NCERT Exemplar Solutions Class 11 Physics Gravitation — Very Short, Short and Long Answer Questions Q17–38

NCERT Exemplar Problems, Class 11 Physics — Gravitation, Very Short Answer Questions 17 to 27, Short Answer Questions 28 to 33 and Long Answer Questions 34 to 38. All 22 questions solved. The book numbers them 8.17 to 8.38.

Three ideas carry this page. A central force conserves angular momentum, so the areal velocity of a planet is constant, and that alone tells us where it moves fastest. For a circular orbit gravity supplies the centripetal force, which fixes $v = \sqrt{GM/r}$ and makes the kinetic energy half the size of the potential energy, $K = GMm/2r$ and $U = -GMm/r$. Energy and angular momentum together settle an elliptical orbit: the speeds at its nearest and farthest points come from those two conservation laws.

Key insight. When a question asks how something changes along an orbit, write it in terms of $r$ using $v = \sqrt{GM/r}$ or $r v = \text{constant}$ and read the answer off the power of $r$. Questions 26, 30, 34 and 38 all go this way.

Very Short Answer Questions

Question 17

Molecules in air in the atmosphere are attracted by gravitational force of the earth. Explain why all of them do not fall into the earth just like an apple falling from a tree.

Solution. Every molecule does feel the downward pull, just like the apple. The difference is that air molecules are not at rest: they have thermal speeds of hundreds of metres per second in random directions, and they keep colliding. Gravity only biases this random motion slightly downwards, so the molecules settle into a balance in which the air is densest near the ground and thins out with height. An apple has no such random motion to oppose its weight, so it simply falls.

Question 18

Give one example each of central force and non-central force.

Solution. A central force acts along the line joining the body to a fixed centre, and its size depends only on the distance. The gravitational force of a point mass and the electrostatic force of a point charge are both of this kind. A non-central force does not point along that line: the magnetic force between two current-carrying loops, or the spin-dependent nuclear force between two nucleons, are examples.

Question 19

Draw areal velocity versus time graph for mars.

Solution. Gravity on Mars from the sun is a central force, so Mars’s angular momentum about the sun is constant. The areal velocity is $\dfrac{dA}{dt} = \dfrac{L}{2m}$, which is therefore also constant (Kepler’s second law). Its graph against time is a horizontal straight line above the time axis.

Question 20

What is the direction of areal velocity of the earth around the sun?

Solution. In a small time $\Delta t$ the radius vector sweeps the area $\Delta \mathbf{A} = \tfrac12\,\mathbf{r}\times\mathbf{v}\,\Delta t$, so the areal velocity is the vector $\tfrac12\,\mathbf{r}\times\mathbf{v}$. A cross product is perpendicular to both vectors, so it points normal to the plane containing the sun and the earth’s orbit, in the direction given by the right-hand rule for the earth’s sense of revolution.

Question 21

How is the gravitational force between two point masses affected when they are dipped in water keeping the separation between them the same?

Solution. Unlike the electric force, the gravitational force $Gm_1m_2/r^2$ contains no property of the medium between the masses. Water between them changes nothing, so the force is unchanged. (Water does add buoyancy and its own small attraction to each mass, but the force between the two masses is the same.)

Question 22

Is it possible for a body to have inertia but no weight?

Solution. Yes. Inertia depends on mass, which a body always has. Weight is the gravitational force on it, which depends on where it is. At the centre of the earth the pulls of the surrounding matter cancel, so the body has no weight but still has its full mass and inertia. The same is true far out in space, away from all attracting bodies.

Question 23

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?

Solution. No. Electric shielding works because a conductor contains positive and negative charges that rearrange themselves to cancel the outside field inside. There is no negative mass, so nothing can rearrange to cancel a gravitational field. A hollow sphere produces no field of its own inside, but the fields of all other masses pass through it unchanged, so gravity cannot be screened.

Question 24

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?

Solution. Yes, in principle. The station as a whole is in free fall, so at its centre of mass the earth’s pull just supplies the centripetal force and nothing seems to be pulling. But $g$ changes with distance from the earth, so a point on the earth side of a large station is pulled slightly more, and a point on the far side slightly less, than the station needs. If the station is large enough for this variation in $g$ to be measurable, the astronaut can detect gravity through it.

Question 25

The gravitational force between a hollow spherical shell (of radius $R$ and uniform density) and a point mass is $F$. Show the nature of $F$ vs $r$ graph where $r$ is the distance of the point from the centre of the hollow spherical shell of uniform density.

