NCERT Solutions for Class 11 Physics Physics Chapter 8 Gravitation includes all the important topics with detailed explanation that aims to help students to understand the concepts better. Students who are preparing for their Class 11 Physics exams must go through NCERT Solutions for Class 11 Physics Physics Chapter 8 Gravitation. Going through the solutions provided on this page will help you to know how to approach and solve the problems.
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NCERT Solutions for Class 11 Physics Physics Chapter 8 Gravitation
NCERT Solutions for Class 11 Physics Physics Chapter 8 Gravitation are been solved by expert teachers of CBSETuts.com. All the solutions given in this page are solved based on CBSE Syllabus and NCERT guidelines.
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?
(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.
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.
(a) Acceleration due to gravity g decreases with increasing altitude as
(b) Acceleration due to gravity g decreases with depth as g′=g
(c) Acceleration due to gravity is, .g=GMe /R2e Therefore, g is independent of mass of body
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?
Using kepler’s third law,
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.
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.
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.
(a) Kinetic energy
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?
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.
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.
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.
Which of the following symptoms is likely to afflict an astronaut in space
(a) swollen feet,
(b) swollen face,
(d) orientational problem.
(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.
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.
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.
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)
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.
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).
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
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
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
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?
Using Kepler’s third law,
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?
Acceleration due to gravity g at height h is given by
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?
Variation of acceleration due to gravity g with depth d is
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
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
Here, υ0 = speed of projectile when far away from the earth
υ = velocity of projection of the body
υE = escape velocity
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.
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).
Mass of each star,M x 2 x 1030 kg
Initial distance between two stars ,r= 109 Km =1012 m
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?
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.
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.
Gravitational potential at height h from the surface of earth is
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).
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.
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.
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
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.
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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