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Question 12 Marks
Choose the correct alternative: If the zero of potential energy is at infinity, the total energy of an orbiting satellite is negative of its kinetic/ potential energy.
Answer
Kinetic energy.Explanation:
Total mechanical energy of a satellite is the sum of its kinetic energy (always positive) and potential energy (may be negative). At infinity, the gravitational potential energy of the satellite is zero. As the Earth-satellite system is a bound system, the total energy of the satellite is negative. Thus, the total energy of an orbiting satellite at infinity is equal to the negative of its kinetic energy.
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Question 22 Marks
Answer the following: 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, Astronaut can hope to detect gravity if the size of th e spaceship is extremely large, then the magnitude of the gravity will become appreciable and hence the gravitational effect of the spaceship may been me measurable.
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Question 32 Marks
Answer the following: 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. Electrical forces depend upon the nature of the intervening medium while the gravitational forces don’t depend upon the nature of the intervening medium. So, such shielding acts are not possible in case of gravitation i.e., gravity screens are not possible.
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Question 42 Marks
Answer the following: 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
Earth moon distance is very small as compared to earth-sun distance. Tidal effect is inversely proportional to the cube of the distance it means it is not governed by inverse square law like the gravitational force (which obeys inverse square law). Hence, tidal effect of moon is larger than that due to the sun.
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Question 52 Marks
Define gravitational field strength. What is the field at a point distance r from a mass M?
Answer
Gravitational field strength is defined as the force experienced by unit mass kept at a point. At a distance r, the force is $\text{F}=\frac{\text{GMm}}{\text{r}^2}$ The gravitational field $\frac{\text{F}}{\text{m}}=\frac{\text{GMm}}{\text{r}^2\text{m}}=\frac{\text{GM}}{\text{r}^2}$
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Question 62 Marks
We can shield a charge from electric fields by putting it inside a hollow conductor. Can we shield a body from the gravitational influence of nearby matter by putting it inside a hollow sphere or by some other means?
Answer
The gravitational force does not depend upon the nature of the medium, but the electric force depends on the intervening medium between the charge and the electric field. So we cannot shield a body from the gravitational force.
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Question 72 Marks
What is a geostationary satellite? Is it same as geo synchronous satellite?
Answer
If the revolution period of a satellite is same as that of the earth, it is called a geo-stationary satellite. Yes, geo-synchronous satellites are geo-stationary satellites.
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Question 82 Marks
Under what circumstances would your weight become zero?
Answer
The weight will become zero under the following circumstances:
  1. During free fall.
  2. At the centre of the earth.
  3. In an artificial satellite.
  4. At a point where gravitational pull of earth is equal to the gravitational pull of the Moon.
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Question 92 Marks
A satellite is revolving just near the Earth's surface. Compute its orbital velocity. Given that radius of Earth $\mathrm{R}= 6400 km $ and $3=9.8 \mathrm{~ms}^{-2}$.
Answer
For a satellite revolving just near the Earth's surface, the orbital velocity has a magnitude given by, $\text{v}_\text{orb}=\sqrt{\text{gR}}$ $\therefore\ \text{v}_\text{orb}=\sqrt{9.8\times6400\times1000}$ $=7.92\times10^3\text{ms}^{-1}$ $=7.92\text{km s}^{-1}$
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Question 102 Marks
If the earth stops rotating about its own axis, what is the effect on the value of weight of any object on the surface of earth?
Answer
If the earth stops rotating, the weight of the object increases due to absence of centrifugal force. However, the weight at the poles remains the same.
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Question 112 Marks
The escape velocity v of a body depends upon:
  1. The acceleration due to gravity $'g\ '$ of the planet.
  2. The radius of the planet $'R\ '.$
Establish dimensionally the relationship between them.
Answer
$\text{v}\propto\text{g}^{\text{a}}\text{R}^{\text{b}}$
$[\text{LT}_{-1}]=\text{K}[\text{LT}^{-2}]^\text{a}[\text{L}]^{\text{b}}$
$\text{a}=\frac{1}{2},\text{b}=\frac{1}{2}$ $\text{K}=\sqrt{2}$
$\therefore\text{v}=\sqrt{2\text{gR}}$
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Question 122 Marks
A spherical planet has mass $\text{M}_p$ and diameter $\text{D}_p$. A particle of mass m falling freely near the surface of this planet will experience an acceleration due to gravity, equal to whom?
Answer
Force is given by, $\text{F}=\frac{\text{GM}_\text{e}}{\text{R}^2}=\frac{\text{GM}_\text{P}\text{m}}{\Big(\frac{\text{D}_\text{P}}{2}\Big)}=\frac{4\text{GM}_\text{P}\text{m}}{\text{D}^2_\text{P}}$ $\frac{\text{F}}{\text{m}}=\frac{4\text{GM}_\text{P}}{\text{D}^2_\text{P}}$
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Question 132 Marks
Why does the earth have an atmosphere and the moon does not?