Solution. Inside a uniform shell the pulls of the different parts cancel exactly, so $F = 0$ for every $r < R$. Outside, the shell acts as if all its mass $M$ were at its centre, so $F = GMm/r^2$ for $r > R$. The graph is therefore zero from the centre up to $r = R$, jumps to its largest value $GMm/R^2$ at the surface, and then falls off as $1/r^2$.

Question 26

Out of aphelion and perihelion, where is the speed of the earth more and why?

Solution. The areal velocity $\tfrac12 r v_\perp$ is constant. At perihelion and aphelion the velocity is perpendicular to the radius, so $r v$ is the same at both. At perihelion $r$ is smallest, so $v$ must be largest: the earth has to cover a greater distance along its orbit in the same time to sweep the same area.

Question 27

What is the angle between the equatorial plane and the orbital plane of (a) Polar satellite? (b) Geostationary satellite?

Solution. A polar satellite passes over both poles, so its orbit lies in a plane containing the earth’s axis, perpendicular to the equator. A geostationary satellite has to stay over one point on the equator, so its orbit lies in the equatorial plane itself.

Short Answer Questions

Question 28

Mean solar day is the time interval between two successive noon when sun passes through zenith point (meridian). Sidereal day is the time interval between two successive transit of a distant star through the zenith point (meridian). By drawing appropriate diagram showing earth’s spin and orbital motion, show that mean solar day is four minutes longer than the sidereal day. In other words, distant stars would rise 4 minutes early every successive day. (Hint: you may assume circular orbit for the earth.)

Solution. Draw the sun at the centre of the earth’s circular orbit and the earth at two positions one day apart, with a fixed point P on the earth facing the sun at noon on the first day. The earth spins and revolves in the same sense.

• Sidereal day. A distant star is so far away that the direction to it is the same from both positions. P faces it again after the earth has turned through exactly $360^\circ$.
• Solar day. In one day the earth moves along its orbit by $360^\circ/365.25 \approx 1^\circ$, so the direction to the sun has turned by about $1^\circ$. For P to face the sun again, the earth must turn through about $361^\circ$.

The extra $1^\circ$ of spin is what makes the solar day longer. Since the earth turns $361^\circ$ in the $24$ hours of a solar day, $1^\circ$ takes $$\frac{24 \times 60}{361}\ \text{min} \approx 3.99\ \text{min} \approx 4\ \text{min}.$$ So the sidereal day is about $23$ h $56$ min, and a given star crosses the meridian (and rises) about $4$ minutes earlier each day.

Question 29

Two identical heavy spheres are separated by a distance 10 times their radius. Will an object placed at the mid point of the line joining their centres be in stable equilibrium or unstable equilibrium? Give reason for your answer.

Solution. Let each sphere have mass $M$ and radius $R$; the centres are $10R$ apart, so the mid point is $5R$ from each. There the two pulls are equal and opposite, so the object is in equilibrium. Now move it a small distance $x$ towards the right-hand sphere. The forces on it are $$\frac{GMm}{(5R – x)^2}\ \text{to the right}, \qquad \frac{GMm}{(5R + x)^2}\ \text{to the left}.$$ The first is larger, so the net force points to the right, away from the mid point, and it grows as $x$ grows. The object keeps moving away, so the equilibrium is unstable. (A sideways displacement, perpendicular to the line, would be pulled back; but an equilibrium that fails for any one direction of displacement is unstable.)

Question 30

Show the nature of the following graph for a satellite orbiting the earth. (a) KE vs orbital radius $R$ (b) PE vs orbital radius $R$ (c) TE vs orbital radius $R$.

Solution. For a circular orbit gravity supplies the centripetal force: $\dfrac{mv^2}{R} = \dfrac{GMm}{R^2}$, so $mv^2 = GMm/R$. Hence $$K = \frac{GMm}{2R}, \qquad U = -\frac{GMm}{R}, \qquad E = K + U = -\frac{GMm}{2R}.$$ All three vary as $1/R$. The KE curve is positive and falls towards zero; the PE curve is negative, twice as deep as the TE curve, and rises towards zero; the TE curve is the mirror image of the KE curve in the $R$ axis.

Question 31

Shown are several curves (Fig. 8.2). Explain with reason, which ones amongst them can be possible trajectories traced by a projectile (neglect air friction).

