. Assume the stars to remain undistorted until they collide. (Use the known value of G).
Answer: Here, mass of each star, M = 2 x 10³⁰ kg
Initial potential between two stars, r = 10⁹ km = 10¹² m.
Initial potential energy of the system = -GMm/r
Total K.E. of the stars = 1/2Mv² + 1/2Mv²
where v is the speed of stars with which they collide. When the stars are about to collide, the distance between their centres, r’ = 2 R.
:. Final potential energy of two stars = -GMm/2R
Since gain in K.E. is at the cost of loss in P.E

Question 8. 21. Two heavy spheres each of mass 100 kg and radius 0.10 mare placed 1.0 m apart on ahorizontal table. What is the gravitational field and potential at the mid point of the line joining the centres of the spheres ? Is an object placed at that point in equilibrium ? If so, is the equilibrium stable or unstable?
Answer: Here G = 6.67 x 10^-11 Nm² kg^-2; M = 100 kg; R = 0.1 m, distance between the two spheres, d = 1.0 m
Suppose that the distance of either sphere from the mid-point of the line joining their centre is r. Then r=d/2=0.5 m. The gravitational field at the mid-point due to two spheres will be equal and opposite.

Question 8. 23. A star 2.5 times the mass of the sun and collapsed to a size of 12 km rotates with a speed of 1.2 rev. per second. (Extremely compact stars of this kind are known as neutron stars. Certain stellar objects called pulsars belong to this category). Will an object placed on its equator remain stuck to its surface due to gravity? (mass of the sun = 2 x 10³⁰ kg).
Answer: Acceleration due to gravity of the star,g= GM/R² …………(i)
Here M is the mass and R is the radius of the star.
The outward centrifugal force acting on a body of mass m at the equator of the star =mv2/R =mR w2——-(ii)
From equation (i), the acceleration due to the gravity of the star

Question 8. 24. A spaceship is stationed on Mars. How much energy must be expended on the spaceship to rocket it out of the solar system? Mass of the spaceship = 1000 kg, Mass of the Sun = 2 x 10³⁰ kg. Mass of the Mars = 6.4 x 10²³ kg, Radius of Mars = 3395 km. Radius of the orbit of Mars = 2.28 x 10¹¹ m, G = 6.67 x 10^-11 N m² kg^-2.
Answer:  Let R be the radius of orbit of Mars and R’ be the radius of the Mars. M be the mass of the Sun and M’ be the mass of Mars. If m is the mass of the space-ship, then Potential energy of space-ship due to gravitational attraction of the Sun = – GM m/R Potential energy of space-ship due to gravitational attraction of Mars = – G M’ m/R’ Since the K.E. of space ship is zero, therefore,