As observed from earth, the sun appears to move in an approximate circular orbit. For the motion of another pl

NCERT Exemplar · Q2

Q. 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

(a) be similarly true.

(b) not be true because the force between earth and mercury is not inverse square law.

(d) not be true because mercury is influenced by forces other than gravitational forces.

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Concept understanding — Newton's Law of Universal Gravitation

Newton's Law of Universal Gravitation

Every particle of matter attracts every other particle with a force directed along the line joining them. For two point masses $m_1$ and $m_2$ separated by a distance $r$ , the magnitude of the attraction is

F = \frac{G\,m_1 m_2}{r^{2}},

where $G = 6.67 \times 10^{-11}\ \text{N m}^{2}\,\text{kg}^{-2}$ is the universal gravitational constant. The force is attractive, acts along the line of centres, and obeys Newton's third law (equal and opposite on the two bodies). Its defining feature is the inverse-square dependence on separation.

Superposition. When several masses act on a body, the net gravitational force is the vector sum of the individual forces. For an extended body — a rod, ring or shell — the body is treated as a collection of point-mass elements and the contributions are integrated. For a uniform ring of mass $M$ and radius $R$ , the force on a point mass $m$ placed on its axis at distance $h$ from the centre is directed along the axis and has magnitude

F = \frac{G M m\, h}{\left(R^{2} + h^{2}\right)^{3/2}},

because the components perpendicular to the axis cancel by symmetry while the axial components add. This axial formula is not a simple $1/h^{2}$ law: as $h$ changes, the force scales as $h / (R^{2}+h^{2})^{3/2}$ , so comparing the force at two axial positions (say $h$ and $2h$ ) means substituting into this expression rather than using the point-mass inverse square directly. Only when $h \gg R$ does the ring behave like a point mass and $F \to GMm/h^{2}$ .

Newton's law of gravitation is one of the earliest topics in the NCERT Class 11 Physics chapter on Gravitation, and "universal law of gravitation formula and examples" or "gravitation class 11 important questions" are searched constantly by CBSE board and NEET/JEE Main aspirants. The extended-body superposition case shown here — finding the force due to a ring on an axial point mass — is a classic JEE Main gravitation numerical built directly on this NCERT Class 11 foundation.

From Earth, the Sun appears to orbit Earth because Earth itself orbits the Sun. However, Mercury's motion as observed from Earth will not be a simple circular orbit around Earth because the Sun's gravitational force is the primary influence on Mercury's orbit, not Earth's. The correct option is (C).

When we observe celestial bodies, their apparent motion depends entirely on our frame of reference. The question asks us to compare the apparent motion of the Sun as seen from Earth with the apparent motion of Mercury as seen from Earth.

Let's break down the underlying physics and observations:

Earth's Orbit Around the Sun:

From a heliocentric (Sun-centered) perspective, Earth orbits the Sun in an approximately elliptical path, which we often simplify to a circle for basic understanding.

Important

The gravitational force between the Sun and Earth is given by Newton's Law of Universal Gravitation:

$F = G \frac{M_{Sun} M_{Earth}}{r^2}$

This force is what keeps Earth in orbit around the Sun.
Apparent Motion of the Sun from Earth:

Because Earth is orbiting the Sun, an observer on Earth perceives the Sun to be moving around Earth. This is a consequence of relative motion. Since Earth's orbit is nearly circular, the Sun's apparent path in the sky over a year is also approximately circular. This is the basis for the "apparent circular orbit" mentioned in the question.
Mercury's Orbit Around the Sun:

Similar to Earth, Mercury also orbits the Sun in an elliptical path. The Sun's gravitational force is the dominant force dictating Mercury's motion.