Answer
The escape velocity from the earth surface is about 11.2km/sec which is greater than the r.m.s. velocity of the air molecules. The escape velocity from the moon's surface is about 2.4km/sec which is less than the r.m.s. velocity of the air molecules. Hence, the earth has an atmosphere and the moon does not.
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Question 142 Marks
A 10kg mass is to be divided into two parts, such that the force of attraction between them is maximum. What is the mass of each portion?
Answer
Let m kg and (10 - m) kg be the mass of two parts separated by a distance r. The force is $\text{F}=\frac{\text{Gm}(10-\text{m})}{\text{r}^2}$ For force to be maximum, $\frac{\text{dF}}{\text{dm}}=0$ $\therefore\frac{\text{G}}{\text{r}^2}[\text{m}(-1)+(10-\text{m}\times1)=0]$ $\therefore\text{m}=10-\text{m}$ $\therefore\text{m}=5\text{kg}$
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Question 152 Marks
Imagine what would happen if the value of $G$ becomes:
  1. $100$ times of its present value.
  2. Times of its present value.
Answer
  1. Earth's attraction would be so large that you would be crushed to the earth.
  2. Earth's attraction would be so less that we can easily jump from the top of a multi$-$storey building.
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Question 162 Marks
If the choice of origin is shifted, what is the change in $(i)$ Gravitational potential $(ii)$ Potential energy?
Answer
  1. Remains dependent on the choice of origin.
  2. Potential energy is independent since it depends only on separation.
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Question 172 Marks
Determine the speed with which the earth would have to rotate on its axis so that a person on the equator would weigh 3/5th as much as at present. Take the equatorial radius as 6400km.
Answer
Acceleration due to gravity at the equator is, $\text{g}_\text{e}=\text{g}-\text{R}\omega^2$ $\text{mg}_\text{e}=\text{mg}-\text{mR}\omega^2$ $\frac35\text{mg}=\text{mg}-\text{mR}\omega^2$ $\Big[\because\text{mg}_\text{e}=\frac35\text{mg}\Big]$ $\therefore\omega=\sqrt{\frac{2\text{g}}{5\text{R}}}=\sqrt{\frac{2\times9.8}{6\times6400\times10^3}}=7.8\times10^{-4}\text{rad/s}$
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Question 182 Marks
Show that the orbital velocity of a satellite revolving the earth is $7.92\text{km s}^{-1}$.
Answer
Orbital velocity, $\text{v}_0=\sqrt{\text{gR}}$ $=\sqrt{9.8\times6.4\times10^6}=7.92\text{km s}^{-1}$
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Question 192 Marks
The radii of two planets are R and 2R respectively and their densities $\rho\text{ and }\frac{\rho}{2}$ respectively. What is the ratio of acceleration due to gravity at their surfaces?
Answer
$\text{g}=\frac{\text{GM}}{\text{R}^2}=\frac{\text{g}\frac{4}{3}\pi\text{R}^3\rho}{\text{R}^2}=\text{g}\frac{4}{3}\pi\text{R}\rho$ $\therefore\frac{\text{g}_1}{\text{g}_2}=\frac{\text{R}}{2\text{R}}.\frac{\rho}{\frac{\rho}{2}}=1$ $\therefore\text{g}_1:\text{g}_2=1:1$
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Question 202 Marks
The escape speed on the earth is 11.2km/ s.What is its value for a planet having double the radius and eight times the mass of the earth?
Answer
$v_p$ (escap speed on a planet) $=\sqrt{\frac{\text{GM}_\text{p}}{\text{R}_\text{p}}}$ $v_e$ (escape speed on the earth) $=\sqrt{\frac{\text{GM}_\text{e}}{\text{R}_\text{e}}}$ Clearly, $\frac{\text{v}_\text{p}}{\text{v}_\text{e}}=\sqrt{\frac{\text{M}_\text{p}}{\text{M}_\text{e}}\times\frac{\text{R}_\text{e}}{\text{R}}_\text{p}}=\sqrt{8\times\frac12}=2$ $\text{v}_\text{p}=2\text{v}_\text{e}22.4\text{km/s}$
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Question 212 Marks
The binding energy of a particle of mass m attached with the earth of mass M and radius R is $\frac{\text{GMm}}{\text{R}}.$ How much energy must be supplied to the particle so that it may escape the gravitational field of the earth? In what form this energy is supplied to the particle?
Answer
The particle will escape the gravitational field of the earth if the energy supplied to it is R equal to its binding energy. Thus, energy supplied to the particle $=\frac{\text{GMm}}{\text{R}}.$ This energy is supplied in the form of the kinetic energy of the particle.
