NCERT Solutions for Class 11 Physics Chapter 8 Gravitation

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Access gravitation class 11 numericals with solutions - Exercise Questions

Q: 8.1 (a) Answer the following questions:

(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?

Answer:

No, because gravitational force doesn't depend upon the material medium. It is independent of the presence of other materials.

Q: 8.1 (b) Answer the following :

(b) An astronaut inside a small space ship 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?

Answer:

Yes if the size of the space station is large then he can detect gravity.

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?

Answer:

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.

Q: 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 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.

Answer:

(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:

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

$g\ =\ \frac{GM}{R^2}$ 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.

Q:8.3 Suppose there existed a planet that went around the sun was twice as fast as the earth. What would be its orbital size as compared to that of the earth?

Answer:

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.

Q: 8.4 Io, one of the satellites of Jupiter, has an orbital period of $1.769$ days and the radius of the orbit is $4.22\times 10^8\hspace{1mm}m$ . Show that the mass of Jupiter is about one-thousandth that of the sun.

Answer:

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.

Q: 8.5 Let us assume that our galaxy consists of $2.5\times 10^1^1$ 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 $10^5$ ly.

Answer:

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 :

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

Q: 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) 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?

Answer:

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

Q: 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:

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

Q: 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) 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.

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

Q: 8.10 In the following two exercises, choose the correct answer from among the given ones: The gravitational intensity at the centre of a hemispherical shell of uniform mass density has the direction indicated by the arrow (see Fig 8.12) (i) a, (ii) b, (iii) c, (iv) 0

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Answer:

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 .

Q: 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.

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Answer:

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 .

Q: 8.12 A rocket is fired from the earth towards the sun. At what distance from the earth’s centre is the gravitational force on the rocket zero? Mass of the sun $=2\times 10^3^0\hspace{1mm} kg$ , mass of the earth $=6\times 10^2^4\hspace{1mm} kg$ . Neglect the effect of other planets etc. (orbital radius $=1.5\times 10^1^1\hspace{1mm} m$ ).

Answer:

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 = $1.5\times 10^{11}$ )

Q: 8.13 How will you ‘weigh the sun’, that is estimate its mass? The mean orbital radius of the earth around the sun is $1.5\times 10^8\hspace{1mm}km$ .

Answer:

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 .

Q: 8.14 A saturn year is $29.5$ times the earth year. How far is the saturn from the sun if the earth is $1.50\times 10^8\hspace{1mm}km$ away from the sun?

Answer:

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?

Answer:

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 $=\ 28\ N$

Q: 8.16 Assuming the earth to be a sphere of uniform mass density, how much would a body weigh half way down to the centre of the earth if it weighed $\small 250\hspace {1mm}N$ on the surface?

Answer:

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 $=6.4 \times10^6\hspace {1mm}m$ ; $G=6.67 \times10^-^1^1\hspace {1mm}Nm^2kg^-^1$ .

Answer:

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

Q: 8.18 The escape speed of a projectile on the earth’s surface is $\small 11.2\hspace {1mm}km\hspace {1mm}s^-^1$ . 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.

Answer:

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 $=\ 31.68\ Km/s$

Q: 8.19 A satellite orbits the earth at a height of $\small 400\hspace {1mm}km$ above the surface. How much energy must be expended to rocket the satellite out of the earth’s gravitational influence? Mass of the satellite $\small =200\hspace {1mm}kg$ ; ! mass of the earth $\small =6.0 \times10^2^4\hspace {1mm}kg$ ; radius of the earth $\small =6.4 \times10^6\hspace {1mm}m$ ; $\small G=6.67 \times10^-^1^1\hspace {1mm}Nm^2 kg^-^2$ .

Answer:

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

Q: 8.20 Two stars each of one solar mass ( $\small =2 \times 10^3^0\hspace {1mm}kg$ ) are approaching each other for a head on collision. When they are a distance $\small 10^9\hspace{1mm}km$ , their speeds are negligible. What is the speed with which they collide? The radius of each star is $\small 10^4\hspace{1mm}km$ . Assume the stars to remain undistorted until they collide. (Use the known value of G).