The figure has six sketches around the earth. (a) A flattened loop leaving the earth whose axis is parallel to, but below, the line through the earth’s centre. (b) A flattened loop leaving the upper right of the earth, tilted, its axis well clear of the centre. (c) An elongated ellipse leaving the earth towards the upper right, whose long axis, extended, passes through the earth’s centre. (d) A spiral winding outwards around the earth. (e) An elongated loop leaving the bottom of the earth, its axis off the centre. (f) A figure-of-eight style curve with loops on either side of the earth.

Solution. Outside the earth a projectile moves under the earth’s gravity alone, which is an inverse-square central force. Its path is therefore a conic section, here part of an ellipse, with the centre of the earth at one focus. A focus lies on the major axis, so the long axis of the ellipse must pass through the earth’s centre. Only curve (c) does this. The loops in (a), (b) and (e) are ellipse-like, but their axes miss the centre; (d) is a spiral, which would need energy to be lost or gained; and (f) is not a conic at all.

Question 32

An object of mass $m$ is raised from the surface of the earth to a height equal to the radius of the earth, that is, taken from a distance $R$ to $2R$ from the centre of the earth. What is the gain in its potential energy?

Solution. The potential energy at distance $r$ is $-GMm/r$, so $$\Delta U = \left(-\frac{GMm}{2R}\right) – \left(-\frac{GMm}{R}\right) = \frac{GMm}{2R}.$$ With $GM = gR^2$ this is $\dfrac{mgR}{2}$. Note that $mgh$ with $h = R$ would give twice this; $mgh$ only works for heights small compared with $R$, because $g$ falls as we go up.

Question 33

A mass $m$ is placed at P a distance $h$ along the normal through the centre O of a thin circular ring of mass $M$ and radius $r$ (Fig. 8.3). If the mass is removed further away such that OP becomes $2h$, by what factor the force of gravitation will decrease, if $h = r$?

The figure shows the ring in a vertical plane with centre O, radius $r$ marked, and the point P on the axis through O at distance $h$ from O.

Solution. Every element of the ring is at the same distance $\sqrt{r^2 + h^2}$ from P. By symmetry the components of their pulls perpendicular to the axis cancel in pairs, and only the components along PO survive. Each is reduced by the factor $h/\sqrt{r^2+h^2}$, so $$F(h) = \frac{GMm\,h}{(r^2 + h^2)^{3/2}}.$$ With $h = r$: $F_1 = \dfrac{GMm\,r}{(2r^2)^{3/2}} = \dfrac{GMm}{2\sqrt2\,r^2}$.

With $OP = 2h = 2r$: $F_2 = \dfrac{GMm\,(2r)}{(5r^2)^{3/2}} = \dfrac{2GMm}{5\sqrt5\,r^2}$.

So $$\frac{F_2}{F_1} = \frac{2}{5\sqrt5}\times 2\sqrt2 = \frac{4\sqrt2}{5\sqrt5} \approx 0.51,$$ and the force decreases by a factor of $\dfrac{5\sqrt5}{4\sqrt2} \approx 1.98$, not by $4$ as a point-mass inverse-square law would suggest.

Long Answer Questions

Question 34

A star like the sun has several bodies moving around it at different distances. Consider that all of them are moving in circular orbits. Let $r$ be the distance of the body from the centre of the star and let its linear velocity be $v$, angular velocity $\omega$, kinetic energy $K$, gravitational potential energy $U$, total energy $E$ and angular momentum $l$. As the radius $r$ of the orbit increases, determine which of the above quantities increase and which ones decrease.

Solution. For a circular orbit, $\dfrac{mv^2}{r} = \dfrac{GMm}{r^2}$, so $v = \sqrt{GM/r}$. Write every quantity in terms of $r$:

Quantity	In terms of $r$	As $r$ increases
$v$	$\sqrt{GM/r} \propto r^{-1/2}$	decreases
$\omega = v/r$	$\sqrt{GM/r^3} \propto r^{-3/2}$	decreases
$K = \tfrac12 mv^2$	$GMm/2r \propto r^{-1}$	decreases
$U$	$-GMm/r$	increases (becomes less negative)
$E = K + U$	$-GMm/2r$	increases (becomes less negative)
$l = mvr$	$m\sqrt{GMr} \propto r^{1/2}$	increases

The energies need care with the sign: $U$ and $E$ are negative, and as $r$ grows they move towards zero, which is an increase. The angular momentum grows because $r$ rises faster than $v$ falls.

Question 35

Six point masses of mass $m$ each are at the vertices of a regular hexagon of side $l$. Calculate the force on any of the masses.

