Which of the following options are correct? (Note: more than one of the given options may be correct.) (a) Acc

NCERT Exemplar · Q9

Q. Which of the following options are correct? (Note: more than one of the given options may be correct.)

(a) Acceleration due to gravity decreases with increasing altitude.

(b) Acceleration due to gravity increases with increasing depth (assume the earth to be a sphere of uniform density).

(d) Acceleration due to gravity is independent of the mass of the earth.

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Concept understanding — Variation Of Gravity

Variation of Gravity: Why Your Weight Changes Even When You Don't

Imagine you step on a weighing scale at sea level in Mumbai, then carry that same scale to the top of Mount Everest. The scale would show a smaller number — you'd weigh less. But you haven't lost any mass. What changed?

The force pulling you down — gravity — is not constant everywhere on Earth. It varies. That's what we mean by variation of gravity.

The Core Idea

Gravity is the force with which the Earth pulls objects toward its centre. The strength of this pull depends on two things: the mass of the Earth and your distance from its centre. Since the Earth is not a perfect sphere and it spins, that distance and the effective pull change from place to place.

The acceleration due to gravity, denoted by $g$ , is approximately $9.8 \, \text{m/s}^2$ at sea level. But that's an average. The actual value can be slightly higher or lower depending on where you are.

Why Does Gravity Vary? Three Main Reasons

1. Altitude (Height Above Sea Level)

This is the most intuitive one. As you go higher, you move farther from the Earth's centre. Gravity follows an inverse-square law: double the distance, and the force becomes one-fourth.

The formula for $g$ at a height $h$ above the Earth's surface (where $R$ is Earth's radius, about 6400 km) is:

g_h = \frac{GM}{(R+h)^2}

For small heights compared to $R$ , we can approximate:

g_h \approx g \left(1 - \frac{2h}{R}\right)

Note

This means for every kilometre you go up, $g$ decreases by roughly $0.003 \, \text{m/s}^2$ . That's why at the top of a tall mountain, you weigh about 0.5% less than at sea level.

2. Depth (Going Underground)

What happens if you go down a mine or into the Earth's crust? Intuition might say gravity increases because you're closer to the centre. But the opposite happens.

Inside the Earth, the mass above you pulls upward, partially cancelling the pull from below. For a uniform Earth, only the mass inside the sphere of radius $r$ (your distance from the centre) contributes to gravity at that point.

g_d = \frac{GM'}{r^2}

Where $M'$ is the mass of the sphere of radius $r$ . If Earth had uniform density $\rho$ , then $M' = \frac{4}{3}\pi r^3 \rho$ , giving:

g_d = \frac{4}{3}\pi G \rho r

This means gravity decreases linearly as you go deeper. At the centre of the Earth, $g = 0$ — you'd be weightless, pulled equally in all directions.

Watch out

This linear decrease assumes uniform density. The real Earth has a dense iron core, so the actual variation is more complicated — gravity actually increases slightly as you go down through the crust before eventually decreasing.

3. Rotation of the Earth (Latitude Effect)

The Earth spins once every 24 hours. This rotation creates a centrifugal force that acts outward, away from the axis of rotation. This force effectively reduces the weight you feel.

The effect is strongest at the equator (where the rotational speed is highest, about 1670 km/h) and zero at the poles (where you're on the axis of rotation).

The effective $g$ at latitude $\phi$ is:

g_{\text{eff}} = g - \omega^2 R \cos^2 \phi

Where $\omega$ is Earth's angular speed ( $7.3 \times 10^{-5} \, \text{rad/s}$ ) and $R$ is Earth's radius.

At the equator ( $\phi = 0^\circ$ ), the reduction is about $0.034 \, \text{m/s}^2$ — roughly 0.35% of $g$ .

Tip

| Location | Approximate $g$ (m/s²) | Why? |

|----------|------------------------|------|

| Equator (sea level) | 9.78 | Fastest rotation + bulging equator |

| 45° latitude | 9.81 | Intermediate |

| North Pole | 9.83 | No rotation effect + closer to centre |

4. Shape of the Earth (Oblateness)

The Earth is not a perfect sphere. Because of its rotation, it bulges at the equator and flattens at the poles. The equatorial radius is about 21 km larger than the polar radius.

This means:

At the poles, you're closer to the Earth's centre → stronger gravity
At the equator, you're farther from the centre → weaker gravity …

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Acceleration due to gravity changes with altitude, depth, and latitude. It decreases with increasing altitude and increases with increasing latitude.

