Assuming the earth to be a sphere of uniform mass density, how much would a body weigh half way down to the ce

Exercises · 7.16

Q. Assuming the earth to be a sphere of uniform mass density, how much would a body weigh half way down to the centre of the earth if it weighed 250 N on the surface?

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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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Inside a uniform sphere, gravitational field strength decreases linearly with distance from the centre. At half the radius, the weight drops to half its surface value: 125 N.

Why weight changes inside the Earth

When you stand on the surface, the entire mass of the Earth pulls you downward. But what happens as you descend into a tunnel toward the centre?

The key insight comes from Newton's shell theorem: only the mass closer to the centre than you are contributes to the gravitational pull you feel. All the mass in the spherical shell outside your position exerts zero net force on you. This is a beautiful consequence of the inverse-square law combined with spherical symmetry.

For a uniform sphere, this means the effective mass pulling you inward shrinks as you go deeper, and so does your weight.

Step-by-step calculation

Set up the geometry

Let the Earth have radius $R$ and uniform density $\rho$ . On the surface, the gravitational field strength is $g = \frac{GM}{R^2}$ , where $M = \frac{4}{3}\pi R^3 \rho$ is the total mass.
Find the effective mass at depth

At a distance $r$ from the centre (where $r < R$ ), only the mass within radius $r$ contributes:

M(r) = \frac{4}{3}\pi r^3 \rho

Calculate the field strength at radius $r$ The gravitational field at distance $r$ from the centre is:

g(r) = \frac{GM(r)}{r^2} = \frac{G \cdot \frac{4}{3}\pi r^3 \rho}{r^2} = \frac{4}{3}\pi G \rho \cdot r

Notice this is linear in $r$ . We can express it in terms of the surface field:

g(r) = g \cdot \frac{r}{R}

because $g = \frac{4}{3}\pi G \rho R$ .

g(r) = g_{\text{surface}} \cdot \frac{r}{R}

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Q.Assuming the earth to be a sphere of uniform mass density, how much would a body weigh half way down to the centre of the earth if it weighed 250 N on the surface?

Why weight changes inside the Earth

Step-by-step calculation

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