Solution. Take the mass at vertex A of the hexagon ABCDEF. By symmetry the net force on it points along AD, towards the centre, so we add the components along AD. The other five masses are at three distances from A:

• B and F, the neighbours, at distance $l$; AB and AF make $60^\circ$ with AD. Together they give $2\cdot\dfrac{Gm^2}{l^2}\cos60^\circ = \dfrac{Gm^2}{l^2}$.
• C and E at distance $\sqrt3\,l$; AC and AE make $30^\circ$ with AD. Together they give $2\cdot\dfrac{Gm^2}{3l^2}\cos30^\circ = \dfrac{Gm^2}{\sqrt3\,l^2}$.
• D, the opposite vertex, at distance $2l$ along AD, giving $\dfrac{Gm^2}{4l^2}$.

Adding, $$F = \frac{Gm^2}{l^2}\left(1 + \frac{1}{\sqrt3} + \frac14\right) \approx 1.83\,\frac{Gm^2}{l^2},$$ directed from the mass towards the centre of the hexagon.

Question 36

A satellite is to be placed in equatorial geostationary orbit around earth for communication. (a) Calculate height of such a satellite. (b) Find out the minimum number of satellites that are needed to cover entire earth so that at least one satellites is visible from any point on the equator. [$M = 6 \times 10^{24}$ kg, $R = 6400$ km, $T = 24$ h, $G = 6.67 \times 10^{-11}$ SI units]

Solution. (a) A geostationary satellite has $T = 24$ h. Gravity supplies the centripetal force, $\dfrac{GMm}{r^2} = m\left(\dfrac{2\pi}{T}\right)^2 r$, so $$r = \left(\frac{GMT^2}{4\pi^2}\right)^{1/3} = \left(\frac{6.67\times10^{-11}\times 6\times10^{24}\times (86400)^2}{4\pi^2}\right)^{1/3} \approx 4.23\times10^{7}\ \text{m}.$$ The height above the surface is $h = r – R = 4.23\times10^7 – 6.4\times10^6 \approx 3.59\times10^{7}$ m, about $35\,900$ km.

(b) From the satellite, the farthest points of the equator it can see are where its line of sight just grazes the earth. At such a point the radius is perpendicular to the line of sight, so the angle $\theta$ at the earth’s centre between the satellite and that point satisfies $$\cos\theta = \frac{R}{r} = \frac{1}{1 + h/R} = \frac{1}{1 + 5.61} \Rightarrow \theta \approx 81.3^\circ.$$ One satellite therefore covers an arc of $2\theta \approx 162.6^\circ$ of the equator. Since $360^\circ/162.6^\circ \approx 2.21$, two satellites are not enough and three are.

Question 37

Earth’s orbit is an ellipse with eccentricity $0.0167$. Thus, earth’s distance from the sun and speed as it moves around the sun varies from day to day. This means that the length of the solar day is not constant through the year. Assume that earth’s spin axis is normal to its orbital plane and find out the length of the shortest and the longest day. A day should be taken from noon to noon. Does this explain variation of length of the day during the year?

Solution. Angular speed along the orbit. Angular momentum is conserved, so $r^2\omega$ is constant. At perihelion $r_p = a(1-e)$ and at aphelion $r_a = a(1+e)$, so $$\frac{\omega_p}{\omega_a} = \left(\frac{1+e}{1-e}\right)^2 = \left(\frac{1.0167}{0.9833}\right)^2 \approx 1.0691.$$ Take the mean angular speed $\omega$, which corresponds to the mean solar day, as the geometric mean of $\omega_p$ and $\omega_a$. Then $\dfrac{\omega_p}{\omega} = \dfrac{\omega}{\omega_a} = \sqrt{1.0691} \approx 1.034$.

Length of the day. On average the earth moves about $1^\circ$ along its orbit each day, so it spins $361^\circ$ from noon to noon, and $361^\circ$ takes $24$ h. Near perihelion it moves $1.034^\circ$ a day, so a solar day needs $361.034^\circ$ of spin; near aphelion it moves $0.967^\circ$, so it needs $360.967^\circ$. Since $1^\circ$ of spin takes $\dfrac{24\times3600}{361} \approx 239$ s, $$\text{longest day} \approx 24\ \text{h} + 0.034 \times 239\ \text{s} \approx 24\ \text{h}\ 8\ \text{s},$$ $$\text{shortest day} \approx 24\ \text{h} – 0.033 \times 239\ \text{s} \approx 23\ \text{h}\ 59\ \text{min}\ 52\ \text{s}.$$ The longest solar day falls near perihelion (early January) and the shortest near aphelion (early July).

