NCERT Solutions for Class 11 Physics Chapter 8 Gravitation

NCERT Solutions for Class 11 Physics Chapter 8 Gravitation: Let's assume that you are standing at the North pole. According to you, a person standing at the south pole is inverted, still, he is not falling back why? We all know the reason that it is due to gravity. The solutions of NCERT class 11 physics chapter 8 gravitation discuss questions based on the mathematical and physical significance of gravitation. CBSE NCERT solutions for class 11 physics chapter 8 gravitation have questions based on Kepler's Law which is important for exams like NEET and JEE Main. Kepler's Law was obtained from the observations of Galileo on planetary motions. Another two important topics are the escape velocity and satellites. Questions based on these concepts are also discussed in the  NCERT solutions for class 11 physics chapter 8 gravitation. You will study the unit electrostatics in NCERT class 12 physics, so, while studying Electrostatistics chapter you can compare the formula with the formulas that you will study in chapter Gravitation. NCERT solutions can make your learning easy if you are want to prepare for other classes too. The main topics of the chapter are listed below

8.1 Introduction

8.2 Kepler’s Laws

8.3 Universal Law Of Gravitation

8.4 The Gravitational Constant

8.5 Acceleration Due To Gravity Of The Earth

8.6 Acceleration Due To Gravity Below And Above The Surface Of Earth

8.7 Gravitational Potential Energy

8.8 Escape Speed

8.9 Earth Satellites

8.10 Energy Of An Orbiting Satellite

8.11 Geostationary And Polar Satellites

8.12 Weightlessness

Q: 8.1 (c)  Answer the following:

(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?

Apart from the gravitational pull, the tidal effect also depends upon the cube of the distance between the two. Since the distance between earth and sun is much larger than the distance between the sun and moon so it not also balances but is more than the effect of gravitational force. Thus the tidal effect of the moon’s pull is greater than the tidal effect of the sun.

(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 $-GMm(1/r_2\hspace{1mm}-\hspace{1mm}1/r_1)$  is more/less accurate than the formula $mg(r_2-r_1)$ for the difference of potential energy                                   between two points $r_2$ and $r_1$  distance away from the centre of the earth.

(a) Acceleration due to gravity decreases with increasing altitude.

The relation between the two is given by :

$g_h\ =\ \left ( 1\ -\ \frac{2h}{R_e} \right )g$

(b) Acceleration due to gravity decreases with increasing depth.

The relation is given below:

$g_d\ =\ \left ( 1\ -\ \frac{d}{R_e} \right )g$

(c)  Acceleration due to gravity is independent of the mass of the body.

$g\ =\ \frac{GM}{R^2}$                               Here M is the mass of the earth.

(d)  The formula $-GMm(1/r_2\hspace{1mm}-\hspace{1mm}1/r_1)$  is more accurate than the formula $mg(r_2-r_1)$ for the difference of potential energy between two points $r_2$ and $r_1$  distance away from the centre of the earth.

Time taken by planet to complete a revolution around sun  =       $\frac{1}{2}T_e$

Using Kepler's law of planetary motion we can write :

$\left ( \frac{R_p}{R_e} \right )^3\ =\ \left ( \frac{T_p}{T_e} \right )^2$

or                                                 $\frac{R_p}{R_e}\ =\ \left ( \frac{\frac{1}{2}}{1} \right )^\frac{2}{3}$

or                                                  $\frac{R_p}{R_e}\ =\ 0.63$

Thus the planet is 0.63 times smaller than earth.

The orbital period in days is   $=\ 1.769\times 24 \times 60\times 60\ s$

Mass is given by :

$M\ =\ \frac{4 \pi ^2 R^3}{GT^2}$

Thus the ratio of the mass of Jupiter and mass of the sun is :

$\frac{M_s}{M_j} =\ \frac{\frac{4 \pi ^2 R_e^3}{GT_e^2}}{\frac{4 \pi ^2 R_{io}^3}{GT_{io}^2}}$

or                                                       $=\ \left ( \frac{1.769 \times 24\times 60\times 60}{365.25\times 24\times 60\times 60} \right )^2\times \left ( \frac{1.496\times 10^{11}}{4.22\times 10^8} \right )^3$

or                                                      $\approx 1045$

Thus the mass of Jupiter is nearly one-thousandth that of the sun.

