Problem 61 - Master Resource Book in JEE Main Physics

Question

A solid sphere is of density \(\rho\) and radius \(R\). The gravitational field at a distance \(r\) from the centre of the sphere, where \(r

Step-by-Step Solution

Answer

The correct answer is (d): \(\frac{4\pi G \rho r}{3}\).

Step 1: Understanding the Problem

We need to find the gravitational field inside a solid sphere of radius \( R \) and density \( \rho \) at a distance \( r \) from its center, where \( r < R \).

Step 2: Applying Gauss's Law for Gravity

Symmetry allows us to use Gauss's Law for Gravity: \( \Phi_g = \oint \mathbf{g} \, d\mathbf{A} = -4\pi G M_{enc} \), where \( M_{enc} \) is the mass enclosed by a Gaussian surface of radius \( r \).

Step 3: Calculating the Enclosed Mass

The mass \( M_{enc} \) within radius \( r \) is \( M_{enc} = \rho \cdot V = \rho \cdot \frac{4}{3} \pi r^3 \).

Step 4: Solving for the Gravitational Field

Use the formula for gravitational potential: \( g = -\frac{4\pi G M_{enc}}{4\pi r^2} \). Substituting \( M_{enc} = \rho \cdot \frac{4}{3} \pi r^3 \) into the equation gives \( g = \frac{4\pi G \rho r}{3} \).

Step 5: Identifying the Correct Answer

The gravitational field inside a solid sphere at a distance \( r \) from its center is given as \( g = \frac{4\pi G \rho r}{3} \). This matches option (d).

Key Concepts

Gauss's Law for Gravity

Gauss's Law for Gravity is a powerful tool that helps us understand how gravitational fields are distributed. It is analogous to Gauss's law in electrostatics. In its simplest form, the law involves the gravitational flux, which is the product of the gravitational field, \( \mathbf{g} \), and the area, \( d\mathbf{A} \), over a closed surface. Mathematically, it can be expressed as \( \Phi_g = \oint \mathbf{g} \, d\mathbf{A} = -4\pi G M_{enc} \), where \( G \) is the gravitational constant and \( M_{enc} \) is the mass enclosed within a Gaussian surface.
To solve problems using Gauss's Law, particularly in cases involving spherical symmetry, you select a Gaussian surface that matches the symmetry, like a sphere, and calculate the gravitational field based on the enclosed mass.
This symmetric distribution makes the analysis straightforward, because the gravitational field at any point on the Gaussian surface is constant in magnitude.

Mass Enclosed in a Sphere

To find the mass enclosed in a sphere, you must consider the volume of the sphere and its density. In scenarios where you're analyzing a sphere with radius \( r \) within a larger sphere of radius \( R \), the enclosed mass \( M_{enc} \) is derived from the volume of the inner sphere and the density.
The formula for the mass enclosed is \( M_{enc} = \rho \times V = \rho \times \frac{4}{3} \pi r^3 \). Here, \( \rho \) represents the density of the sphere, and \( \frac{4}{3} \pi r^3 \) is the volume of a sphere with radius \( r \).
This calculation is crucial because it tells us how much mass contributes to the gravitational field at distance \( r \) from the sphere's center.

Density of Sphere

Density is a measure of how much mass is contained in a given volume. For a uniform solid sphere, like the one in our exercise, the density \( \rho \) is constant throughout the sphere.
When calculating gravitational properties, knowing the density allows you to determine the mass contained within any portion of the sphere. Since the sphere is uniform, the density simplifies calculations as the same value of \( \rho \) applies no matter which spherical slice or section you're analyzing.
Without this uniformity, calculating the gravitational field inside the sphere would become significantly more complex.

Gravitational Potential

Gravitational potential refers to the potential energy per unit mass at a point in a field. It is linked to the gravitational field, \( g \), via the formula \( g = -\frac{4\pi G M_{enc}}{4\pi r^2} \). When calculating inside a sphere, you substitute the enclosed mass to find the gravitational field.
After substituting \( M_{enc} = \rho \cdot \frac{4}{3} \pi r^3 \) into the equation, you get \( g = \frac{4\pi G \rho r}{3} \).
This gives a clear formula for the gravitational field as a function of the distance from the center of the sphere, demonstrating how the potential changes with \( r \).

Sphere Radius and Gravitational Field

The sphere's radius influences the gravitational field experienced inside it. In the scenario where \( r The gravitational field inside the sphere is determined by both this radius and the amount of mass it encloses. As shown in the provided solution, the gravitational field formula \( g = \frac{4\pi G \rho r}{3} \) involves the radius \( r \), indicating that the strength of the gravitational field is directly proportional to the distance from the center of the sphere within these confines.
Therefore, as you move further from the center, up to the radius \( R \), the gravitational field strengthens linearly with \( r \), underlining the role of the spherical radius in gravitational calculations.