Does it explain the seasons’ day lengths? No. These changes are only about $8$ seconds either way, whereas the hours of daylight change by several hours over the year. That large variation comes from the tilt of the earth’s spin axis to its orbital plane, which this question has assumed away.

Question 38

A satellite is in an elliptic orbit around the earth with aphelion of $6R$ and perihelion of $2R$ where $R = 6400$ km is the radius of the earth. Find eccentricity of the orbit. Find the velocity of the satellite at apogee and perigee. What should be done if this satellite has to be transferred to a circular orbit of radius $6R$? [$G = 6.67 \times 10^{-11}$ SI units and $M = 6 \times 10^{24}$ kg]

Solution. Eccentricity. $r_a = a(1+e) = 6R$ and $r_p = a(1-e) = 2R$. Dividing, $\dfrac{1+e}{1-e} = 3$, so $e = \dfrac12$ (and $a = 4R$).

Speeds. Angular momentum is conserved, and at both ends the velocity is perpendicular to the radius, so $v_p r_p = v_a r_a$, giving $v_a = v_p/3$. Energy is conserved: $$\tfrac12 v_p^2 – \frac{GM}{2R} = \tfrac12 v_a^2 – \frac{GM}{6R}.$$ Substituting $v_a = v_p/3$: $\tfrac12 v_p^2\left(1 – \tfrac19\right) = GM\left(\dfrac{1}{2R} – \dfrac{1}{6R}\right) = \dfrac{GM}{3R}$, so $$v_p = \sqrt{\frac{3GM}{4R}} = \sqrt{\frac{3\times6.67\times10^{-11}\times6\times10^{24}}{4\times6.4\times10^{6}}} \approx 6.85\ \text{km/s}, \qquad v_a = \frac{v_p}{3} \approx 2.28\ \text{km/s}.$$

Transfer to a circular orbit of radius $6R$. A circular orbit at $6R$ needs $v_c = \sqrt{\dfrac{GM}{6R}} \approx 3.23$ km/s. At apogee the satellite is already at $6R$ and moving perpendicular to the radius, but at only $2.28$ km/s, too slow to stay at that distance. So when it reaches apogee, rockets should be fired along its direction of motion to raise its speed by $3.23 – 2.28 \approx 0.95$ km/s.

Common mistakes

• Thinking gravity can be screened in Question 23. A hollow shell has no field of its own inside, but it cannot cancel the fields of other masses.
• Saying the areal velocity lies along the orbit in Question 20. It is $\tfrac12\,\mathbf{r}\times\mathbf{v}$, perpendicular to the orbital plane.
• Drawing $F$ rising smoothly from zero in Question 25. For a thin shell the force is zero right up to $r = R$ and jumps there.
• Using $mgh$ for a height of one earth radius in Question 32. $g$ falls with height; the correct gain is $mgR/2$, half of $mgR$.
• Using the inverse-square law for the ring in Question 33. Near a ring the force is $GMmh/(r^2+h^2)^{3/2}$, so doubling $h$ does not quarter it.
• Calling $E$ “decreasing” in Question 34 because $|E|$ falls. $E$ is negative and moves towards zero, which is an increase.
• Forgetting the $\cos$ factors in Question 35. Only the components along the line to the centre add; the others cancel in pairs.
• Firing the rockets at perigee in Question 38. The orbit must be circularised at $6R$, which is where the satellite is at apogee.

Practise next

• Exemplar Gravitation: Multiple Choice Questions — Questions 1 to 16, on Kepler’s laws, the variation of $g$ and what changes if the law of gravitation is altered.
• NCERT textbook Chapter 7 — Gravitation for the theory of gravitational potential energy, escape speed and orbiting satellites.
• A planet has twice the mass of the earth and twice its radius. If the escape speed from the earth is $11.2$ km/s, what is the escape speed from the planet?

Show answer

$11.2$ km/s. Escape speed is $\sqrt{2GM/R}$; doubling both $M$ and $R$ leaves $M/R$, and hence the escape speed, unchanged.

• A satellite orbits the earth in a circle at a height equal to the earth’s radius. Taking $g = 9.8$ m/s$^2$ and $R = 6400$ km, find its period.

Show answer

About $1.44 \times 10^{4}$ s, i.e. $4.0$ h. The orbital radius is $2R$, so $T = 2\pi\sqrt{(2R)^3/GM} = 2\pi\sqrt{8R/g} = 2\pi\sqrt{8\times6.4\times10^6/9.8} \approx 14\,360$ s.