A solid sphere of mass M and radius a is surrounded by a uniform concentric spherical shell of thick

Q: Why can we treat the thick shell as a point mass at the center?

According to Newton's Shell Theorem, any spherically symmetric mass distribution (like a uniform shell) creates a gravitational field outside of it that is mathematically identical to the field created if all its mass were concentrated at its geometric center.

Q: What if the point was inside the thick shell, say at $r = 2a$?

If the point is inside the shell, the Shell Theorem states that the portion of the shell outside the radius $r$ contributes zero net gravitational field. You would only consider the mass of the solid sphere and the portion of the shell that is enclosed within the radius $r = 2a$.

Q: Does the thickness of the shell matter for a point on its outer surface?

No, as long as the point is on or outside the outer surface, the only things that matter are the total mass of the shell and the distance from the center. The internal distribution (thickness) doesn't change the external field.

Q: Why do we simply add $E_1$ and $E_2$?

Gravitational field is a vector quantity. Since both the solid sphere and the shell are centered at the same origin, their gravitational pulls at any point will both point directly towards the center. Because they point in the exact same direction, we can simply add their magnitudes.

Q: Would the answer change if the shell was not uniform?

Yes! The Shell Theorem strictly requires the mass distribution to be spherically symmetric (uniform density at any given radius). If the shell had more mass on one side than the other, we could no longer treat it as a point mass at the center.

Question

A solid sphere of mass $M$ and radius $a$ is surrounded by a uniform concentric spherical shell of thickness $2a$ and $2M$. The gravitational field at distance $3a$ from the centre will be

Options:

(A) $\frac{GM}{9a^2}$

(B) $\frac{2GM}{9a^2}$

(C) $\frac{GM}{3a^2}$

(D) $\frac{2GM}{3a^2}$

Sigma Percentile · Accepted Answer

### Visualizing the Mass Distribution

Imagine you are standing in space, looking at a fascinating celestial setup. At the very center, there is a solid sphere of mass $M$ and radius $a$. Surrounding this sphere is a thick, uniform spherical shell. The inner radius of this shell perfectly hugs the solid sphere at $a$, and it has a thickness of $2a$. This means the outer boundary of the shell is at a distance of $a + 2a = 3a$ from the center. The mass of this thick shell is $2M$.

Our mission is to find the net gravitational field at a point exactly $3a$ away from the center. Notice that this point lies right on the outer surface of the spherical shell.

### The Magic of the Shell Theorem

To solve this, we rely on one of the most elegant principles in physics: **Newton's Shell Theorem**. The theorem states two powerful facts:
1. For a point outside a spherically symmetric mass distribution, the entire mass can be treated as if it were concentrated at the exact center.
2. For a point inside a uniform spherical shell, the net gravitational field due to the shell is exactly zero.

Since our point of interest is at $r = 3a$, it is outside the inner solid sphere and exactly on the outer surface of the spherical shell. Therefore, we can treat both masses as if they are point masses located at the center!

### Calculating the Individual Fields

Let's break the problem down by calculating the gravitational field contributed by each object independently.

**1. Field due to the Solid Sphere ($E_1$):**
The solid sphere has a mass $M$. Since our point is outside it, we use the standard formula for the gravitational field of a point mass:

$$E_1 = \frac{GM}{r^2}$$

Substituting $r = 3a$, we get:

$$E_1 = \frac{GM}{(3a)^2} = \frac{GM}{9a^2}$$

**2. Field due to the Spherical Shell ($E_2$):**
The spherical shell has a mass $2M$. Because our point is on its outer surface, the Shell Theorem allows us to treat its entire mass as being at the center as well:

$$E_2 = \frac{G(2M)}{r^2}$$

Substituting $r = 3a$, we get:

$$E_2 = \frac{G(2M)}{(3a)^2} = \frac{2GM}{9a^2}$$

### The Final Superposition

Gravitational fields are vectors. Because both the solid sphere and the spherical shell pull an object towards the center, their fields point in the exact same direction. To find the net gravitational field, we simply add their magnitudes together:

$$E_{net} = E_1 + E_2$$

$$E_{net} = \frac{GM}{9a^2} + \frac{2GM}{9a^2}$$

$$E_{net} = \frac{3GM}{9a^2}$$

Simplifying the fraction by dividing the numerator and denominator by 3, we arrive at our final, elegant answer:

$$E_{net} = \frac{GM}{3a^2}$$

This simple addition is a beautiful demonstration of the principle of superposition in physics!

Animated Solution for Physics - Gravitation: A solid sphere of mass $M$ and radius $a$ is surrounded by a uniform concentric spherical shell of thickness $2 a$ and $2 M$ . The gravitational field at distance $3 a$ from the centre will be

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Visualized Solution

\text{System Setup}

\text{Point of Interest}

E_1 = \frac{GM}{r^2}

E_2 = \frac{G(2M)}{r^2}

E_{net} = E_1 + E_2

\text{What if } r < 3a?

The Sigma Insight: Gravitational Field

Visualizing the Mass Distribution

The Magic of the Shell Theorem

Calculating the Individual Fields

The Final Superposition