Question

Two bodies of mass m and 9m are placed at a distance of R. The gravitational potential on the line joining the bodies where the gravitational fied equals zero, will be (G = gravitational contant):

• A. \(-\frac{20 Gm}{R}\)
• B. \(-\frac{8Gm}{R}\)
• C. \(-\frac{12Gm}{R}\)
• D. \(-\frac{16Gm}{R}\)

Answer 1

The Correct Option is D

To find the gravitational potential on the line joining two masses where the gravitational field equals zero, we need to first determine the point where the gravitational field due to both masses is equal and opposite, thus cancelling each other out.

Let's assume the two masses, \( m \) placed at \( A \) and \( 9m \) at \( B \), are separated by a distance \( R \). Let the point where the gravitational field equals zero be at a distance \( x \) from mass \( m \). Thus, it is \( (R - x) \) from mass \( 9m \).

The gravitational field due to a mass \( M \) at a distance \( r \) is given by:

\(E = \frac{GM}{r^2}\)

For the net gravitational field to be zero at point \( P \), the magnitudes should be equal:

\(\frac{Gm}{x^2} = \frac{G \times 9m}{(R-x)^2}\)

By simplifying:

\(\frac{1}{x^2} = \frac{9}{(R-x)^2}\)

Taking the square root of both sides, we get:

\(\frac{1}{x} = \frac{3}{R-x}\)

Rearranging gives:

\(R-x = 3x\)

Solve for \( x \):

\(R = 4x \Rightarrow x = \frac{R}{4}\)

Now, calculate the gravitational potential at this point. The potential \( V \) due to a mass \( M \) at a distance \( r \) is:

\(V = -\frac{GM}{r}\)

The total potential \( V \) at point \( P \) is the sum of the potentials due to both masses:

1. Potential due to mass \( m \): \(V_1 = -\frac{Gm}{x} = -\frac{Gm}{R/4} = -\frac{4Gm}{R}\)
2. Potential due to mass \( 9m \): \(V_2 = -\frac{G \times 9m}{(R-x)} = -\frac{G \times 9m}{(3R/4)} = -\frac{12Gm}{R}\)

Total potential at point \( P \):

\(V = V_1 + V_2 = -\frac{4Gm}{R} - \frac{12Gm}{R} = -\frac{16Gm}{R}\)

Therefore, the correct answer is:

\(-\frac{16Gm}{R}\)