NCERT Solutions Class 11 Physics Chapter 8 Gravitation – Here are all the NCERT solutions for Class 11 Physics Chapter 8. This solution contains questions, answers, images, explanations of the complete chapter 8 titled Of Gravitation taught in Class 11. If you are a student of Class 11 who is using NCERT Textbook to study Physics, then you must come across chapter 8 Gravitation After you have studied the lesson, you must be looking for answers of its questions. Here you can get complete NCERT Solutions for Class 11 Physics Chapter 8 Gravitation in one place.
NCERT Solutions Class 11 Physics Chapter 8 Gravitation
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For a better understanding of this chapter, you should also see summary of Chapter 8 Gravitation , Physics, Class 11.
| Class | 11 |
| Subject | Physics |
| Book | Physics Part I |
| Chapter Number | 8 |
| Chapter Name |
Gravitation |
NCERT Solutions Class 11 Physics chapter 8 Gravitation
Class 11, Physics chapter 8, Gravitation solutions are given below in PDF format. You can view them online or download PDF file for future use.
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Question & Answer
Q.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 ?
Ans : (a) No (b) Yes (a) Gravitational influence of matter on nearby objects cannot be screened by any means. This is because gravitational force unlike electrical forces is independent of the nature of the material medium. Also, it is independent of the status of other objects. (b) If the size of the space station is large enough, then the astronaut will detect the change in Earth's gravity (c) Tidal effect depends inversely upon the cube Of the distance while, gravitational force depends inversely on the square of the distance. Since the distance between the Moon and the Earth is smaller than the distance between the Sun and the Earth, the tidal effect of the Moon's pull is greater than the tidal effect of the Sun's pull.
Q.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 \( -G M m\left(1 / r_{2}-1 / r_{1}\right) \) is more/less accurate than the formula \( \left(r_{2}-r_{t}\right) \) for the difference of potential energy between two points \( r_{2} \text { and } r_{1} \) distance away from the centre of the earth.
Ans : (a) Decreases (b) Decreases (c) Mass Of the body (d) More Explanation (a) Acceleration due to gravity at depth h is given by the relation. \( \mathbf{g}_{n}=\left(1-\frac{2 h}{R_{e}}\right) g \) Where, R= Radius of the Earth g = Acceleration due to gravity on the surface of the Earth It is clear from the given relation that acceleration due to gravity decreases with an increase in height. (b) Acceleration due to gravity at depth d is given by the relation: \( \mathbf{g}_{d}=\left(1-\frac{d}{R_{e}}\right) g \) It is clear from the given relation that acceleration due to gravity decreases with an increase in depth. (c) Acceleration due to gravity of body of mass m is given by the relation: \( \mathrm{g}=\frac{\mathrm{G} M}{R^{2}} \) Where, G = Universal gravitational constant M = Mass of the Earth R= Radius of the Earth Hence, it can be inferred that acceleration due to gravity is independent of the mass of the body. (d) Gravitational potential energy of two points \( r_{2} \text { and } r_{1} \) distance away from the centre of the Earth is respectively given by: \( \begin{array}{l}{V\left(r_{1}\right)=-\frac{G m M}{r_{1}}} \\ {V\left(r_{2}\right)=-\frac{\operatorname{Gin} M}{r_{2}}}\end{array} \) Difference in potential energy, \( V=V\left(r_{2}\right)-V\left(r_{1}\right)=-\operatorname{Gim} M\left(\frac{1}{r_{2}}-\frac{1}{r_{1}}\right) \) Hence, this formula is more accurate than the formula \( m g\left(r_{2}-r_{1}\right) \)
Q.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?
Ans : Lesser by a factor of 0.63 Time taken by the Earth to complete one revolution around the Sun, \( T_{\mathrm{e}}= \) 1 year Orbital radius of the Earth in its orbit, \( R_{\mathrm{e}}=1 \mathrm{AU} \) Time taken by the planet to complete one revolution around the Sun,\( T_{P}=\frac{1}{2} T_{e}=\frac{1}{2} \) Orbital radius of the planet \(=R_{p} \) From Kepler's third law of planetary motion, we can write: \( \begin{array}{l}{\left(\frac{R_{r}}{R_{e}}\right)^{3}=\left(\frac{T_{r}}{T_{e}}\right)^{2}} \\ {\frac{R_{p}}{R_{v}}=\left(\frac{T_{P}}{T_{e}}\right)^{\frac{2}{3}}} \\ {\quad=\left(\frac{2}{1}\right)^{3}=(0.5)^{\frac{2}{3}}=0.63}\end{array} \) Hence, the orbital radius of the planet will be 0.63 times smaller than that of the Earth.