Watch out

It's a common misconception to think that planets orbit each other. While all celestial bodies exert gravitational forces on each other, the dominant force determines the primary orbit. For Mercury, the Sun's gravity is vastly stronger than Earth's gravity.
Apparent Motion of Mercury from Earth:

Now, consider Mercury's motion as observed from Earth. Both Earth and Mercury are orbiting the Sun.
- Mercury is orbiting the Sun.
- Earth is also orbiting the Sun.
- Therefore, Mercury's apparent path as seen from Earth will be a combination of its own motion around the Sun and Earth's motion around the Sun. This results in a complex path, often appearing as loops or retrograde motion against the background stars, rather than a simple, continuous circular orbit around Earth.
Evaluating the Options:
- (A) be similarly true. This would imply Mercury appears to orbit Earth in an approximate circle. This is incorrect because the Sun, not Earth, is the primary gravitational influence on Mercury.
- (B) not be true because the force between earth and mercury is not inverse square law. This statement is factually incorrect. The gravitational force between any two masses, including Earth and Mercury, is an inverse square law.
- (C) not be true because the major gravitational force on mercury is due to sun. This is the correct explanation. Mercury's primary motion is governed by the Sun's gravity. Earth's gravity is a minor perturbation. Therefore, Mercury does not orbit Earth, and its apparent motion from Earth will not be a simple circular path around Earth.
- (D) not be true because mercury is influenced by forces other than gravitational forces. While other forces (like solar radiation pressure) exist, they are generally negligible compared to gravitational forces for planetary orbits. This is not the primary reason for the complex apparent motion.

The key takeaway is that the apparent motion of a planet as seen from another planet is a result of both planets orbiting their common gravitational center (the Sun), and the dominant gravitational force on each planet.

✓Final answer

The correct option is (C) .

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:

$F_w = G \frac{m_1 m_2}{r^2} = F_v$

Compute the ratio

$\frac{F_v}{F_w} = \frac{F_v}{F_v} = 1$
So the ratio is $1:1$ .

Watch out

A common mistake is to treat gravity like electrostatics and assume water reduces the force. But gravity is not a dipole force; it is always attractive and unaffected by the medium. The “buoyancy” or “drag” you feel in water is not gravity changing — it’s a separate effect from fluid pressure and viscosity.

Tip

If the problem had asked about electrostatic force between two charges, the ratio would be the dielectric constant of water (about 80), giving a huge difference. That contrast is exactly why this question tests whether you know gravity is medium-independent.

✓Final answer
$The correct option is (B) . ANSWER: B$
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).

Rearranging for $m$ with $r=50\ \text{cm}=0.5\ \text{m}$ and $a=5\times10^{-10}\ \text{m s}^{-2}$ :
$m=\frac{a\,r^2}{G}=\frac{(5\times10^{-10})(0.5)^2}{6.674\times10^{-11}}.$ $m=\frac{1.25\times10^{-10}}{6.674\times10^{-11}}\approx1.873\ \text{kg}.$
✓Final answer
$\approx1.8728\ \text{kg}$
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

When the space around the masses is filled with a liquid of density $\rho$ , the masses themselves remain the same. The distance $d$ between their centers also remains the same. The liquid does not change $G$ , $M_1$ , $M_2$ , or $d$ . Hence, the gravitational force between them is still

$F' = \frac{G M_1 M_2}{d^2} = F.$

Address a common confusion

Watch out

A classic pitfall is to think the liquid “shields” gravity, similar to how a dielectric reduces the electric force between charges. But gravity has no “anti-gravity” material; the gravitational constant $G$ is truly universal. The liquid’s density only affects buoyant forces on the masses, not the gravitational attraction between them.

Conclude

Since the force is unchanged, the correct choice is (B).

✓Final answer
$The correct option is (B). ANSWER: B$
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?

Einstein revolutionized gravity with general relativity (1915), but he never measured $G$ .

Galileo studied falling bodies and inertia, but gravity as a universal force was unknown in his time.

Watch out

A common mistake is to think Newton “determined” $G$ because he wrote the equation. In fact, Newton only knew the relative strength of gravity (e.g., that the Sun’s pull on Earth is much larger than Earth’s pull on a stone), but he had no absolute value for $G$ .

Tip

Cavendish’s experiment is often called “weighing the Earth” because once you know $G$ and the acceleration due to gravity $g$ , you can solve for Earth’s mass: $M_\text{Earth} = \frac{g R_\text{Earth}^2}{G}$ .

✓Final answer
$The correct option is (B) . ANSWER: B$
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:

Its mass: same density, volume ratio (R/2)^3 / R^3 = 1/8, so m_cut = M/8.

Its distance from m: 3R - R/2 = 5R/2.

F_cut = G (M/8) m / (5R/2)^2 = G (M/8) m / (25 R^2 / 4) = G M m / (50 R^2).

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.

✓Final answer
$The correct option is (C) — 41F/50$

ANSWER: C

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