Choose the correct alternative: (a) Acceleration due to gravity increases/decreases with increasing altitude. (b) Acceleration due to gravity increases/decreases with increasing depth (assume the earth to be a sphere of uniform density). (c) Acceleration due to gravity is independent of mass of the earth/mass of the body. (d) The formula $-GMm\left(\dfrac{1}{r_2} - \dfrac{1}{r_1}\right)$ is more/less accurate than the formula $mg(r_2 - r_1)$ for the difference of potential energy between two points $r_2$ and $r_1$ distance away from the centre of the earth.

> [!TLDR] > Gravity decreases with altitude and with depth, depends only on Earth's mass (not the body's mass), and the inverse‑square formula for potential energy is more accurate than the constant‑$g$ approximation. --- ### The core idea The acceleration due to gravity $g$ is not a universal constant — it changes as you move away from or into the Earth. The reason is simple: gravity depends on distance from the centre of mass, and on how much mass is “below” you. For a spherical Earth of uniform density, we can derive exact expressions for $g$ at any point. --- ### 1. Effect of altitude (height above surface) At the surface, $g = \frac{GM}{R^2}$ where $R$ is Earth’s radius and $M$ its mass. At a height $h$ above the surface, distance from centre is $r = R + h$. The gravitational acceleration is: $$g_h = \frac{GM}{(R+h)^2}$$ Since the denominator is larger, $g_h [!TIP] > For $h \ll R$, you can approximate: $g_h \approx g\left(1 - \frac{2h}{R}\right)$. This is handy in multiple‑choice problems. --- ### 2. Effect of depth (inside the Earth) Assume uniform density $\rho$. At a depth $d$ below the surface, your distance from the centre is $r = R - d$. Only the mass inside the sphere of radius $r$ pulls you inward — the outer shell contributes zero net force (shell theorem). Mass inside radius $r$: $M_r = \rho \cdot \frac{4}{3}\pi r^3 = M \cdot \frac{r^3}{R^3}$. So the acceleration at depth $d$ is: $$g_d = \frac{G M_r}{r^2} = \frac{G M}{R^3} \cdot r = g \cdot \frac{r}{R}$$ Since $r [!WARNING] > A common mistake is to think gravity increases inside the Earth because you’re “closer to the centre”. But less mass is pulling you — and the net effect is a decrease. --- ### 3. Dependence on mass of the body Newton’s law: $F = \frac{GMm}{r^2}$. The acceleration of a body of mass $m$ is $a = F/m = \frac

Choose the correct alternative: (a) Acceleration due to gravity increases/decreases with increasing altitude.

Exercises · 7.2

Q. Choose the correct alternative:

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

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

(d) The formula

-GMm\left(\dfrac{1}{r_2} - \dfrac{1}{r_1}\right)

is more/less accurate than the formula

mg(r_2 - r_1)

for the difference of potential energy between two points

r_2

and

r_1

distance away from the centre 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

This shape effect combines with the rotation effect to give the latitude variation shown in the table above.

The Complete Picture

Putting it all together, the variation of gravity with latitude $\phi$ and height $h$ is given by:

g(\phi, h) = g_0 \left(1 - \frac{2h}{R}\right) \left(1 - \frac{\omega^2 R}{g_0} \cos^2 \phi\right)

Where $g_0 \approx 9.806 \, \text{m/s}^2$ is the standard value at 45° latitude at sea level.

Important

The variation of gravity is small — typically less than 0.5% across the Earth's surface. But it matters for precise measurements, satellite orbits, and even for defining the kilogram (since a spring scale calibrated in Mumbai would read differently in London).

A Quick Summary

Factor	Effect on $g$	Why
Going up (altitude)	Decreases	Farther from Earth's centre
Going down (depth)	Decreases (linearly for uniform Earth)	Less mass below you
Moving to equator	Decreases	Rotation + bulge
Moving to poles	Increases	No rotation + closer to centre

The key takeaway: gravity is not a fixed number. It's a local property that depends on where you are on (or inside) the Earth. The $9.8 \, \text{m/s}^2$ you memorise is just a convenient average — the real story is richer, and now you know why.

