Statement I : For a mass M kept at the centre of a cube of side a, the flux of gravitational field p

Q: Why is the gravitational flux $4\pi GM$ and not just $M/G$?

In electrostatics, the proportionality constant is $1/(4\pi\epsilon_0)$. In gravitation, the constant is $G$. If we equate their roles, $1/(4\pi\epsilon_0) \equiv G$, which means $1/\epsilon_0 \equiv 4\pi G$. Thus, the electrostatic flux $q/\epsilon_0$ translates directly to $M \cdot 4\pi G$ in gravitation.

Q: Does the mass have to be exactly at the center of the cube for the flux to be $4\pi GM$?

Not at all! As long as the mass is completely enclosed anywhere inside the closed surface, the total flux will remain exactly $4\pi GM$. The position of the mass only affects the flux through individual faces, not the total sum.

Q: What if the surface was a sphere instead of a cube?

The total flux would still be exactly $4\pi GM$. This is the magic of Gauss's law—it is completely independent of the size or shape of the enclosing surface.

Q: Why is there sometimes a negative sign in the formal definition of gravitational flux?

The negative sign indicates that the gravitational field is always attractive (pointing inwards towards the mass), whereas the standard area vector of a closed surface points outwards. The magnitude of the flux, however, is $4\pi GM$.

Q: Can we use this gravitational Gauss's law to find the gravitational field of a solid sphere?

Absolutely! Just like in electrostatics, you can use a spherical Gaussian surface to easily derive the gravitational field both inside and outside a uniform solid sphere without doing complex integrations.

Question

Statement I : For a mass $M$ kept at the centre of a cube of side $a$, the flux of gravitational field passing through its sides is $4\pi GM$.

Statement II : If the direction of a field due to a point source is radial and its dependence on the distance $r$ from the source is given as $1/r^2$, its flux through a closed surface depends only on the strength of the source enclosed by the surface and not on the size or shape of the surface.

Options:

(A) Statement I is true, Statement II is true; Statement II is the correct explanation of Statement I

(B) Statement I is true, Statement II is true; Statement II is not the correct explanation of Statement I

(C) Statement I is true, Statement II is false

(D) Statement I is false, Statement II is true

Sigma Percentile · Accepted Answer

### The Symphony of Physics

One of the most beautiful aspects of physics is how different phenomena, seemingly unrelated, share the exact same mathematical skeleton. When you study electrostatics, you learn about Coulomb's Law. When you study gravitation, you learn about Newton's Law of Universal Gravitation.

Notice the striking similarity:
$$F_E = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r^2}$$
$$F_G = G \frac{m_1 m_2}{r^2}$$

Both forces are radial, and both follow the **inverse-square law** (they weaken proportionally to $1/r^2$). Because of this identical mathematical structure, any theorem that applies to the electric field will have a direct, perfectly analogous theorem in the gravitational field.

### The Concept of Flux

Flux, in simple terms, is a measure of how many "field lines" pierce through a given surface. Mathematically, the gravitational flux $\Phi_g$ through a closed surface is defined as the surface integral of the gravitational field $\vec{g}$:

$$\Phi_g = \oint \vec{g} \cdot d\vec{A}$$

### Gauss's Law: The Shape-Shifter

In electrostatics, Gauss's Law states that the total electric flux through any closed surface is equal to the enclosed charge divided by $\epsilon_0$.

$$\oint \vec{E} \cdot d\vec{A} = \frac{q_{	ext{enclosed}}}{\epsilon_0}$$

Why does the shape of the surface not matter? Imagine a point charge radiating field lines. As you move further away, the surface area of a sphere increases by $r^2$. However, the field strength decreases by exactly $1/r^2$. When you multiply them together to find the flux, the $r^2$ terms perfectly cancel out!

This cancellation is the pure magic of the inverse-square law. As long as the field is radial and proportional to $1/r^2$, the flux through *any* closed surface depends **only** on the source enclosed, not on the size or shape of the surface. This is exactly what **Statement II** asserts, making it fundamentally true.

### The Gravitational Equivalent

To find the gravitational equivalent of Gauss's Law, we just need to map the constants. By comparing the force equations, we can see that the electrostatic constant $\frac{1}{4\pi\epsilon_0}$ is perfectly analogous to the gravitational constant $G$.

This means that $\frac{1}{\epsilon_0}$ is analogous to $4\pi G$.

If we replace $q$ with $M$ and $\frac{1}{\epsilon_0}$ with $4\pi G$ in the electrostatic Gauss's Law, we get the Gravitational Gauss's Law:

$$\oint \vec{g} \cdot d\vec{A} = -4\pi G M_{	ext{enclosed}}$$

*(Note: The negative sign simply indicates that the gravitational field is always attractive, pointing inwards, while the area vector points outwards. The magnitude of the flux is $4\pi GM$.)*

### Concluding the Problem

Now, let's look at **Statement I**. It claims that the flux through a cube enclosing a mass $M$ is $4\pi GM$.

Since a cube is a closed surface, and we have just established that the shape of the surface does not matter (thanks to the inverse-square nature of the field), the total flux must indeed be $4\pi GM$. Therefore, **Statement I** is true.

Furthermore, the reason *why* Statement I is true is exactly the principle described in Statement II. The radial, inverse-square nature of the field guarantees that the flux is independent of the cubic shape.

Thus, both statements are true, and Statement II is the correct, fundamental explanation for Statement I.

Select Answer:

Visualized Solution

$Visualizing the Setup$

$Gauss’s Law Analogy$

$Analyzing Statement II$

$Analyzing Statement I$

$Final Conclusion$

The Sigma Insight: Gravitational Force

The Symphony of Physics

The Concept of Flux

Gauss's Law

The Gravitational Equivalent

Concluding the Problem

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Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Visualizing the Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Gauss’s Law Analogy

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Analyzing Statement II

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Analyzing Statement I

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Conclusion