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Question 222 Marks
A planet moving along an elliptical orbit is closest to the Sun at a distance $r_1$ and farthest away at a distance of $r_2$. If $v_1$ and $v_2$ are the linear velocities at these points respectively, then find the ratio $\frac{\text{v}_1}{\text{v}_2}.$
Answer
From the law of conservation of angular momentum, $\text{mr}_1\text{v}=\text{mr}_2\text{v}_2$ $\Rightarrow\text{r}_1\text{v}_1=\text{r}_2\text{v}_2$ $\frac{\text{v}_1}{\text{v}_2}=\frac{\text{r}_2}{\text{r}_2}$
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Question 232 Marks
Jf the kinetic energy of a satellite revolving around the earth in any orbit is doubled, then what will happen to it?
Answer
The total energy of a satellite in any orbit, E = -K, where K is its KE in that orbit. If its kinetic energy is doubled, i.e. an additional kinetic energy (K) is given to it, E = -K + K = 0 and the satellite will leave its orbit and go to infinity.
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Question 242 Marks
Discuss the variation of $'g\ '$
  1. Due to height.
  2. Due to depth.
Answer
  1. $\text{g}=\text{g}\Big(1-\frac{2\text{h}}{\text{R}}\Big)$
  2. $\text{g}'=\text{g}\Big(1-\frac{\text{d}}{\text{R}}\Big)$

The value of $g$ decreases both on moving up and on moving down from the surface of earth. This can be shown graphically as above.
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Question 252 Marks
What is the acceleration due to gravity at the bottom of a sea 30km deep taking radius of the earth as 6.3 × 10km?
Answer
$\text{g}'=\text{g}\Big(1-\frac{\text{d}}{\text{R}}\Big)$ $=9.8\Big(1-\frac{30\times1000}{6.3\times(1000)^2}\Big) $ $=9.8\Big(1-\frac{1}{210}\Big)$ $=9.8\Big(\frac{209}{210}\Big)=9.75\text{ms}^{-2}$
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Question 262 Marks
What are the conditions under which a rocket fired from the earth, launches an artificial satellite of earth?
Answer
  1. Velocity acquired is more than the escape velocity to go out in space.
  2. Provide sufficient velocity to move in a path of its own.
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Question 272 Marks
The masses M in the two figures are identical. Which of the two systems has the smaller gravitational potential energy? Why?
Answer
The first system has the outer most masses closer to each other than the second system. Hence the gravitational P.E. of the second system is lesser than that of the first system.
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Question 282 Marks
Determine the speed with which the earth has to rotate on its axis so that a person on the equator weigh $\Big(\frac{3}{5}\Big)^{\text{th}}$ as much as at present. Take the equatorial radius as 6400km.
Answer
$\text{Here, W = mg}$ $\text{and }\text{W}'=\Big(\frac{3}{5}\Big)\text{mg}$ $\text{As W}'=\text{W}-\text{mR}\omega^2$ $\text{So},\frac{3}{5}\text{mg}=\text{mg}-\text{m}\text{R}\omega^2$ $\text{or }\text{mR}\omega^2=\text{mg}-\frac{3}{5}\text{mg}=\frac{2}{5}\text{mg}$ $\text{or }\omega=\sqrt{\frac{2\text{g}}{5\text{R}}}=\sqrt{\frac{2\times9.8}{5\times6400\times10^3}}$ $=4.315\times10^{-4}\text{rad/s}.$
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Question 292 Marks
The asteroid pallas has an orbital period of 4.62 year. Find the semi-major axis of its orbit. Given, $\mathrm{G}=6.67 \times 10^{-}$ ${ }^{11} \mathrm{Nm}^2 \mathrm{~kg}^{-2}$, Mass of the sun $=1.99 \times 10^{30} \mathrm{~kg}$ and 1 year $=3.156 \times 10^7 \mathrm{~s}$
Answer
$\text{T}^2=\frac{4\pi^2\text{a}^3}{\text{GM}_\text{s}}$$\text{a}=\Big[\frac{\text{GM}_\text{s}\text{T}^2}{4\pi^2}\Big]^\frac{1}{3}$
Substituting values, we get
$a = 4.15 \times 10^{11}m$
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Question 302 Marks
The earth is acted upon by the gravitational attraction of the sun. Why don't the earth fall into the sun?
Answer
The earth is orbiting round the sun in a stable orbit (nearly circular) such that the gravitational attraction of the sun just provides the required centripetal force to the earth for its orbital motion. So, net force on the earth is zero and consequently, the earth does not fall into the sun.
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Question 312 Marks
The orbiting velocity of an earth-satellite is 8km s. What will be the escape velocity?
Answer
Escape velocity, $\text{v}_\text{e}=\sqrt{2}\text{v}_0$ $\text{v}_\text{e}=\sqrt2\times8=11.31\text{km s}^{-1}$
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Question 322 Marks
Calculate the gravitational force which Sun exerts on Jupiter.