We know that one light year is $9.45\times 10^{15}\ m$.

The time period of rotation is given by :

$T\ =\ \left ( \frac{4 \pi r^3}{GM} \right )^\frac{1}{2}$

Putting all the value (in SI units) in the above equation we get :

$=\ \left ( \frac{4 \times \left ( 3.14 \right )^2\times (4.73)^3\times 10^{60}}{6.67\times 10^{-11} \times 5\times 10^{41}} \right )^\frac{1}{2}$

or                                                  $=\ 1.12\ \times 10^{16}\ s$

In years :

$=\ \frac{1.12\ \times 10^{16}}{365 \times 24 \times 60 \times 60}\ =\ 3.55\times 10^8\ years$

(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) The total energy will be negative of its kinetic energy. Since at infinity potential energy is zero and total energy is negative.

(b) The energy required will be less as the stationary object on earth has no energy initially whereas the satellite has gained energy due to rotational motion.

Q: 8.7 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?

The escape velocity from the earth is given by :

$V_{esc}\ =\ \sqrt{2gR}$

Since the escape velocity depends upon the reference (potential energy), so it only depends upon the height of location.

(i)   No

(ii)   No

(iii)  No

(iv)  Yes

Among all only angular momentum and total energy of the comet will be constant other all factors given will vary from point to point.

(a) In space we have state of weightlessness, so the swollen feet does not affect astronaut as there is no gravitational pull (so cannot stand).

(b) The swollen face will be affected as the sense organs such as eyes, ears etc. will be affected.

(c) It will affect the astronaut as it may cause mental strain.

(d) It will affect the astronaut as space has different orientations.

We know that the gravitational potential (V) in a sphere is constant throughout. Also, the gravitational intensity will act in the downward direction as its upper half is cut (gravitational force will also act in a downward direction). Hence the required direction of gravitational intensity is shown by arrow C.

As stated in the previous question, the gravitational potential in a spherical shell is constant throughout. So the gravitational intensity will act in downward direction (as upper half is cut). So the required direction is shown by arrow e.

Let the distance where the gravitational force acting on satellite P becomes zero be x from the earth.

Thus we can write :

$\frac{GmM_s}{(r-x)^2}\ =\ \frac{GmM_e}{r^2}$

or                                                 $\left ( \frac{r-x}{x} \right )\ =\ \left ( \frac{2\times 10^{30}}{60\times 10^{24}} \right )^\frac{1}{2}$

or                                                                      $=\ 577.35$

Hence :

$x\ =\ \frac{1.5\times 10^{11}}{578.35}\ =\ 2.59\times 10^8\ m$                                        (Since r  =  $1.5\times 10^{11}$)

Mass of sun can be calculated by using the following formula:-

$M\ =\ \frac{4 \times \pi^2 \times r^3 }{GT^2}$

Putting the known values in the above formula, we obtain :

$=\ \frac{4 \times (3.14)^2 \times (1.5\times 10^{11})^3 }{6.67\times 10^{-11}\times (365.25\times 24\times 60 \times 60)^2}$

or                                                $=\ \frac{133.24\times 10}{6.64\times 10^4}\ =\ 2\times 10^{30}\ Kg$

Thus the mass of the sun is nearly  $2\times 10^{30}\ Kg$.

Kepler's third law give us the following relation :

$T\ =\ \left ( \frac{ \pi^2 r^3}{GM} \right )^ \frac{1}{2}$

Thus we can write :

$\frac{r_s^3}{r_e^3}\ =\ \frac{T_s^2}{T_e^2}$

or                                                       $r_s\ =\ r_e\left ( \frac{T_s}{T_e} \right )^ \frac{2}{3}$

or                                                               $=\ 1.5 \times 10^{11}\times \left ( \frac{29.5 T_e}{T_e} \right )^ \frac{2}{3}$

or                                                               $=\ 14.32 \times 10^{11}\ m$

Thus the distance between the sun and Saturn is  $14.32 \times 10^{11}\ m$.

