Question

The gravitational field due to a mass distribution is \( E = \frac{K}{x^3} \) along x-direction (K is a constant). Taking the gravitational potential to be zero at infinity, its value at a distance \( x \) is:

• A. \( \frac{K}{x} \)
• B. \( \frac{K}{2x} \)
• C. \( \frac{K}{x^2} \)
• D. \( \frac{K}{2x^2} \)

Hint:

Gravitational potential is the negative integral of the gravitational field.

Answer 1

The Correct Option is D

The gravitational potential \( V \) is related to the gravitational field \( E \) by: \[ E = -\frac{dV}{dx} \] We are given that \( E = \frac{K}{x^3} \), and we need to find \( V \). 
Integrating \( E \) with respect to \( x \), we get: \[ V = -\int E \, dx = -\int \frac{K}{x^3} \, dx = \frac{K}{2x^2} \] Thus, the gravitational potential at distance \( x \) is \( \frac{K}{2x^2} \). 
Final Answer: \( \frac{K}{2x^2} \)