Different points in earth are at slightly different distances from the sun and hence experience different forc

NCERT Exemplar · Q3

Q. 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)

(a) the torque is zero.

(b) the torque causes the earth to spin.

(d) the torque causes the earth to move around the sun.

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Concept understanding — Center of Mass

What is the Center of Mass?

Imagine you pick up a broom by the handle and try to balance it horizontally on one finger. You instinctively slide your finger along the handle until the broom stays level. That point — the one where the broom doesn't tip — is its center of mass.

Now think about throwing a cricket bat. It spins and wobbles in the air, but there is one point on the bat that follows a smooth, parabolic path, as if all the bat's mass were concentrated there. That point is also the center of mass.

The core idea is simple: the center of mass is the average position of all the mass in an object. It's the point where you could imagine the entire mass of the object being concentrated, and the object would behave the same way under the influence of external forces.

Why does this matter?

When you push an object at its center of mass, it moves in a straight line without rotating. Push it anywhere else, and it will both move and spin. This is why:

A car's stability depends on where its center of mass is (lower = safer).
A tightrope walker holds a long pole — moving the pole shifts their combined center of mass back over the rope.
In projectile motion, the center of mass of a system (like an exploding firework) continues along the original parabolic path, even though the fragments scatter.

The precise definition

For a system of particles, the center of mass is the weighted average of their positions, where the weight is the mass of each particle.

\vec{R}_{\text{CM}} = \frac{m_1 \vec{r}_1 + m_2 \vec{r}_2 + \cdots + m_n \vec{r}_n}{m_1 + m_2 + \cdots + m_n} = \frac{\sum m_i \vec{r}_i}{\sum m_i}

Here:

$\vec{R}_{\text{CM}}$ is the position vector of the center of mass
$m_i$ is the mass of the $i$ -th particle
$\vec{r}_i$ is the position vector of that particle

For a continuous object (like a rod or a sphere), the sum becomes an integral:

\vec{R}_{\text{CM}} = \frac{1}{M} \int \vec{r} \, dm

where $M$ is the total mass and $dm$ is an infinitesimal mass element.

Breaking it down with an example

Take two masses on a light rod: $m_1 = 2 \text{ kg}$ at $x = 0$ , and $m_2 = 3 \text{ kg}$ at $x = 5 \text{ m}$ .

The center of mass is:

x_{\text{CM}} = \frac{(2)(0) + (3)(5)}{2 + 3} = \frac{0 + 15}{5} = 3 \text{ m}

So the center of mass is at $x = 3 \text{ m}$ , closer to the heavier mass. That makes intuitive sense — the heavier mass "pulls" the average toward itself.

Note

The center of mass does not have to be inside the object. A ring or a hollow sphere has its center of mass at the geometric center, which is empty space.

Key properties to remember

External forces only — Internal forces (like collisions between parts of the system) do not affect the motion of the center of mass. Only external forces can change its velocity.
If no external force acts, the center of mass moves with constant velocity (or stays at rest). This is the law of conservation of momentum applied to the whole system.
For symmetric objects with uniform density, the center of mass coincides with the geometric center. For irregular shapes, it shifts toward the region with more mass.
In a uniform gravitational field, the center of mass and the center of gravity are the same point. (They differ only if gravity varies significantly across the object — not something you'll see in school problems.)

A final intuition

Think of the center of mass as the balance point of an object. If you could place a tiny, invisible support exactly at that point, the object would be perfectly balanced in any orientation. Every piece of mass on one side is exactly counterbalanced by the pieces on the other side.

That's why, when you jump off a boat, the boat moves backward — your center of mass and the boat's center of mass shift relative to each other, but the center of mass of the whole system (you + boat) stays put (if no external horizontal force acts). This is the heart of why the center of mass concept is so powerful: it lets you treat a complicated, spinning, wobbling object as a single point for many problems.

Looking up "Center of Mass: definition, formula & real-world examples" is a good habit before an exam, and it is worth knowing that Center of Mass is a core, NCERT-aligned topic from the System of Particles and Rotational Motion portion of the Class 11 Physics curriculum, and it is tested regularly in CBSE board exams as well as in JEE Main and NEET. Cross-checking this explanation against the relevant NCERT Physics chapter and solving a few past-year questions will round out your preparation.

The gravitational force from the Sun varies across Earth's diameter, but because Earth is spherically symmetric the net torque about Earth's center of mass is zero; the tidal force gradient does not produce rotation of Earth around its own axis or orbital motion—those have independent causes.

Why the torque vanishes: the Equivalence Principle and spherical symmetry

The Sun pulls harder on the near side of Earth than the far side because gravity weakens with distance. You might think this differential force—called a tidal force—would twist Earth, producing a torque about its center of mass. But torque depends not just on force magnitude but on where the force acts and in what direction.