Q.4: In, one of the satellites of Jupiter, has an orbital period of 1.769 days and the radius of the orbit is \( 4.22 \times 10^{8} \mathrm{m}\). Show that the mass of Jupiter is about one-thousandth that of the sun
Ans : Orbital period of \( I_{0}, T_{l o}=1.769 \text { days }=1.769 \times 24 \times 60 \times 60 \mathrm{s} \) Orbital radius of \(I_{0}, R_{l o}=4.22 \times 10^{8} \mathrm{m} \) Satellite \( I_{0} \) is revolving around the jupiter Mass of the latter is given by the relation: \( M_{j}=\frac{4 \pi^{2} R_{\pm}^{3}}{G T_{l i 0}^{2}} \) Where \( M_{J}= \) mass of jupiter G= Universal gravitational constant Orbital period of the Earth, \( T_{e}=365.25 \text { days }=365.25 \times 24 \times 60 \times 60 \mathrm{s} \) Orbital radius of the earth , \( R_{e}=1 \mathrm{AU}=1.496 \times 10^{11} \mathrm{m} \) Mass of Sun is given as: \( M_{\mathrm{s}}=\frac{4 \pi^{2} R_{\mathrm{r}}^{3}}{\mathrm{G} T_{c}^{2}} \) \( \therefore \frac{M_{i}}{M_{J}}= \frac{4 \pi^{2} R_{e}^{3}}{G T_{e}^{2}} \times \frac{G T_{b}^{2}}{4 \pi^{2} R_{l u}^{3}} =\frac{R_{c}^{3}}{R_{b}^{3}} \times \frac{T_{l o}^{2}}{T_{c}^{2}} \) \( \begin{array}{l}{=\left(\frac{1.769 \times 24 \times 60 \times 60}{365.25 \times 24 \times 60 \times 60}\right)^{2} \times\left(\frac{1.496 \times 10^{11}}{4.22 \times 10^{8}}\right)^{3}} \\ {=1045.04}\end{array} \) \(\begin{array}{l}{\because \frac{M_{s}}{M_{J}} \sim 1000} \\ {M_{s} \sim 1000 \times M_{J}}\end{array} \) Hence, it can be inferred that the mass of Jupiter is about one-thousandth that of the Sun.
Q.5: Let us assume that our galaxy consists of \( 2.5 \times 10^{11}\) stars each of one solar mass. How long will a star at a distance of 50,000 ly from the galactic centre take to complete one revolution ? Take the diameter of the Milky Way to be \( 10^{5} \mathrm{ly} \) .
Ans : Mass of our galaxy Milky Way, \( M=2.5 \times 10^{11} \) Solar mass = Mass of Sun \( =2.0 \times 10^{36} \mathrm{kg} \) Mass of our galaxy, \( M=2.5 \times 10^{11} \times 2 \times 10^{36}=5 \times 10^{41} \mathrm{kg} \) Diameter Of Milky Way, \( d=10^{5} \mathrm{ly} \) Radius of Milky Way, \( r=5 \times 10^{4} \mathrm{ly} \) \( \begin{array}{l}{1 \mathrm{ly}=9.46 \times 10^{15} \mathrm{m}} \\ {\therefore r=5 \times 10^{4} \times 9.46 \times 10^{15}} \\ {=4.73 \times 10^{20} \mathrm{m}}\end{array} \) Since a star revolves around the galactic centre of the Milky Way, its time period is given by the relation: \( \begin{aligned} T &=\left(\frac{4 \pi^{2} r^{3}}{G M}\right)^{\frac{1}{2}} \\ &=\left(\frac{4 \times(3.14)^{2} \times(4.73)^{3} \times 10^{60}}{6.67 \times 10^{-11} \times 5 \times 10^{41}}\right)^{\frac{1}{2}}=\left(\frac{39.48 \times 105.82 \times 10^{30}}{33.35}\right)^{\frac{1}{2}} \\ &=\left(125.27 \times 10^{30}\right)^{\frac{1}{2}}=1.12 \times 10^{16} \mathrm{s} \end{aligned} \) \( \begin{array}{l}{1 \text { year }=365 \times 324 \times 60 \times 60 \mathrm{s}} \\ {1 \mathrm{s}=\frac{1}{365 \times 24 \times 60 \times 60} \mathrm{years}}\end{array} \) \( \begin{aligned} \therefore 1.12 \times 10^{16} \mathrm{s} &=\frac{1.12 \times 10^{16}}{365 \times 24 \times 60 \times 60} \\ &=3.55 \times 10^{8} \text { years } \end{aligned} \)
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