Answer
Given: $G = 6.67 \times 10^{-11}Nm^2/ kg^2 M_J = 1.9 \times 10^{27}kg M_s = 1.99 \times 10^{30}kg$ Mean distance = $7.8 \times 10^{+11}m$
$\text{F}_\text{g}=\frac{\text{GM}_{\text{S}}\text{M}_\text{J}}{\text{r}^2}$ $=\frac{6.67\times10^{-11}\times1.99\times10^{30}\times1.9\times10^{27}}{(7.8\times10^{11})^2}$ $=4.145\times10^{23}\text{N}$
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Question 332 Marks
Is it possible to place a satellite, so that it is always over New Delhi? Why?
Answer
No, generally the geo-stationary satellites will be in the equatorial plane. Since New Delhi does not lie in this plane, it is not possible.
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Question 342 Marks
Two masses each of 100kg are separated by 1 metre. Find the gravitational (i) field (ii) potential at the mid-point of the line joining them.
Answer
Gravitational field strength is the force experienced by unit mass kept at a point. Since it is a vector, we have to add vectorially.
$\therefore\text{E}_{\text{g0}}=\text{E}_{\text{gA}}+\text{E}_{\text{gB}}=0,$ Since the masses are equal and separation is same.
But gravitational potential $=-\frac{\text{Gm}}{\text{r}}$
Potential at O (mid-point) $=-\frac{\text{G}\times100}{0.5}-\frac{\text{G}\times100}{0.5}$
$=-\frac{200\text{G}}{0.5}=-400\text{G }\text{J/m}$
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Question 352 Marks
If suddenly the gravitational force of attraction between the earth and a satellite revolving around it becomes zero, what will happen to the satellite?
Answer
Satellite will move tangentially with constant speed equal to its orbital velocity at the time the force ceases to act.
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Question 362 Marks
How can you find the mass of earth, starting from the law of gravitation?
Answer
From law of gravitation, $\text{F} =\frac{\text{GMm}}{\text{r}^2}$If m is a mass on the surface of earth of radius R. The acceleration of the mass m on the surface of earth is g. $\therefore$ Force experienced F = mg Using the two forces we have, $\text{M}=\frac{\text{gR}^2}{\text{G}}$
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Question 372 Marks
A small mass is released by an astronaut in a satellite in space. Will it fall on the earth?
Answer
The orbit of the satellite is independent of its mass. The released mass will continue to orbit the earth as a satellite and hence, will not fall on the earth.
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Question 382 Marks
How far above the earth's surface does the value of g becomes 20% of its value on the surface?
Answer
$\frac{\text{g}'}{\text{g}}=\frac{20}{100},\text{g}'=\text{g}\frac{20}{100}$ $=\text{g}\Big(1-\frac{2\text{h}}{\text{R}}\Big)$ $\frac{20}{100}=\Big(1-\frac{2\text{h}}{\text{R}}\Big),\frac{1}{5}$ $=1-\frac{2\text{h}}{\text{R}}$ $\frac{2\text{h}}{\text{R}}=\frac{4}{5},\text{h}=\frac{2}{5}\text{R}.$
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Question 392 Marks
A satellite does not need any fuel to circle around the earth. Why?
Answer
The gravitational force between satellite and the earth provides the centripetal force required by the satellite to move in a circular orbit. The satellite orbits around earth at such a higher height where air friction is neglible.
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Question 402 Marks
For a satellite orbiting in an orbit, close to the surface of earth, to escape, what is the percentage increase in the kinetic energy required?
Answer
Orbital velocity for close to earth orbits $=\sqrt{\text{gR}}$ Escape velocity required $=\sqrt{2\text{gR}}$ % Increase required $=\frac{\text{v}_{\text{e}}-\text{v}_{\text{o}}}{\text{v}_{\text{o}}}\times100$ $=\frac{\sqrt{\text{gR}(\sqrt{2}-1)}}{\sqrt{\text{gR}}}\times100=41.4\%$
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Question 412 Marks
Choose the correct alternative: If the zero of potential energy is at infinity, the total energy of an orbiting satellite is negative of its kinetic/ potential energy.
Answer
Kinetic energy.Explanation:
Total mechanical energy of a satellite is the sum of its kinetic energy (always positive) and potential energy (may be negative). At infinity, the gravitational potential energy of the satellite is zero. As the Earth-satellite system is a bound system, the total energy of the satellite is negative. Thus, the total energy of an orbiting satellite at infinity is equal to the negative of its kinetic energy.
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Question 422 Marks
The speed of a plane at the perihelion is v, and the radius at perihelion is $r_p$. At aphelion $r_a$, what will be the speed?
Answer
Since angular momentum L has to be conserved, L = mvr = constant. For a given planet, $\text{v}_{\text{p}}\text{r}_{\text{p}}=\text{v}_{\text{a}}\text{r}_{\text{a}}$ $\therefore\text{v}_{\text{a}}=\frac{\text{v}_{\text{p}}\text{r}_{\text{p}}}{\text{r}_{\text{a}}}$
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Question 432 Marks
A mass of 1g is separated from another mass of 1g by a distance of 1cm. How many g-wt of force exists between them?