"Variation Of Gravity derivation" and "Variation Of Gravity numerical problems" are two of the most common searches tied to this topic, and Variation Of Gravity is a core, NCERT-aligned topic from the Gravitation portion of the Class 11 Physics curriculum, and it is tested regularly in CBSE board exams as well as in JEE Main and NEET. Pairing this explanation with NCERT Physics textbook practice and previous years' questions is the surest way to lock the concept in before an exam.

Gravity decreases with altitude and with depth, depends only on Earth's mass (not the body's mass), and the inverse‑square formula for potential energy is more accurate than the constant‑ $g$ approximation.

The core idea

The acceleration due to gravity $g$ is not a universal constant — it changes as you move away from or into the Earth. The reason is simple: gravity depends on distance from the centre of mass, and on how much mass is “below” you. For a spherical Earth of uniform density, we can derive exact expressions for $g$ at any point.

1. Effect of altitude (height above surface)

At the surface, $g = \frac{GM}{R^2}$ where $R$ is Earth’s radius and $M$ its mass.

At a height $h$ above the surface, distance from centre is $r = R + h$ . The gravitational acceleration is:

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

Since the denominator is larger, $g_h < g$ . So gravity decreases with increasing altitude.

Tip

For $h \ll R$ , you can approximate: $g_h \approx g\left(1 - \frac{2h}{R}\right)$ . This is handy in multiple‑choice problems.

2. Effect of depth (inside the Earth)

Assume uniform density $\rho$ . At a depth $d$ below the surface, your distance from the centre is $r = R - d$ . Only the mass inside the sphere of radius $r$ pulls you inward — the outer shell contributes zero net force (shell theorem).

Mass inside radius $r$ : $M_r = \rho \cdot \frac{4}{3}\pi r^3 = M \cdot \frac{r^3}{R^3}$ .

So the acceleration at depth $d$ is:

g_d = \frac{G M_r}{r^2} = \frac{G M}{R^3} \cdot r = g \cdot \frac{r}{R}

Since $r < R$ , we have $g_d < g$ . Gravity decreases linearly with depth, reaching zero at the centre.

Watch out

A common mistake is to think gravity increases inside the Earth because you’re “closer to the centre”. But less mass is pulling you — and the net effect is a decrease.

3. Dependence on mass of the body

Newton’s law: $F = \frac{GMm}{r^2}$ . The acceleration of a body of mass $m$ is $a = F/m = \frac{GM}{r^2}$ . The $m$ cancels out. So $g$ is independent of the mass of the body — it depends only on Earth’s mass and distance.

g = \frac{GM}{r^2}

4. Accuracy of potential energy formulas

The exact gravitational potential energy difference between two points at distances $r_1$ and $r_2$ from Earth’s centre is:

\Delta U = -GMm\left(\frac{1}{r_2} - \frac{1}{r_1}\right)

The approximate formula $mg(r_2 - r_1)$ assumes $g$ is constant — which is only true when $r_2 - r_1 \ll R$ . The exact formula works for any separation. So the inverse‑square formula is more accurate.

Note

| Formula | When valid | Accuracy |

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

| $-GMm\left(\frac{1}{r_2} - \frac{1}{r_1}\right)$ | Always (for a spherical Earth) | Exact |

| $mg(r_2 - r_1)$ | Only for small height changes | Approximate |

✓Final answer

decreases. decreases. mass of the body. more.

←Previous question7.1 — Answer the following: (a) You can shield a charge from electrical forc…→Next question7.3 — Suppose there existed a planet that went around the Sun twice as fast …

Q.Choose the correct alternative:

The core idea

1. Effect of altitude (height above surface)

2. Effect of depth (inside the Earth)

3. Dependence on mass of the body

4. Accuracy of potential energy formulas

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