Answer:

The 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 :

Q: 8.21 Two heavy spheres each of mass $\small 100\hspace {1mm}kg$ and radius $\small 0.10\hspace {1mm}m$ are placed $\small 1.0\hspace {1mm}m$ apart on a horizontal table. What is the gravitational force 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:

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.

NCERT Solutions for Class 11 Physics Chapter 8 Gravitation Additional Exercise

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 $\small =6400\hspace{1mm}km$ .

Answer:

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 _.

Q: 8.23 A star $\small 2.5$ times the mass of the sun and collapsed to a size of $\small 12 \hspace{1mm}km$ rotates with a speed of $\small 1.2 \hspace{1mm}$ 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 $\small =2 \times 10^3^0 \hspace{1mm}kg$ ).

Answer:

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.

Q: 8.24 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 $\small =1000\hspace{1mm}kg$ ; mass of the sun $\small =2 \times 10^3^0\hspace{1mm}kg$ ; mass of mars $\small =6.4 \times 10^2^3\hspace{1mm}kg$ ; radius of mars $\small =3395\hspace{1mm}km$ ;radius of the orbit of mars $\small =2.28 \times 10^8\hspace{1mm}km$ ; $\small G=6.67 \times 10^-^1^1\hspace{1mm}Nm^2kg^-^2$

Answer:

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

Q: 8.25 A rocket is fired ‘vertically’ from the surface of mars with a speed of $\small 2\hspace{1mm}km\hspace{1mm}s^-^1$ . If $\small 20\%$ 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 $\small =6.4\times 10^2^3\hspace{1mm}kg$ ; radius of mars $\small =3359\hspace{1mm}km$ ; $\small G=6.67\times 10^-^1^1\hspace{1mm}Nm^2kg^-^2$ .

Answer:

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.

Gravitation solutions class 11 consist of a total of twenty-five questions, in which questions number 22,23,24 and 25 belong to the additional exercises. NCERT Solutions for Class 11 Physics Chapter 8 Gravitation provides step-by-step explanations for all the questions and exercises given in the textbook. These gravitation class 11 ncert solutions are prepared by subject matter experts who have extensive knowledge of the concepts covered in the chapter.

More About Gravitation Class 11

You will study the unit electrostatics in NCERT Class 12 physics, so, while studying Electrostatics chapter you can compare the formula with the formulas that you will study in Gravitation Class 11. NCERT solutions can make your learning easy if you are want to prepare for other classes too. The main NCERT topics of the chapter are listed below

NCERT Solutions for Class 11 Physics Chapter Wise

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

Physics class 11 Gravitation: Important Formulas and Diagrams

Some of the important gravitation chapter class 11 formula are given below:

Gravitational Force-

$F\; = \frac{G\, m_{1}\, m_{2}}{r^{2}}$

Acceleration Due to Gravity (g)-

$g=\frac{GM}{R^{2}}$

Gravitational Potential-

$V=-\int \overrightarrow{I}\cdot \overrightarrow{dr}$ or $V=-\frac{W}{m}=-\int \frac{\overrightarrow{F}\cdot \overrightarrow{dr}}{m}$

Work Done Against Gravity-

$W=\Delta U=GMm\left [ \frac{1}{r_{1}}-\frac{1}{r_{2}} \right ]$

Escape Velocity-

$V_{c}=\sqrt{\frac{2GM}{R}}$

Kepler's Laws-

$\frac{dA}{dt}= \frac{L}{2m}$

$T^{2}\: \alpha\: a^{3}$

Energy of Satellite-

$E=-\frac{GMm}{2r}$

Highlight of NCERT Solutions for Class 11 Physics Chapter 8 Gravitation

Here are some of the highlights of the gravitation class 11:

Newton's law of gravitation: class 11 physics chapter 8 begins with an introduction to the concept of gravity and Newton's law of gravitation, which states that every particle in the universe attracts every other particle with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.
Gravitational field: chapter 8 physics class 11 goes on to explain the concept of a gravitational field, which is the region in space around a massive object where the force of gravity is felt. The strength of the gravitational field is proportional to the mass of the object and decreases with distance from the object.
Gravitational potential energy: The gravitation chapter class 11 also covers the concept of gravitational potential energy, which is the energy possessed by an object due to its position in a gravitational field.
Kepler's laws of planetary motion: ncert gravitation class 11 includes an introduction to Kepler's laws of planetary motion, which describe the motion of planets around the sun.
Escape velocity: Class 11 physics gravitation exercise concludes with the concept of escape velocity, which is the minimum velocity an object needs to escape the gravitational pull of a massive object.