Q: 8.15  A body weighs $63\hspace {1mm}N$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 at height h from the surface of eath is :

$g'\ =\ \frac {g}{\left ( 1\ +\ \frac{h}{R} \right )^2}$

For     $h\ =\ \frac{R}{2}$    we have :

$g'\ =\ \frac {g}{\left ( 1\ +\ \frac{\frac{R}{2}}{R} \right )^2}$

or                                                          $=\ \frac{4}{9}g$

Thus the weight of the body will be :

$W\ =\ mg'$

or                                                         $=\ m\times \frac{4}{9}g\ =\ \frac{4}{9}mg$

or                                                          $=\ 28\ N$

Position of the body is (depth)  :

$=\ \frac{1}{2}R_e$

Acceleration due to gravity at depth d is given by :

$g'\ =\ \left ( 1\ -\ \frac{d}{R_e} \right )g$

or                                                $=\ \left ( 1\ -\ \frac{\frac{R_e}{2}}{R_e} \right )g$

or                                                $=\ \frac{1}{2}g$

Thus the weight of the body is:-

$W\ =\ mg'$

or                                                 $=\ m\times \frac{1}{2}g\ =\ \frac{mg}{2}$

or                                                 $=\ 125\ N$

Thus the weight of the body is 125 N.

Q: 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 $=6.0 \times10^2^4\hspace {1mm}kg$; mean radius of the earth $=6.4 \times10^6\hspace {1mm}m$; $G=6.67 \times10^-^1^1\hspace {1mm}Nm^2kg^-^1$.

The total energy is given by :

Total energy  =  Potential energy  +  Kinetic energy

$=\ \left ( \frac{ -GmM_e}{R_e} \right )\ +\ \frac{1}{2}mv^2$

At the highest point velocity will be zero.

Thus the total energy of the rocket is :

$=\ \left ( \frac{ -GmM_e}{R_e\ +\ h} \right )\ +\ 0$

Now we will use the conservation of energy :

Total energy initially (at earth's surface)    =    Total energy at height h

$\left ( \frac{ -GmM_e}{R_e} \right )\ +\ \frac{1}{2}mv^2\ =\ \left ( \frac{ -GmM_e}{R_e\ +\ h} \right )$

or                                                       $\frac{1}{2}v^2\ =\ \frac{ gR_eh}{R_e\ +\ h}$

or                                                     $h\ =\ \frac{R_e v^2}{2gR_e\ -\ v^2}$

or                                                           $=\ \frac{6.4\times 10^{6}\times (5\times 10^3)^2}{2g\times 6.4\times 10^6\ -\ (5\times 10^3)^2}$

or                                                           $=\ 1.6\times 10^6\ m$

Hence the height achieved by the rocket from earth's centre =   R  +  h

$=\ 6.4\times 10^6\ +\ 1.6 \times 10^6$

or                                                                                             $=\ 8\times 10^6\ m$

Let us assume the speed of the body far away from the earth is  $v_f$.

Total energy on earth is :

$=\ \frac{1}{2}mv_p^2\ -\ \frac{1}{2}mv_{esc}^2$

And the total energy when the body is far from the earth is :

$=\ \frac{1}{2}mv_f^2$

(Since the potential energy at far from the earth is zero.)

We will use conservation of energy :  -

$\frac{1}{2}mv_p^2\ -\ \frac{1}{2}mv_{esc}^2\ =\ \frac{1}{2}mv_f^2$

or                                          $v_f\ =\ \sqrt{\left ( v_p^2\ -\ v_{esc}^2 \right )}$

or                                                  $=\ \sqrt{\left ( \left ( 3v_{esc} \right )^2\ -\ v_{esc}^2 \right )}$

or                                                  $=\ \sqrt{8}v_{esc}$

or                                                 $=\ 31.68\ Km/s$

The total energy of the satellite at height h is given by :

$=\ \frac{1}{2}mv^2\ +\ \left ( \frac{-GM_em}{R_e\ +\ h} \right )$

We know that the orbital speed of the satellite is :