Answer
$\text{F}=\text{G}\frac{\text{m}_1\text{m}_2}{\text{r}^2}$ $=(6.67\times10^{-8})\Big(\frac{1\times1}{1^2}\Big)\text{dyne}$ $=6.67\times10^{-8}\text{dyne}=\frac{6.67\times10^{-8}}{980}$ $=7\times10^{-11}\text{g-wt}$
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Question 442 Marks
If apple and earth pull each other with equal force, why does earth not move towards apple?
Answer
Acceleration caused on earth due to apple is negligible due to large mass of earth and because the observer is in the same frame so he cannot see motion.
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Question 452 Marks
Does the change in gravitational potential energy of a body between two given points depend upon the nature of path followed, why?
Answer
The change in gravitational potential energy of a body between two given points depends only upon the position of the given points and is independent of the path followed. is due to the fact that the gravitational force is a conservative force and work done by a conservative force depends only on the position of initial and final points and is independent of path followed.
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Question 462 Marks
The time period of a satellite of earth is 5h. If the separation between the earth and the satellite is increased to four times the previous value, then what will be the new time period?
Answer
$\text{T}_2=\text{T}_1\Big[\frac{\text{R}_2}{\text{R}_1}\Big]^{\frac32}$ $=\text{T}_1(4)^\frac32=\text{8T}_1=40\text{h}$
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Question 472 Marks
What is the value of escape velocity from the surface of $(i)$ earth, $(ii)$ moon?
Answer
  1. $\text{v}_{\text{e}}=\sqrt{2\text{gR}}=11.2\text{km/s.}$
  2. $\text{v}_{\text{m}}=\sqrt{\frac{2\text{g}\times\text{R}_{\text{m}}}{6}}=2.38\text{km/s.}$
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Question 482 Marks
What is the gravitational potential energy of a body at heighth from the earth surface?
Answer
Gravitational potential energy, i.e., $\text{U}_\text{h}=-\frac{\text{GMm}}{\text{R}+\text{h}}=-\frac{\text{gR}^2\text{m}}{\text{R}+\text{h}}$ $\Big[\text{where},\text{g}=\frac{\text{GM}}{\text{R}^2}\Big]$ $=-\frac{\text{gR}^2\text{m}}{\text{R}\Big(1+\frac{\text{h}}{\text{R}}\Big)}=-\frac{\text{mgR}}{1+\frac{\text{h}}{\text{R}}}$
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Question 492 Marks
The acceleration due to gravity on a planet is $1.96 \mathrm{~m} \mathrm{~s}^{-2}$. If it is safe to jump from a height of 2m on the earth, then what will be the corresponding safe height on the planet?
Answer
The safety of a person depends upon the momentum with which the person hits the planet. Since, the mass of the person is constant, therefore the maximum velocity v is the limiting factor.
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Question 502 Marks
Define Escape velocity and find its value.
Answer
The velocity with which a body is escaped vertically upwards and it just cross the gravitational field is called escape velocity of the body. Its value is, $\text{V}_\text{e}=\sqrt{2\text{gR}_\text{e}}=11.2\text{km/ s}$
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Question 512 Marks
An astronaut, by mistake, drops his food packet from an artificial satellite orbiting around the earth. Will it reach the surface of the earth? Why?
Answer
The food packet will not fall on the earth. As the satellite as well as astronaut were in a state of weightlessness, hence, the food packet, when dropped by mistake, will also start moving with the same velocity as that of satellite and will continue to move along with the satellite in the same orbit.
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Question 522 Marks
Earth's radius is about $6370\ km.$ A mass of $20\ kg$ is taken to a height of $160\ km$ above the earth's surface.
  1. What is the mass of the objects at that height?
  2. How much does the object weigh at this height?
Answer
  1. The mass of the object remain $20\ kg$ at that height.
  2. $\text{W}=\frac{\text{GMm}}{\text{r}^2}$
$\Rightarrow\frac{\text{W}_2}{\text{W}_1}=\frac{\text{r}^2_1}{\text{r}^2_2}$
since $G, M$ and $m$ are constant.
$\therefore\text{W}_2=9.8\times20\times\Big(\frac{6370}{6370+160}\Big)^2=186.5\text{N}$
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Question 532 Marks
A mass $M_1$ revolves around another mass $M_2$ in a path of radius r, what is the angular momentum associated with $M_1$?
Answer
Angular momentum L is the product of twice the mass and aerial velocity. $\therefore\text{L}=2\text{M}\frac{\text{dA}}{\text{dt}}=2\text{M}_1\times\frac{\pi\text{r}^2}{\text{T}}$ $=2\text{M}_1\frac{\pi\text{r}^2\text{v}}{2\pi\text{r}}$ $=\text{M}_1\text{r}\sqrt{\frac{\text{GM}_2}{\text{r}}}$ $=\text{M}_1\sqrt{\text{GM}_2\text{r}}$
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Question 542 Marks
An astronaut, by mistake, drops his food packet from an artificial satellite orbiting around the Earth. Will it reach the surface of Earth? Why?
