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

NCERT Exemplar · Q6

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

(a) will not move around the sun since they have very small masses compared to sun.

(b) will move in an irregular way because of their small masses and will drift away into outer space.

(d) will move in orbits like planets and obey Kepler's laws.

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Kepler's laws apply to any body orbiting the Sun under gravity, regardless of mass. Asteroids, despite their small size, orbit the Sun in closed elliptical paths and obey all three of Kepler's laws.

The heart of this question is understanding what Kepler's laws actually depend on. Many students mistakenly think that only "large" objects like planets follow these laws, or that an object's mass determines whether it can orbit at all. This is a fundamental misunderstanding of gravitational motion.

Kepler's laws describe the motion of bodies under the Sun's gravitational influence. The key insight is that these laws emerge from Newton's law of universal gravitation:

F = \frac{GMm}{r^2}

where $M$ is the Sun's mass, $m$ is the orbiting body's mass, $G$ is the gravitational constant, and $r$ is the separation. When you work through the orbital mechanics, the orbiting body's mass $m$ cancels out of the equations of motion. The trajectory depends only on the Sun's mass, the initial position, and the initial velocity—not on whether the orbiting object is Jupiter or a pebble.

Let's examine each option:

Option (A) claims asteroids won't move around the Sun because of small mass.

This confuses mass with the ability to orbit. The gravitational force on an asteroid is $F = \frac{GMm}{r^2}$ , and its acceleration is $a = F/m = \frac{GM}{r^2}$ . The mass cancels! A grain of sand and Jupiter experience the same acceleration at the same distance from the Sun. Small mass does not prevent orbital motion.
Option (B) suggests irregular motion leading to drift into space.

Irregular motion would require either non-gravitational forces or chaotic interactions with multiple massive bodies. In the Sun's gravitational field alone, any object follows a conic section (ellipse, parabola, or hyperbola) determined by its energy. Asteroids in bound orbits (negative total energy) move in closed ellipses, just like planets. Their small mass doesn't make them "drift away."
Option (C) proposes closed orbits but no obedience to Kepler's laws. …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

COMEDK 2026Set 2026-A1 markMCQ
$m_1$
›Reveal solutionSolution
Gravitational force between two masses depends only on their masses and the distance between them, not on the surrounding medium. Therefore the force remains unchanged in water, so the ratio is 1:1.

Concept & Intuition

The key idea is that gravity is a fundamental, long-range force that acts directly between masses. Unlike electric or magnetic forces, gravity is not affected by the presence of a dielectric or any other medium — it does not get “screened” or “reduced” by water, air, or vacuum. This is because the gravitational interaction is mediated by spacetime curvature, not by particles that interact with the medium. So, moving the setup from vacuum to water changes nothing about the gravitational force.

Step-by-step reasoning

Recall Newton’s law of universal gravitation In a vacuum, the gravitational force between two point masses $m_1$ and $m_2$ separated by distance $r$ is given by

$F_v = G \frac{m_1 m_2}{r^2}$
where $G$ is the universal gravitational constant.

Consider the effect of a medium For forces like electrostatic force, the presence of a medium introduces a dielectric constant $\varepsilon_r$ that reduces the force:

$F_{\text{medium}} = \frac{1}{\varepsilon_r} F_{\text{vacuum}}$
However, gravity has no such “permittivity” or “permeability” — there is no gravitational analogue of a dielectric constant. The constant $G$ is universal and does not change with the medium.

Apply to the water medium Water is a material medium, but it does not alter the gravitational interaction. The masses remain the same, the distance remains the same, and $G$ remains the same. Therefore: $F_w = G \frac{m_1 m_2}{r^2} = F_v$ …
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COMEDK 2026Set 2026-A1 markMCQ
$5 \times 10^{-10} \mathrm{~ms}^{-2}$
›Reveal solutionSolution

The heavier particle's acceleration is $a=\dfrac{Gm}{r^2}$ , so $m=\dfrac{a r^2}{G}=\dfrac{(5\times10^{-10})(0.5)^2}{6.674\times10^{-11}}\approx1.87\ \text{kg}$ — option (D).

