Class 11th Chapter -8 Gravitation |Physics| NCERT Solution | Edugrown

NCERT Exercises

Question 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, because the gravitational force of attraction on a point mass situated inside a spherical shell is zero.
(b) Yes, an astronaut inside the spaceship can detect the variation in gravity, if the size of the spaceship is large enough.
(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 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 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 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 center of the earth.
Answer:
(a) Acceleration due to gravity g decreases with increasing altitude as
g′=g

²
(b) Acceleration due to gravity g decreases with depth as g′=g 
(c) Acceleration due to gravity is, .g=GM_e /R²_e Therefore, g is independent of mass of body
(d) More

Question 3.
Suppose there existed a planet that went around the sun twice as fast as the earth. What would be its orbital size as compared to that of the earth?
Answer:
Using kepler’s third law,


Question 4.
One of the satellites of Jupiter, has an orbital period of 1.769 days and the radius of the orbit is 4.22 x 10⁸ m. Show that the mass of Jupiter is about one-thousandth that of the sun.
Answer:

Question 5.
Let us assume that our galaxy consists of 2.5 x 10¹¹ stars each of one solar mass. How long will a star at a distance of 50,000 ly from the galactic center take to complete one revolution? Take the diameter of the Milky Way to be 10⁵ ly.
Answer:

Question 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) Kinetic energy
(b) less

Question 7.
Does the escape speed of a body from the earth depend on
(a) the mass of the body,
(b) the location from where it is projected,
(c) the direction of projection,
(d) the height of the location from where the body is launched?
Answer:
The escape velocity does not depend upon the mass of body, the direction of projection and the angle of projection. It depends upon the gravitational potential at the point from where the body is launched. Since this potential depends slightly on the latitude and height of the point, therefore, the escape velocity depends slightly on these factors.

Question 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 comet while going on elliptical orbit around the Sun has constant angular momentum and total energy at all locations but other quantities vary with locations.

Question 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) We know that the legs carry the weight of the body in the normal position due to gravity pull. The astronaut in space is in  weightlessness state. Hence, swollen feet may not affect his working.

(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. As eyes, ears, nose, mouth etc. are all embedded in the face, hence, swoden face may affect to great extent the seeing/hearing/smelling/eating capabilities of the astronaut in space.

(c) Headache is due to mental strain. It will persist whether a person is an astronaut in space or he is on earth. It means headache will have the same effect on the astronaut in space as on a person on earth.

(d) Space also has orientation. We also have the frames of reference in space. Hence, orientational problem will affect the astronaut in space.

Question 10.
The gravitational intensity at the center of a hemispherical shell of uniform mass density has the direction indicated by the arrow (see figure) (i) a, (ii) b, (iii) c, (iv) 0.

Answer:
The gravitational potential is constant at all points inside a spherical shell. Therefore, the gravitational potential gradient at all points inside the spherical shell is zero [i.e. as V is constant, dV/dr = 0]. Since gravitational intensity is equal to negative of the gravitational potential gradient, hence the gravitational intensity is zero at all points inside a hollow spherical shell.

This indicates that the gravitational forces acting on a particle at any point inside a spherical shell, will be symmetrically placed. Therefore if we remove the upper hemi­spherical shell, the net gravitational force acting on the particle at the center Q or at some other point P will be acting downwards which will also be the direction of gravitational intensity. It is so because, the gravitational intensity at a point is the gravitational force per unit mass at that point. Hence the gravitational intensity at the center Q will be along c, i.e., option (iii) is correct.

Question 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)
Answer:
As per explanation given in the answer of Q.10, the direction of gravitational intensity at P will be along e, i.e., the option (ii) is correct.

Question 12.
A rocket is fired from the earth towards the sun. At what distance from the earth’s center is the gravitational force on the rocket zero? Mass of the sun = 2 x 10³⁰ kg, mass of the earth = 6 x 10²⁴ kg. Neglect the effect of other planets etc. (orbital radius = 1.5 x 10¹¹ m).
Answer:
Let d be the distance of a point from the earth where gravitational forces on the rocket due to the sun and the earth become equal and opposite. Then distance of rocket from the sun = (r – d). If m is the mass of rocket then

Question 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
Answer:
The gravitational force acting on Earth due to Sun is

radius of the earth around the sun. Now, the centripetal force acting on the earth due to the Sun is