Answer
The food packet will not fall on the Earth. As the satellite as well as astronaut were in a state of weightlessness, hence the food packet, when dropped by mistake, will also start moving with the same velocity as that of satellite and will continue to move along with the satellite in the same orbit.
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Question 552 Marks
On what factor does the escape speed from a surface depend?
Answer
Value of escape speed at the surface of a planet is given by the relation, $\text{v}_\text{es}=\sqrt{\frac{2\text{GM}}{\text{R}}}=\sqrt{2\text{gR}}$ Thus, the value of escape speed from the surface of a planet depends upon (i) value of acceleration due to gravity gat the surface and (ii) the size (i.e. radius) R of the planet only. It is independent of all other factors. e.g. The mass and size of the body to be projected, angle of projection, etc.
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Question 562 Marks
Give one example each of central force and non - central force.
Answer
Gravitational force, electrostatic force due to point mass and point charges are the examples of central force. Spin dependent nuclear force, magnetic force between two current carrying loops are the examples of non - central forces.
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Question 572 Marks
The value of acceleration due to gravity at the moon is $\frac{1}{6}\text{th}$ of the value of g at the surface of the earth, and the diameter of the moon is $\frac{1}{4}\text{th}$ of the diameter of the earth. Compare the ratio of the escape velocities.
Answer
$\text{v}_{\text{e}}=\sqrt{\frac{2\text{GM}}{\text{R}}}=\sqrt{2\text{gR}}$ $=\sqrt{\text{gD}}$ $\frac{(\text{v}_{\text{e}})_{\text{moon}}}{(\text{v}_{\text{e}})_{\text{earth}}}=\frac{\sqrt{(\text{gD})_{\text{moon}}}}{\sqrt{(\text{gD})_{\text{earth}}}}$ $\sqrt{\frac{1}{6}\times\frac{1}{4}}=\frac{1}{4.9}$
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Question 582 Marks
A planet reduces its radius by 1% with its mass remaining same. How acceleration due to gravity varies?
Answer
When mass is same, $\text{g}\propto\frac{1}{\text{R}^2}.$$\therefore\frac{\Delta\text{g}}{\text{g}}=2\frac{\Delta\text{R}}{\text{R}}$
% variation of g is 2%.
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Question 592 Marks
A spacecraft consumes more fuel in going from earth to moon than it does on the return trip. Comment on this.
Answer
In going from earth to moon, the spacecraft has to do more work against the greater gravitational attraction of the earth. For the return journey, moon's gravitational force is much less, hence less work is done and less fuel is consumed.
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Question 602 Marks
What is the gravitational force on a body inside a spherical shell? Why is it so?
Answer
Inside a spherical shell, gravitational force is zero. Since there is no mass inside, the gravitational field and thereby the force is zero.
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Question 612 Marks
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?
Answer
Astronaut inside a small spaceship experience a very small negligible constant acceleration and hence astronaut feel weight less ness. If the space station has too much large mass and size then he can only experience acceleration due to gravity in the proximity of moon or on the moon.
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Question 622 Marks
Why a tennis ball bounces higher on hills than on plains?
Answer
As the acceleration due to gravity on hills is less than that on the surface of the earth (effect of height), therefore, a tennis ball bounces higher on hills than on plains.
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Question 632 Marks
What is the height at which the value of g is the same as at a depth of $\frac{\text{R}}{2}?$
Answer
At depth $\frac{\text{R}}{2},\text{g}'=\text{g}\Big(1-\frac{\text{R}}{2\text{R}}\Big)=\frac{\text{g}}{2}$At height $\text{x},\text{g}'=\text{g}\Big(1-\frac{2\text{x}}{\text{R}}\Big)$
$\therefore\text{g}\Big(1-\frac{2\text{x}}{\text{R}}\Big)=\frac{\text{g}}{2}$
$\frac{1}{2}=\frac{2\text{x}}{\text{R}}$
$\therefore\text{x}=\frac{\text{R}}{4}$
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Question 642 Marks
A body has a sense of weightlessness in a satellite revolving around the earth, why?
Answer
The astronauts and the satellite require the centripetal force to revolve around the earth. Their weight is used up in providing the necessary centripetal force. Hence an astronaut feels weightlessness in the space.
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Question 652 Marks
State Newton's law of gravitation in vector form.
Answer
The force of attraction between a pair of masses $m_1$ and $m_2$ separated by a length 'r' is given by. $\overrightarrow{\text{F}_{12}}=-\frac{\text{Gm}_1\text{m}_2}{\text{r}_2}\hat{\text{r}}_{21}$ $[\text{F}_{12}\rightarrow\text{force on }1\text{ due to }2].$ $\hat{\text{r}}\rightarrow\text{Points from }2\text{ to }1.$ -ve sign shows that the force is attractive.
