NCERT Exemplar Solutions Class 11 Physics Gravitation — Multiple Choice Questions Q1–16

NCERT Exemplar Problems, Class 11 Physics — Gravitation, MCQ I and MCQ II. All 16 questions solved.

The book numbers these questions 8.1 to 8.16; here they are Questions 1 to 16. Questions 1 to 8 are MCQ I (exactly one option is correct) and Questions 9 to 16 are MCQ II (one or more options may be correct). Four ideas carry the whole page:

• The inverse-square law is special. Only a force falling off as $1/r^2$ gives closed elliptical orbits and a field-free interior inside a uniform shell.
• Kepler’s laws have different sources. The second law needs only a central force; the first needs an inverse-square central force; the third needs the force per unit mass to be the same for every planet.
• Equality of gravitational and inertial mass is an experimental fact, and it is why every body falls with the same acceleration.
• Real orbits are perturbed. Air drag, a second attracting body or a slowly changing $G$ all spoil the perfect ellipse.

Key insight. In most of these questions, ask what the force law actually is (central? inverse-square? proportional to the mass being moved?) and the right options follow. Questions 10, 12 and 14 are built entirely on that.

Choose the correct answer from the given options. (MCQ I: one correct. MCQ II: one or more correct.)

Question 1

The earth is an approximate sphere. If the interior contained matter which is not of the same density everywhere, then on the surface of the earth, the acceleration due to gravity

Solution. If the density varies from place to place without spherical symmetry, the earth no longer acts like a point mass at its centre. The field at the surface then varies in magnitude, and its direction need not point exactly at the centre, so (a), (b) and (c) all claim more than we can say. What we can say is that a point on the surface has the whole earth on one side of it: every bit of matter pulls it roughly inwards, and these pulls cannot cancel. So $g$ cannot be zero anywhere on the surface.

Question 2

As observed from earth, the sun appears to move in an approximate circular orbit. For the motion of another planet like mercury as observed from earth, this would

Solution. The sun appears to go round the earth only because the earth goes round the sun; the relative motion of two bodies bound to each other is a simple orbit. Mercury is not bound to the earth. Its motion is governed by the sun, and seen from a moving earth it traces a complicated path with loops (apparent retrograde motion), not a circle. The earth–mercury force does follow the inverse-square law, which rules out (b), and no non-gravitational force matters, which rules out (d).

Question 3

Different points in earth are at slightly different distances from the sun and hence experience different forces due to gravitation. For a rigid body, we know that if various forces act at various points in it, the resultant motion is as if a net force acts on the c.m. (centre of mass) causing translation and a net torque at the c.m. causing rotation around an axis through the c.m. For the earth-sun system (approximating the earth as a uniform density sphere)

Solution. A uniform sphere attracts an outside body as if all its mass were at its centre, so the force of the earth on the sun acts along the line joining the two centres. By Newton’s third law the total force of the sun on the earth acts along the same line, which passes through the earth’s centre of mass. A force through the c.m. has no moment about it, so the torque is zero. The earth’s spin is a leftover from its formation, not something the sun’s pull produces, and its orbital motion is caused by the net force, not by a torque.

Question 4

Satellites orbiting the earth have finite life and sometimes debris of satellites fall to the earth. This is because,

Solution. Gravitation alone gives a closed orbit that repeats for ever, so (b) is wrong, and a satellite needs no power to stay in orbit, so (a) is irrelevant. In a low orbit there is still a trace of atmosphere. Its drag removes mechanical energy, the orbit shrinks, and the satellite drops into denser air, which speeds the process until it falls. Collisions do happen but are rare; they are not the general reason.

Question 5

Both earth and moon are subject to the gravitational force of the sun. As observed from the sun, the orbit of the moon

Solution. An exact ellipse about the sun needs a force always directed towards the sun. The moon feels the sun’s pull and also the earth’s pull, which points towards the earth, not the sun. Their sum is not a central force about the sun, so the moon’s path seen from the sun is not strictly an ellipse. Nothing forces it to be a closed curve either, which rules out (c), and the other planets are far too distant to cause a considerable deviation, which rules out (d).

Question 6

In our solar system, the inter-planetary region has chunks of matter (much smaller in size compared to planets) called asteroids. They

Solution. The sun’s pull on any body is proportional to that body’s mass, so the acceleration it produces, $GM_s/r^2$, does not depend on the body’s mass at all. A small asteroid is therefore governed by exactly the same inverse-square central force per unit mass as a planet. It moves in an ellipse with the sun at a focus, sweeps equal areas in equal times and obeys $T^2 \propto a^3$.

