Animated Solution for Physics - Gravitation: The magnitudes of the gravitational field at distance $r_1$ and $r_2$ from the centre of a uniform sphere of radius $R$ and mass $M$ are $F_1$ and $F_2$, respectively. Then

Imagine standing on the surface of a massive, uniform planet.

The gravitational pull you feel is a consequence of the entire mass of the planet acting upon you.

But what happens if you start tunneling deep towards the center of this planet?

Or what if you blast off in a rocket, soaring high above its atmosphere?

This classic problem from the 1994 JEE paper invites us to explore how the gravitational field intensity, $F$, behaves in these two fundamentally different environments: deep inside a solid sphere and far out in space.

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Let us define our system.

We have a uniform solid sphere of mass $M$ and radius $R$.

We want to compare the gravitational field strengths $F_1$ and $F_2$ at two different radial distances, $r_1$ and $r_2$, from the center of the sphere.

To solve this, we must divide our analysis into two distinct physical domains:

1. The Interior Domain ($r<R$): Where we are deep inside the bulk of the sphere.

2. The Exterior Domain ($r>R$): Where we are completely outside the sphere.

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Deep Inside the Sphere ($r<R$)

When you are inside a uniform solid sphere at a distance $r$ from the center, a fascinating physical phenomenon occurs.

According to Newton's Shell Theorem, the spherical shell of mass that lies at radii greater than $r$ exerts zero net gravitational force on you!

All the gravitational pulls from the outer layers cancel each other out perfectly.

Therefore, the only mass that contributes to the gravitational field at your location is the mass enclosed within a smaller sphere of radius $r$.

Let's write this mathematically. The enclosed mass $M_{\text{enc}}$ is:

Now, applying Newton's law of gravitation as if this enclosed mass were concentrated at the center:

This is a beautiful result!

It tells us that inside a uniform solid sphere, the gravitational field intensity is directly proportional to the distance from the center:

If we have two points $r_1$ and $r_2$ both lying inside the sphere ($r_1<R$ and $r_2<R$), their fields are:

Taking the ratio of these two fields, the constants $G$, $M$, and $R^3$ cancel out beautifully:

This perfectly matches Option (a)!

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Far Out in Space ($r>R$)

Now, let's blast off and look at the sphere from the outside.

According to the second part of the Shell Theorem, for any point outside a spherically symmetric mass distribution, the entire mass behaves as if it were concentrated at a single point at the geometric center.

Thus, the gravitational field intensity simply follows the classic inverse-square law:

Here, the field is inversely proportional to the square of the distance from the center:

If we have two points $r_1$ and $r_2$ both lying outside the sphere ($r_1>R$ and $r_2>R$), their fields are:

Taking the ratio of these two fields:

This perfectly matches Option (b)!

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By carefully analyzing both domains, we have shown that:

Inside the sphere ($r<R$), the field is linear: $\frac{F_1}{F_2}=\frac{r_1}{r_2}$. This makes Option (a)* correct and rules out Options (c) and (d).

Outside the sphere ($r>R$), the field follows the inverse-square law: $\frac{F_1}{F_2}=\frac{r_2^2}{r_1^2}$. This makes Option (b)* correct.

Thus, the correct options are (a) and (b).