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Question 662 Marks
The moon takes about 27 days to complete one orbit around the earth. The orbit is nearly a circle of radius $3.8 \times 10^8 \mathrm{~m}$. Calculate the mass of the earth from this data.
Answer
Let M and m be the mass of the earth and the moon respectively. The gravitational force of attraction provides the centripetal force. $\frac{\text{GMm}}{\text{r}^2}=\frac{\text{mv}^2}{\text{r}}$ $\Rightarrow\text{M}=\frac{\text{v}^2\text{r}}{\text{G}}=\frac{\omega^2\text{r}^3}{\text{G}}$ $=\Big(\frac{2\pi}{27\times24\times3600}\Big)^2\times\frac{(3.8\times10^8)^3}{6.67\times10^{-11}}$ $=5.968\times10^{24}\text{kg}$
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Question 672 Marks
A rocket is fired with a velocity 0.6 times the escape velocity on the surface of earth. How high will it go from the surface?
Answer
Velocity provided $= 0.6v_e$, $\text{v}_{\text{e}}=\sqrt{\frac{2\text{GM}}{\text{R}}}$ Applying conservation, $-\frac{\text{GMm}}{\text{R}}+\frac{1}{2}\text{m}(6\text{v}_{\text{e}})^2=-\frac{\text{GMm}}{\text{R}+\text{h}}$ Solve for h to get $\text{h}=\frac{72}{28}\text{R}.$
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Question 682 Marks
Answer the following: 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, Astronaut can hope to detect gravity if the size of th e spaceship is extremely large, then the magnitude of the gravity will become appreciable and hence the gravitational effect of the spaceship may been me measurable.
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Question 692 Marks
If a planet existed whose mass and radius were both half of those of the earth, what would be the value of the acceleration due to gravity on its surface as compared to what it is on the earth's surface?
Answer
We know, $\text{g}=\frac{\text{GM}}{\text{R}^2}\dots(1)$ $\text{g}'=\frac{\text{GM}'}{(\text{R}')^2}\Big[\text{M}'=\frac{1}{2}\text{M,}\text{R}'=\frac{1}{2}\text{R}\Big]\dots(2)$ Form (1) and (2), we have $\frac{\text{g}'}{\text{g}}=\frac{1}{2}2^2=2$ $\therefore\text{g}'=2\text{g}$
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Question 702 Marks
Show graphically how g varies as you move from the centre of earth to great heights above the surface.
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Question 712 Marks
A mass M is broken into two parts, m and ( M - m ). How is m related to M so that the gravitational force between two parts is maximum?
Answer
Let $m_1 = m, m_2 = M - m \text{F}=\text{G}=\frac{\text{m}(\text{M}-\text{M})}{\text{r}^2}=\frac{\text{G}}{\text{r}}(\text{Mm}-\text{m}^2)$
Differentiating w.r.t. $\text{m},\frac{\text{dF}}{\text{dm}}=\frac{\text{G}}{\text{r}^2}(\text{M})-2\text{m})$
For F to be maximum, $\frac{\text{dF}}{\text{dm}}=0$
$\frac{\text{G}}{\text{r}^2}(\text{M}-2\text{m})=0$
$\text{M}=2\text{m},$ or $\text{m}=\frac{\text{M}}{2}$
$\therefore\text{m}_1=\text{m}_2=\frac{\text{M}}{2}$
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Question 722 Marks
Two particles of equal mass 'm' go round a circle of radius R, under the action of their mutual gravitational attraction. What is the speed of each particle?
Answer

The gravitational force of attraction provides the necessary centripetal force.
$\frac{\text{mv}^2}{\text{R}}=\frac{\text{G}(\text{m})(\text{m})}{(2\text{R})^2}$
$\Rightarrow\text{v}^2=\frac{\text{Gm}}{4\text{R}}$
$\Rightarrow\text{v}=\frac{1}{2}\sqrt{\frac{\text{Gm}}{\text{R}}}$
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Question 732 Marks
A black hole is a body from whose surface nothing may ever escape. What is the condition for a uniform spherical mass M to be a black hole? What should be the radius of such a black hole if its mass is the same as that of the Earth?
Answer
For a body to be a black hole, the escape velocity should be such that even light cannot escape. The limiting case for escape velocity is, $\sqrt{\frac{2\text{GM}}{\text{R}}}\leq\text{c}(\text{speed of light})$ For our earth, $M = M_e = 6 \times 10^{24}kg$, $\text{R}=\frac{2\text{GM}}{\text{c}^2}=9\times10^{-2}\text{m or }9\text{cm}$
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Question 742 Marks
Does the concentration of the earth's mass near its centre change the variation of g with height compared with a homogeneous sphere, how?
Answer
Any change in the distribution of the earth's mass will not affect the variation of acceleration due to gravity with height. This is because for a point outside the earth, the whole mass of the earth is effective and the earth behaves as a homogeneous sphere.