Question 7

Choose the wrong option.

Solution. Statements (a), (b) and (c) are the standard facts. (a) defines the two masses. (b) is right: their equality has been tested to very high precision but is not derived from anything more basic in Newtonian mechanics. (c) follows directly: $g = F/m_i = (GM/R^2)(m_g/m_i)$ is the same for all bodies only because $m_g = m_i$. Statement (d) is wrong. Gravitational mass is an intrinsic property of the particle; neighbouring objects change the force on it, not its mass. The question asks for the wrong option.

Question 8

Particles of masses $2M$, $m$ and $M$ are respectively at points A, B and C with $AB = \tfrac{1}{2}(BC)$. $m$ is much-much smaller than $M$ and at time $t = 0$, they are all at rest (Fig. 8.1). At subsequent times before any collision takes place:

The figure shows A, B and C on a straight line in that order, with $2M$ at A, $m$ at B and $M$ at C, B being nearer A.

Solution. Let $AB = d$, so $BC = 2d$. The pull of $2M$ on $m$ is $\dfrac{G(2M)m}{d^2}$, towards A. The pull of $M$ is $\dfrac{GMm}{(2d)^2} = \dfrac{GMm}{4d^2}$, towards C. The first is eight times the second, so the net force on $m$ is towards A. Because $m$ is tiny it accelerates far faster than the heavy masses, and the force on it only grows as it approaches A, so it moves steadily towards $2M$.

Question 9

Which of the following options are correct?

Solution. Above the surface $g = GM/(R+h)^2$, which falls as $h$ grows: (a) is true. Inside a uniform earth only the mass within radius $R-d$ attracts, giving $g = g_0(1 – d/R)$, which decreases with depth: (b) is false. The earth’s rotation reduces the effective $g$ by $\omega^2R\cos^2\lambda$, which is largest at the equator and zero at the poles, and the earth is also flatter at the poles; both make $g$ grow with latitude $\lambda$: (c) is true. Since $g = GM/R^2$, it plainly depends on the earth’s mass: (d) is false.

Question 10

If the law of gravitation, instead of being inverse-square law, becomes an inverse-cube law-

Solution. Closed elliptical orbits with the sun at a focus come only from an inverse-square force, so with an inverse-cube law they are lost: (a) is true. A circular orbit needs only that the attraction supply the centripetal force, $mv^2/r = k/r^3$, which can be met at any radius by the right speed, so circular orbits remain possible (though unstable): (b) is false. Over the few metres of a thrown stone the force hardly changes, so it is still a nearly constant downward force and the path is nearly a parabola: (c) is true. The cancellation inside a uniform shell depends on the force falling exactly as $1/r^2$, matching the way the area of a patch of shell grows as $r^2$; with $1/r^3$ the nearer side wins and the force inside is not zero: (d) is false.

Question 11

If the mass of sun were ten times smaller and gravitational constant G were ten times larger in magnitudes-

Solution. The mass of the sun does not enter $g$ on earth; $g = GM_e/R^2$, so making $G$ ten times larger makes $g$ ten times larger, and (b) is false. Our weight becomes ten times larger, so lifting our legs and supporting ourselves while walking is far harder: (a). A raindrop’s terminal speed is reached when drag balances weight; heavier weight means a higher terminal speed: (c). An aeroplane’s lift must balance its weight, and lift grows with speed, so it must fly much faster: (d).

Question 12

If the sun and the planets carried huge amounts of opposite charges,

Solution. The force on a planet becomes $\dfrac{GM_sm}{r^2} + \dfrac{kQ_sq}{r^2}$: still attractive, still central, still inverse-square. A central force conserves angular momentum, so equal areas are swept in equal times: the second law holds. An inverse-square central attraction gives elliptical orbits: the first law holds.

The third law is different. For a circular orbit $\dfrac{mv^2}{r} = \dfrac{GM_sm + kQ_sq}{r^2}$, which gives $$T^2 = \frac{4\pi^2 r^3}{GM_s + kQ_s q/m}.$$ The constant now contains $q/m$, which differs from planet to planet, so $T^2/r^3$ is no longer the same for all planets. The third law fails, which rules out (a) and (b).