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These are some of the key highlights of the gravitation class 11 exercise solutions.

Key features of gravitation class 11 questions and answers

Comprehensive Coverage: The class 11 physics gravitation exercise solutions cover all the important topics and questions presented in the chapter, ensuring a thorough understanding of the subject matter.

Step-by-Step Explanations: Each question is accompanied by a detailed step-by-step solution, making it easier for students to follow and learn the concepts.

Clarity and Simplicity: The gravitation class 11 ncert solutions are explained in clear and simple language, making complex concepts more accessible to students.

Diagrams and Graphs: When necessary, diagrams and graphs are included to visually illustrate concepts and enhance comprehension.

Numerical Problem Solving: Practice with solving numerical problems related to gravitational forces, orbital mechanics, and escape velocities, helping students develop problem-solving skills.

Conceptual Clarity: The gravitation class 11 questions and answers aim to build a strong conceptual foundation, allowing students to grasp the fundamental principles governing the motion of celestial bodies.

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 NCERT solutions for Class 11 will help in solving the questions asked in such exams.

NCERT Solutions for Class 11 Subject wise

NCERT solutions for Class 11 Biology

NCERT solutions for Class 11 Maths

NCERT solutions for Class 11 Chemistry

Also Check NCERT Books and NCERT Syllabus here

Subject wise NCERT Exemplar solutions

Frequently Asked Questions (FAQs)

1. What is the weightage of the chapter Gravitation for NEET

One question can be expected from the NCERT chapter Gravitation for NEET exam. The concepts studied till chapter 7 of Class 11 NCERT books may be used to solve problems of chapter Gravitation. Study all the formulas from Gravitation NCERT Notes. And solve more questions using NEET previous year papers and NCERT exemplar.

2. What is the weightage of the chapter gravitation for JEE Main

one question from the chapter gravitation can be expected for JEE Main exam. Some times it can be two also. Grasp all the concepts and formulas of Mechanics well to solve the problems in JEE Main.

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Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

Option 1) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

Option 1) $11.2\, L\, H_{2(g)}$ at STP is produced for every mole $HCL_{(aq)}$ consumed	Option 2) $6L\, HCl_{(aq)}$ is consumed for ever $3L\, H_{2(g)}$ produced
Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ at STP is produced for every mole Al that reacts .

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

Option 1) decrease twice	Option 2) increase two fold
Option 3) remain unchanged	Option 4) be a function of the molecular mass of the substance.

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

Option 1) twice that in 60 g carbon	Option 2) 6.023 × 10²²
Option 3) half that in 8 g He	Option 4) 558.5 × 6.023 × 10²³

Option 1) less than 3	Option 2) more than 3 but less than 6
Option 3) more than 6 but less than 9	Option 4) more than 9

NCERT Solutions for Class 11 Physics Chapter 8 Gravitation

NCERT Solutions for Class 11 Physics Chapter 8 Gravitation

NCERT Solutions for Class 11 Physics Chapter 8 Gravitation Additional Exercise

More About Gravitation Class 11

NCERT Solutions for Class 11 Physics Chapter Wise

Physics class 11 Gravitation: Important Formulas and Diagrams

Highlight of NCERT Solutions for Class 11 Physics Chapter 8 Gravitation

Key features of gravitation class 11 questions and answers

Benefits of NCERT Solutions for Class 11 Physics Chapter 8 Gravitation:

NCERT Solutions for Class 11 Subject wise

Also Check NCERT Books and NCERT Syllabus here

Subject wise NCERT Exemplar solutions

Frequently Asked Questions (FAQs)

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