$v\ =\ \sqrt{\left ( \frac{GM_e}{R_e\ +\ h} \right )}$

Thus the total energy becomes :

$=\ \frac{1}{2}m\times \left ( \frac{GM_e}{R_e\ +\ h} \right )\ +\ \left ( \frac{-GM_em}{R_e\ +\ h} \right )$

or                                                 $=\ - \frac{1}{2} \left ( \frac{GM_em}{R_e\ +\ h} \right )$

Thus the required energy is negative of the total energy :

nbsp;          $E_{req}\ =\ \frac{1}{2} \left ( \frac{GM_em}{R_e\ +\ h} \right )$

or                                                 $=\ \frac{1}{2} \left ( \frac{6.67\times 10^{-11}\times 6 \times 10^{24}\times 200}{6.4\times 10^6\ +\ 0.4\times 10^6} \right )$

or                                                 $=\ 5.9\times 10^9\ J$

Total energy of stars is given by :

$E\ =\ \frac{-GMM}{r}\ +\ \frac{1}{2}mv^2$

or                                            $=\ \frac{-GMM}{r}\ +\ 0$

or                                            $=\ \frac{-GMM}{r}$

Now when starts are just to collide the distance between them is 2R.

The total kinetic energy of both the stars is :

$=\ \frac{1}{2}mv^2\ +\ \frac{1}{2}mv^2\ =\ mv^2$

And the total energy of both the stars is :

$=\ mv^2\ +\ \frac{-GMM}{2r}$

Using conservation of energy we get :

$mv^2\ +\ \frac{-GMM}{2r}\ =\ \frac{-GMM}{r}$

or                                              $v^2\ =\ GM \left ( \frac{-1}{r}\ +\ \frac{1}{2R} \right )$

or                                                      $=\ 6.67\times 10^{-11}\times 2\times 10^{30} \left ( \frac{-1}{10^{12}}\ +\ \frac{1}{2\times 10^7} \right )$

or                                                       $=\ 6.67\times 10^{12}$

Thus the velocity is  :              $\sqrt{6.67\times 10^{12}}\ =\ 2.58\times 10^6\ m/s$

Gravitational force at the midpoint will be zero. This is because both spheres are identical and their forces will be equal but opposite in direction.

The gravitational potential is given by :

$=\ \frac {-GM}{\frac{r}{2}}\ -\ \frac{-GM}{\frac{r}{2}}\ =\ \frac{-4GM}{r}$

$=\ \frac{-4\times 6.67\times 10^{-11}\times 100}{1}$

or                                                                                              $=\ -2.67\times 10^{-8}\ J/Kg$

At the midpoint, we have equal forces in the opposite direction so it is in equilibrium but if we move the body slightly then the particle will move in one direction (as one force will be greater). So it is an unstable equilibrium.

Q: 8.22  As you have learnt in the text, a geostationary satellite orbits the earth at a height of nearly $\small 36,000\hspace{1mm}km$  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 $\small =6.0\times 10^2^4\hspace{1mm}kg$, radius $\small =6400\hspace{1mm}km$.

Height of satellite from earth's surface :    $3.6\times 10^7\ m$

Gravitational potential is given by :

$=\ \frac{-GM}{R\ +\ h}$

or                                                       $=\ \frac{-6.67\times 10^{-11}\times 6\times 10^{24}}{3.6\times 10^7\ +\ 0.64\times 10^7 }$

or                                                       $=\ -\ 9.4\times 10^6\ J/Kg$

Thus potential due to earth gravity is   $-\ 9.4\times 10^6\ J/Kg$.

A body will get stuck at the star's surface if the centrifugal force of star is less than the gravitational force.

The gravitational force is given by :

$F_g\ =\ \frac{GMm}{r^2}$

or                                                             $=\ \frac{6.67\times 10^{-11}\times 5\times 10^{30}\times m}{(1.2\times 10^4)^2}\ =\ 2.31\times 10^{12}\ m\ N$

The centrifugal force is given by :

$F_c\ =\ mr\omega ^2$

or                                                               $=\ mr(2\pi v) ^2$$=\ m\times 1.2\times 10^4\times (2\pi \times 1.2) ^2$

or                                                               $=\ 6.8\times 10^5\ m\ N$

As we can see that the gravitational force is greater than the centrifugal force thus the body will remain at the star.