ω → angular Velocity
radius of the earth around the sun. Now, the centripetal force acting on the earth due to the Sun is
Since, this centripetal force is provided by the gravitational pull of the Sun on the Earth, so,
∴ r = 1.5 x 10⁸ km = 1.5 ^x 10¹¹ m
T = 365 days = 365 x 24 x 60 x 60 s

Question 14.
A saturn year is 29.5 times the earth year. How far is the saturn from the sun if the earth is 1.50 x 10⁸ km away from the sun?
Answer:
Using Kepler’s third law,

Question 15.
A body weighs 63 N on the surface of the earth. What is the gravitational force on it due to the earth at a height equal to half the radius of the earth?
Answer:
Acceleration due to gravity g at height h is given by

Question 16.
Assuming the earth to be a sphere of uniform mass density, how much would a body weigh half way down to the center of the earth if it weighed 250 N on the surface?
Answer:
Variation of acceleration due to gravity g with depth d is

Question 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 10¹¹  Nm² kg²
Answer:

Question 18.
The escape speed of a projectile on the earth’s surface is 11.2 km s^-1. A body is projected out with thrice this speed. What is the speed of the body far away from the earth? Ignore the presence of the sun and other planets. Using law of conservation of energy
Answer:

Here, υ₀ = speed of projectile when far away from the earth
υ = velocity of projection of the body
υ_E = escape velocity

Question 19.
A satellite orbits the earth at a height of 400 km above the surface. How much energy must be expended to rocket the satellite out of the earth’s gravitational influence? Mass of the satellite = 200 kg; mass of the earth = 6.0 x 10²⁴ kg; radius of the earth = 6.4 x 10⁶ m; G = 6.67 x 10^-11 N m² kg^-2.
Answer:

Question 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 of 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:
Mass of each star,M x 2 x 10³⁰ kg
Initial distance between two stars ,r= 10⁹ Km =10¹² m

Question 21.
Two heavy spheres each of mass 100 kg and radius 0.10 m are placed 1.0 m apart on a horizontal table. What is the gravitational field and potential at the mid point of the line joining the centers of the spheres? Is an object placed at that point in equilibrium? If so, is the equilibrium stable or unstable?
Answer:
Gravitational field at the mid-point of the line joining the centers of the two spheres

From equation (i) effective force on the body placed at mid-point is zero, so the body is in equilibrium. If the body is displaced a little towards either mass body from its equilibrium position, it will not return back to its initial position of equilibrium. Hence, the body is in unstable equilibrium.

Question 22.
As you have learnt in the text, a geostationary satellite orbits the earth at a height of nearly 36,000 km from the surface of the earth. What is the potential due to earth’s gravity at the site of this satellite? (Take the potential energy at infinity to be zero). Mass of the earth = 6.0 x 10²⁴ kg, radius = 6400 km.
Answer:
Gravitational potential at height h from the surface of earth is

Question 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:
The object will remain stuck to the surface of star due to gravity, if the acceleration due to gravity is more than the centrifugal acceleration due to its rotation.

Question 24.
A spaceship is stationed on Mars. How much energy must be expended on the spaceship to launch it out of the solar system? Mass of the space ship = 1000 kg; mass of the Sun = 2 x 10³⁰ kg;
mass of Mars = 6.4 x 10²³ kg; radius of Mars = 3395 km; radius of the orbit of Mars = 2.28 x 10⁸ km; G = 6.67 x 10^-11 N m² kg^-2.
Answer:

Total energy of the spaceship, i.e.,
E = K.E. + U = 2.925 x 10¹¹ J – 5.98 x 10n J
= -3.1 x 1011 J
Negative energy denotes that the spaceship is bound to the solar system.
Thus, energy needed by the spaceship to escape from the solar system = 3.1 x 10¹¹ J

Question 25.
A rocket is fired ‘vertically’from the surface of Mars with a speed of 2 km s^-1 If 20% of its initial energy is lost due to Martian atmospheric resistance, how far will the rocket go from the surface of mars before returning to it? Mass of mars = 6.4 x 10²³ kg; radius of Mars = 3395 km; G = 6.67 x 10^-11 Nm²kg-².
Answer:
Let m = mass of the rocket, M = mass of the Mars and R = radius of Mars. Let v be the initial velocity of rocket

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