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Question 752 Marks
If earth be at half of its present distance from the sun, how many days will there be in a year?
Answer
$\text{T}^2\propto\text{r}^3,\text{Since r}\rightarrow\frac{\text{r}}{2},\frac{\text{T}^2_1}{\text{T}^2}=\Big(\frac{1}{2}\Big)^3$ $\therefore\text{T}_1=\Big(\frac{1}{8}\Big)^{\frac{3}{2}}\text{T};\text{T}_1=\Big(\frac{1}{2\sqrt{2}}\Big)^3\text{T}$ $\text{i.e.}\text{T}_1=\frac{\text{T}}{16\sqrt{2}}$
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Question 762 Marks
Why are space crafts usually launched from west to east? Why is it more advantageous to launch rockets in the equatorial plane?
Answer
Earth rotates on its axis from west to east. A satellite launched from west to east will have the advantage of the additional velocity of the earth's rotation. The effect is maximum at the equator, hence it is most advantageous to launch the satellite from west to east on the equatorial plane.
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Question 772 Marks
Answer the following: 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. Electrical forces depend upon the nature of the intervening medium while the gravitational forces don’t depend upon the nature of the intervening medium. So, such shielding acts are not possible in case of gravitation i.e., gravity screens are not possible.
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Question 782 Marks
Answer the following: 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
Earth moon distance is very small as compared to earth-sun distance. Tidal effect is inversely proportional to the cube of the distance it means it is not governed by inverse square law like the gravitational force (which obeys inverse square law). Hence, tidal effect of moon is larger than that due to the sun.
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Question 792 Marks
Define gravitational potential at a point.
Answer
The work done in carrying unit mass from infinity to a point in gravitational field is gravitational potential. $\text{V}_{\text{g}}=-\frac{\text{GM}}{\text{r}}$ where r is the distance of the point from M.
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Question 802 Marks
Two satellites A and B go around a planet Pin circular orbits having radius 4R and R respectively. If the speed of the satellite A is 3V, find the speed of the satellite B.
Answer
$\text{As},\text{v}_{\text{o}}=\sqrt{\frac{\text{GM}}{\text{R}}}$ $\text{So},3\text{v}=\sqrt{\frac{\text{GM}}{4\text{R}}}\text{ and }\text{v}'=\sqrt{\frac{\text{GM}}{\text{R}}}$ $\therefore\frac{\text{v}'}{3\text{v}}=2\text{ or }\text{v}'=6\text{v}$
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Question 812 Marks
If you compare the gravitational force on the earth due to the sun to that of due to the moon, you would find that the sun's pull is greater than the moon's pull. However, the tidal effect of the moon's pull is greater than the tidal effect of the sun. Why?
Answer
Whereas the gravitational force depends inversely on the square of the distance, tidal effect depends inversely on the cube of the distance.
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Question 822 Marks
How is the gravitational force between two point masses affected when they are dipped in water keeping the separation between them the same?
Answer
The Newton’s Universal law of gravitational force of attraction (F) between two bodies of masses $\mathrm{m}_1, \mathrm{~m}_2$ separated by distance r is $\text{F}=\frac{\text{Gm}_1\text{m}_2}{\text{r}^2}$ G does not depend upon the medium. So force of attraction does not change if the masses are kept in water or any other medium.
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Question 832 Marks
Objects at rest on the earth's surface move in circular paths with a period of 24 hours. Are they in orbit in the same sense that an earth's satellite is in orbit? Why not? What would the length of the day have to be to put such objects in a true orbit?
Answer
The objects on the earth's surface are not in orbital motion w.r.t. the earth. In order that an object has an orbital motion close to the earth's surface, its orbital velocity, and period of motion must be, $\text{v}_{\text{o}}=\sqrt{\text{gR}};$ $\text{T}=\frac{2\pi\text{R}}{\text{v}_{\text{o}}}=2\pi\sqrt{\frac{\text{R}}{\text{g}}}=1.4\text{hrs.}$ Therefore, the length of the day has to be 1.4hrs in case the objects on the earth's surface are in the true orbital motion like that of the earth satellite.
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Question 842 Marks
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
In case the size of the space-station becomes large, the forces of gravity will become appreciable, and the astronaut can hope to detect it.
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Question 852 Marks
Shown are several curves. Explain with reason, which ones amongst them can be possible trajectories traced by a projectile (neglect air friction).
Answer
The trajectory of a projectile under gravitational force of earth is a conic section or parabolic or elliptical or its part whose focus must be the centre of the earth. Only (c) option amongst the given curves show the centre of the earth as the focus of trajectory.
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Question 862 Marks
Why do different planets have different escape velocities?
Answer
Escape velocity, $\text{v}=\sqrt{2\text{gR}}=\sqrt{\frac{2\text{GM}}{\text{R}}}.$ Thus escape velocity of a planet depends upon (i) its mass (M) and (ii) its size (R). As different planets have different masses and sizes, so they have different escape velocities.
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