Firstly, the potential energy of spaceship due to the sun is given by :

$=\ \frac{-GMm_s}{r}$

and the potential energy of spaceship due to mars is given by :

$=\ \frac{-GMm_m}{R}$

It is given that the spaceship is stationary so its kinetic energy is zero.

Thus the total energy of spaceship is :

$=\ \frac{-GMm_s}{r}\ +\ \frac{-GMm_m}{R}$

Thus the energy needed to launch the spaceship is :

$=\ \frac{GMm_s}{r}\ +\ \frac{GMm_m}{R}$

or                                                              $=\ 6.67\times 10^{-11}\times 10^3 \left ( \frac{2\times 10^{30}}{2.28\times 10^{11}}\ +\ \frac{6.4\times 10^{23}}{3.395\times 10^6} \right )$

or                                                             $=\ 596.97\times 10^9\ J$

or                                                             $=\ 6\times 10^{11}\ J$

The kinetic energy of the rocket is (initial):-

$=\ \frac{1}{2}mv^2$

And the initial potential energy is :

$=\ \frac{-GMm}{R}$

Thus total initial energy is given by :

$=\ \frac{1}{2}mv^2\ +\ \frac{-GMm}{R}$

Further, it is given that 20 per cent of kinetic energy is lost.

So the net initial energy is :

$=\ 0.4mv^2\ -\ \frac{GMm}{R}$

The final energy is given by :

$=\ \frac{GMm}{R\ +\ h}$

Using the law of energy conservation we get :

$0.4mv^2\ -\ \frac{GMm}{R}\ =\ \frac{GMm}{R\ +\ h}$

Solving the above equation we get :

$h\ =\ \frac{R}{\frac{GM}{0.4v^2R}\ -\ 1}$

or                                               $=\ \frac{0.4R^2v^2}{GM\ -\ 0.4v^2R}$

or                                              $=\ \frac{18.442\times 10^{18}}{42.688\times 10^{12}\ -\ 5.432\times 10^{12}}$

or                                               $=\ 495\times 10^3\ m$

or                                              $=\ 495\ Km$

Thus the required distance is 495 Km.

NCERT solutions for class 11 physics chapter wise

 Chapter 1 NCERT solutions for class 11 physics chapter 1 Physical world Chapter 2 Solutions of NCERT for class 11 physics chapter 2 Units and Measurement Chapter 3 CBSE NCERT solutions for class 11 physics chapter 3 physics Motion in a straight line Chapter 4 NCERT solutions for class 11 physics chapter 4 Motion in a Plane Chapter 5 Solutions of NCERT for class 11 physics chapter 5 Laws of Motion Chapter 6 CBSE NCERT solutions for class 11 physics chapter 6 Work, Energy and Power Chapter 7 NCERT solutions for class 11 physics chapter 7 System of Particles and Rotational motion Chapter 8 NCERT solutions for class 11 physics chapter 8 Gravitation Chapter 9 CBSE NCERT solutions for class 11 physics chapter 9 Mechanical Properties of Solids Chapter 10 NCERT solutions for class 11 physics chapter 10 Mechanical Properties of Fluids Chapter 11 Solutions of NCERT for class 11 physics chapter 11 Thermal Properties of Matter Chapter 12 CBSE NCERT solutions for class 11 physics chapter 12 Thermodynamics Chapter 13 NCERT solutions for class 11 physics chapter 13 Kinetic Theory Chapter 14 Solutions of NCERT for class 11 physics chapter 14 Oscillations Chapter 15 CBSE NCERT solutions for class 11 physics chapter 15 Waves

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Benefits of NCERT solutions for class 11 physics chapter 8 gravitation:

Numerical problems and derivations of the chapter are important. The solutions of NCERT class 11 physics chapter 8 gravitation will help in the final exam and also in competitive exams. For exams like NEET and JEE Main, one or two questions are expected from the chapter gravitation and the CBSE NCERT solutions for class 11 physics chapter 8 gravitation will help in solving the questions asked such exams.