Question 13

There have been suggestions that the value of the gravitational constant G becomes smaller when considered over very large time period (in billions of years) in the future. If that happens, for our earth,

Solution. The force is still central, so angular momentum is conserved, but the attraction weakens slowly. With the same angular momentum and less pull, the earth drifts outwards a little on each revolution, so its path is a slowly opening spiral rather than a closed orbit: (c). As $G$ keeps shrinking, the binding energy $-GM_sm/2r$ heads towards zero, and in the long run the earth is no longer bound: (d). Moving further from the sun would make us colder, not hotter, so (b) is false.

Question 14

Supposing Newton’s law of gravitation for gravitation forces $\mathbf{F}_1$ and $\mathbf{F}_2$ between two masses $m_1$ and $m_2$ at positions $\mathbf{r}_1$ and $\mathbf{r}_2$ read $$\mathbf{F}_1 = -\mathbf{F}_2 = -\frac{\mathbf{r}_{12}}{r_{12}^3}\,GM_0^2\left(\frac{m_1m_2}{M_0^2}\right)^n$$ where $M_0$ is a constant of dimension of mass, $\mathbf{r}_{12} = \mathbf{r}_1 – \mathbf{r}_2$ and $n$ is a number. In such a case,

Solution. The force is still central and inverse-square, but its strength goes as $(m_1m_2)^n$ instead of $m_1m_2$.

• (a) The acceleration of a body of mass $m$ near the earth is $F/m \propto m^{n}/m = m^{n-1}$, which depends on $m$ unless $n = 1$. True.
• (b), (c) An inverse-square central force still gives ellipses and equal areas, so the first and second laws hold. For a circular orbit, $T^2 \propto r^3/m^{n-1}$, so $T^2/r^3$ differs from planet to planet and the third law fails. So (c) is true and (b) false.
• (d) Take a body of mass $m$ and the water it displaces, of mass $m_w$. Its weight goes as $m^n$ and the buoyant force, the weight of the displaced water, as $m_w^n$. If the body is lighter than water, $m < m_w$; with $n$ negative, $m^n > m_w^n$, so the weight exceeds the upthrust and the body sinks. True.

Question 15

Which of the following are true?

Solution. A polar satellite’s orbit passes over (or near) both poles, so it travels north–south while the earth turns beneath it: (a) true, (d) false. A geostationary satellite must keep pace with the earth’s rotation, which is from west to east, so it orbits in the equatorial plane from west to east with a period of 24 hours: (c) true, (b) false.

Question 16

The centre of mass of an extended body on the surface of the earth and its centre of gravity

Solution. The centre of gravity coincides with the centre of mass only where $g$ is the same at every part of the body. For a body up to about $100$ m across, $g$ varies by a tiny fraction (it changes by about $3\times10^{-5}$ of itself over $100$ m of height), so the two points are very close: (d). For a large body $g$ varies appreciably, so (a) fails; and even a sphere’s two points need not coincide in a non-uniform field, so (b) fails. In a uniform field they coincide exactly, so (c) fails. Option (e) says both change: the centre of gravity can shift as the field over the body changes, but the centre of mass depends only on how the body’s mass is arranged, and moving the body does not change that. So (e) is false.

Common mistakes

• Saying $g$ points at the centre in Question 1. That needs spherical symmetry, which a non-uniform interior breaks.
• Ticking “the torque causes the earth to spin” in Question 3. The sun’s pull on a uniform sphere passes through its centre; the spin is primordial.
• Blaming gravity for falling satellites in Question 4. Gravity alone gives a closed orbit; it is air drag that brings them down.
• Forgetting that the distance is $2d$, not $d$, for $M$ in Question 8. The near heavy mass wins by a factor of eight.
• Saying $g$ increases with depth in Question 9. Inside a uniform earth it falls linearly to zero at the centre.
• Ruling out circular orbits under an inverse-cube law in Question 10. They exist, though they are unstable.
• Ticking “all three laws still hold” in Question 12. The extra electric force spoils the planet-independence of $T^2/r^3$.
• Ticking (e) in Question 16. The centre of mass is fixed by the mass distribution and does not move with the body’s surroundings.

Practise next

• Exemplar Gravitation: Very Short, Short and Long Answer Questions — Questions 17 to 38, on areal velocity, satellite energies, geostationary orbits and the length of the day.
• NCERT textbook Chapter 7 — Gravitation for the theory of Kepler’s laws, the shell theorem and the variation of $g$.
• At what height above the earth’s surface does the acceleration due to gravity fall to one quarter of its value at the surface? Give your answer in terms of the earth’s radius $R$.

• Treating the earth as a uniform sphere of radius $R$, at what depth below the surface is the acceleration due to gravity half its surface value?