For a spherically symmetric body like Earth (approximated as uniform density), every mass element on the near hemisphere has a partner on the far hemisphere arranged symmetrically about the center. The Sun's force on the near side is stronger and points slightly inward; the force on the far side is weaker and points slightly outward. When you sum the torques from all these pairs about Earth's center of mass, the angular contributions cancel exactly. The net effect is a stretching along the Earth–Sun line (the tidal bulge) but no net torque about the center.

This is a beautiful consequence of spherical symmetry: the gradient of a central force field (the Sun's gravity) produces no torque on a spherically symmetric mass distribution about its own center.

Step-by-step reasoning

Set up the problem. Earth is a sphere of radius $R$ centered at position $\mathbf{r}_{\text{CM}}$ relative to the Sun. A small mass element $dm$ at position $\mathbf{r}$ experiences gravitational force

d\mathbf{F} = -\frac{GM_{\odot} \, dm}{|\mathbf{r} - \mathbf{r}_{\odot}|^3} (\mathbf{r} - \mathbf{r}_{\odot}),

where $\mathbf{r}_{\odot}$ is the Sun's position. The distance $|\mathbf{r} - \mathbf{r}_{\odot}|$ varies across Earth's diameter by roughly $2R \approx 1.3 \times 10^4 \, \text{km}$ , a tiny fraction of the Earth–Sun distance $d \approx 1.5 \times 10^8 \, \text{km}$ .

Expand the force in a Taylor series. Because $R \ll d$ , write $\mathbf{r} = \mathbf{r}_{\text{CM}} + \boldsymbol{\delta}$ where $|\boldsymbol{\delta}| \leq R$ . Expanding the gravitational force to first order in $\boldsymbol{\delta}/d$ gives

d\mathbf{F} \approx -\frac{GM_{\odot} \, dm}{d^2} \hat{\mathbf{d}} + \text{(tidal terms)},

where $\hat{\mathbf{d}}$ points from Earth's center toward the Sun. The leading term is the same for every $dm$ ; integrating over Earth's mass $M$ yields the familiar orbital force $\mathbf{F}_{\text{net}} = -GM_{\odot}M/d^2 \, \hat{\mathbf{d}}$ acting at the center of mass.

Compute the torque about the center of mass. The torque on element $dm$ is

d\boldsymbol{\tau} = \boldsymbol{\delta} \times d\mathbf{F}.

The uniform part of $d\mathbf{F}$ (the orbital force) contributes

\int \boldsymbol{\delta} \times \left(-\frac{GM_{\odot} \, dm}{d^2} \hat{\mathbf{d}}\right) = -\frac{GM_{\odot}}{d^2} \hat{\mathbf{d}} \times \int \boldsymbol{\delta} \, dm = \mathbf{0},

because $\int \boldsymbol{\delta} \, dm = \mathbf{0}$ by definition of the center of mass.

What about the tidal (gradient) terms? The next-order terms in the expansion are proportional to $\boldsymbol{\delta}$ and produce the tidal force. For a spherically symmetric mass distribution, the integral

\int \boldsymbol{\delta} \times (\text{tidal force on } dm)

also vanishes. Physically, for every element at $+\boldsymbol{\delta}$ there is a symmetric partner at $-\boldsymbol{\delta}$ (or more generally, the angular average over the sphere is zero), so the torques cancel in pairs.

Interpret the options.
- (A) The torque is zero. ✓ This is correct.
- (B) The torque causes Earth to spin. ✗ Earth's rotation (its day) is a relic of the angular momentum from the solar nebula; the Sun's tidal torque on a perfectly spherical Earth is zero. (Real tidal torques arise from Earth's equatorial bulge and the Moon, gradually slowing Earth's spin.)
- (C) Rigid-body result not applicable. ✗ Earth is elastic and deforms under tidal stress, but the center-of-mass theorem and torque calculation for external forces apply perfectly well to any body, rigid or deformable.
- (D) The torque causes Earth to move around the Sun. ✗ Orbital motion is caused by the net force (which acts at the c.m. and produces centripetal acceleration), not by torque. Torque would cause rotation about the c.m., not translation.

Tip

Tidal forces stretch a body along the line joining the two masses (creating ocean tides) but produce no net torque on a sphere. Torques appear only when the body is asymmetric (e.g., the Moon's torque on Earth's equatorial bulge causes precession).

Watch out

Do not confuse the force gradient (which is nonzero and causes tides) with torque (which is zero for a symmetric body). Also, orbital motion is due to the net force, not torque—torque rotates a body about its own axis, force translates the center of mass.

✓Final answer

The correct option is (A) : the torque is zero.

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Why the torque vanishes: the Equivalence Principle and spherical symmetry

Step-by-step reasoning

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