The gravitational force between the two particles is $F=\dfrac{GMm}{r^2}$ . This same force accelerates the heavier particle (mass $M=4\ \text{kg}$ ), so its acceleration is
$a=\frac{F}{M}=\frac{Gm}{r^2},$
which depends only on the lighter mass $m$ and the separation $r$ (the heavier mass cancels). …
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COMEDK 2026Set 2026-M1 markMCQ
$M_1$
›Reveal solutionSolution
The gravitational force between two masses depends only on their masses and the distance between them, not on the surrounding medium. Therefore, the force remains unchanged, so the correct option is (B).

Concept and Intuition

The key idea is that gravitational force is a universal interaction that does not depend on the medium between the masses. Unlike electrostatic or magnetic forces, which can be screened or altered by a dielectric or magnetic material, gravity passes through any substance unchanged. The density of the liquid is irrelevant here — it only affects buoyancy, not gravity. Many students mistakenly think the liquid reduces the force, but that’s a confusion with electrostatic forces in a dielectric medium.

Step-by-step reasoning

Recall Newton’s law of universal gravitation The gravitational force between two point masses $M_1$ and $M_2$ separated by a distance $d$ in vacuum (or air) is given by

$F = \frac{G M_1 M_2}{d^2},$
where $G$ is the universal gravitational constant.

Identify what the law depends on

Notice that $F$ depends only on the masses and the distance — not on any property of the medium (like density, refractive index, or permittivity). This is because gravity is a fundamental force that couples directly to mass-energy, and no intervening material can “block” or “alter” it.

Consider the effect of the liquid …
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COMEDK 2025Set 2025-A1 markMCQ
$Q. The value of universal gravitational constant was first determined by (A) Newton (B) Cavendish (C) Einstein (D) Galileo$
›Reveal solutionSolution
The universal gravitational constant $G$ was first measured experimentally by Henry Cavendish using a torsion balance, not by Newton who only theorized gravity. The correct option is (B).

The key idea here is the difference between theorizing a constant and measuring it. Newton’s law of universal gravitation ( $F = G \frac{m_1 m_2}{r^2}$ ) was published in 1687, but Newton had no way to measure $G$ — he could calculate the product $GM_\text{Earth}$ from the Moon’s orbit, but not $G$ alone. The constant $G$ remained unknown until an experiment could detect the tiny gravitational force between everyday objects.

Why not Newton?

Newton formulated the law, but he lacked the technology to measure the extremely weak attraction between laboratory masses. He famously estimated that the gravitational force between two 1 kg spheres 1 m apart is about $6.7 \times 10^{-11}$  N — far too small for 17th-century instruments. So he never determined $G$ numerically.

Cavendish’s breakthrough (1798)

Henry Cavendish used a torsion balance — a horizontal rod with small lead spheres suspended by a thin wire. When large lead spheres were brought near, the tiny gravitational twist of the wire was measured. From the twist angle and the known properties of the wire, he calculated the force and thus $G$ . His result was within 1% of today’s accepted value ( $6.674 \times 10^{-11} \, \text{N·m}^2/\text{kg}^2$ ).

Why not Einstein or Galileo? …
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COMEDK 2023Set 2023-E1 markMCQ
$M$
›Reveal solutionSolution

Step 3 - remaining body: F_remaining = F_full - F_cut = G M m [ 1/(9R^2) - 1/(50 R^2) ] = (G M m / R^2) * (50 - 9)/450 = (G M m / R^2) * 41/450 = 9F * 41/450 = 41F/50.

Concept: superposition (the 'cavity' trick). Remaining solid = full sphere MINUS the small sphere that was removed. Forces subtract likewise. Both the full sphere and the removed sphere are spherically symmetric, so each attracts the external mass m as if its mass were at its own centre.

Geometry from the figure: the cavity has diameter R (radius R/2) and is internally tangent to the sphere on the side facing m, so its centre lies at distance R/2 from the sphere's centre, ON the line towards m. The mass m is at 3R from the sphere's centre.

Step 1 - full sphere:

F = G M m / (3R)^2 = G M m / (9 R^2). So G M m / R^2 = 9F.

Step 2 - removed portion: …
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Q.In our solar system, the inter-planetary region has chunks of matter (much smaller in size compared to planets) called asteroids. They

Concept understanding — Newton's Law of Universal Gravitation

Newton's Law of Universal Gravitation

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

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