The correct options are (A) and (C).

The acceleration due to gravity, denoted by $g$ , is not constant across the Earth's surface or within its interior. It varies depending on several factors, including altitude, depth, the Earth's rotation (which depends on latitude), and the distribution of mass within the Earth. Understanding these variations is crucial for many applications in physics and geophysics.

Let's analyze each option based on the fundamental principles governing gravitational acceleration.

1. Analyzing Option (A): Acceleration due to gravity decreases with increasing altitude.

The acceleration due to gravity at the Earth's surface is given by $g = \frac{GM}{R^2}$ , where $G$ is the universal gravitational constant, $M$ is the mass of the Earth, and $R$ is the radius of the Earth.

When an object is at an altitude $h$ above the Earth's surface, its distance from the center of the Earth becomes $r = R+h$ . Assuming the Earth is a perfect sphere and its mass is concentrated at its center, the acceleration due to gravity at altitude $h$ , denoted as $g_h$ , is:

g_h = \frac{GM}{(R+h)^2}

We can rewrite this in terms of $g$ :

g_h = \frac{GM}{R^2 \left(1 + \frac{h}{R}\right)^2} = g \left(1 + \frac{h}{R}\right)^{-2}

From this formula, it is clear that as $h$ increases, the denominator $(R+h)^2$ increases, and thus $g_h$ decreases.

Tip

For altitudes much smaller than the Earth's radius ( $h \ll R$ ), we can use the binomial approximation $(1+x)^n \approx 1+nx$ .

$g_h \approx g \left(1 - \frac{2h}{R}\right)$

This approximation clearly shows the decrease in $g$ with increasing altitude.

Therefore, option (A) is correct.

2. Analyzing Option (B): Acceleration due to gravity increases with increasing depth (assume the earth to be a sphere of uniform density).

When we consider a point at a depth $d$ below the Earth's surface, its distance from the center of the Earth is $r = R-d$ . To calculate the acceleration due to gravity at this depth, we only consider the mass of the Earth contained within a sphere of radius $r$ . This is because the gravitational forces due to the spherical shell of matter outside this radius cancel out.

Let $\rho$ be the uniform density of the Earth.

The total mass of the Earth is $M = \frac{4}{3}\pi R^3 \rho$ .

The mass of the inner sphere of radius $r = R-d$ is $M' = \frac{4}{3}\pi r^3 \rho$ .

The acceleration due to gravity at depth $d$ , denoted as $g_d$ , is:

g_d = \frac{GM'}{r^2} = \frac{G \left(\frac{4}{3}\pi r^3 \rho\right)}{r^2} = G \frac{4}{3}\pi \rho r

Substitute $r = R-d$ :

g_d = G \frac{4}{3}\pi \rho (R-d)

We know that $g = \frac{GM}{R^2} = \frac{G (\frac{4}{3}\pi R^3 \rho)}{R^2} = G \frac{4}{3}\pi R \rho$ .

So, we can write $G \frac{4}{3}\pi \rho = \frac{g}{R}$ .

Substituting this into the expression for $g_d$ :

g_d = \frac{g}{R} (R-d) = g \left(1 - \frac{d}{R}\right)

This formula shows that $g_d$ decreases linearly with increasing depth $d$ .

At the surface ( $d=0$ ), $g_d = g$ .
At the center of the Earth ( $d=R$ ), $g_d = 0$ .

Watch out

A common misconception is that gravity increases as one goes deeper into the Earth because one is getting closer to the center. However, the mass contributing to the gravitational force also decreases, and this effect dominates, leading to a decrease in $g$ .

Therefore, option (B) is incorrect.

3. Analyzing Option (C): Acceleration due to gravity increases with increasing latitude.

The Earth rotates about its axis. Due to this rotation, objects on the surface experience a centrifugal force directed outwards from the axis of rotation. This centrifugal force reduces the effective acceleration due to gravity. …

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Q.Which of the following options are correct? (Note: more than one of the given options may be correct.)

1. Analyzing Option (A): Acceleration due to gravity decreases with increasing altitude.

2. Analyzing Option (B): Acceleration due to gravity increases with increasing depth (assume the earth to be a sphere of uniform density).

3. Analyzing Option (C): Acceleration due to gravity increases with increasing latitude.

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