1. Interpret the question

We need the graph of acceleration due to gravity $g$ versus distance $d$ from the centre of the Earth.

Let Earth’s radius be $R$.

So there are two regions:

• Inside the Earth: $0 \le d \le R$
• Outside the Earth: $d \ge R$

2. Expression for $g$ outside the Earth

For a point outside a spherically symmetric Earth,

$g = \frac{GM}{d^2}$

where $M$ is the mass of Earth.

Thus for $d > R$,

$g \propto \frac{1}{d^2}$

So outside the Earth, the graph is a decreasing inverse-square curve.

At the surface $d=R$,

$g = \frac{GM}{R^2} = g_0$

3. Expression for $g$ inside the Earth

Assuming Earth has uniform density, only the mass enclosed within radius $d$ contributes:

$M_d \propto d^3$

Hence,

$g = \frac{G M_d}{d^2} \propto \frac{d^3}{d^2} = d$

So inside the Earth,

$g \propto d$

That means:

• At the centre, $d=0$, so $g=0$
• As $d$ increases from $0$ to $R$, $g$ increases linearly
• At $d=R$, $g$ reaches maximum value $g_0$

So for $0 \le d \le R$,

$g = g_0\frac{d}{R}$

4. Overall variation

Combining both regions: