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 point where the gravitational field equals zero between two masses, we can use the principle of superposition. Let the two masses be \(m\) and \(9m\), placed at a distance \(R\) apart. Let the gravitational field be zero at a point at a distance \(x\) from mass \(m\). The distance from mass \(9m\) then becomes \(R-x\).

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

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

For the gravitational fields to cancel out:

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

Simplifying, we get: 

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

Taking the square root on both sides:

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

Cross-multiplying 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 is the sum of the potentials due to both masses:

Potential due to mass \(m\) at distance \(\frac{R}{4}\):

\(V_1 = -\frac{Gm}{\frac{R}{4}} = -\frac{4Gm}{R}\)

Potential due to mass \(9m\) at distance \(R - \frac{R}{4} = \frac{3R}{4}\):

\(V_2 = -\frac{G(9m)}{\frac{3R}{4}} = -\frac{12Gm}{R}\)

Total potential:

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

Thus, the correct answer is:

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

Answer 2

position of neutral point(zero gravitational field)
\(r_1=\frac{\sqrt(m_1)R}{\sqrt{m_1+\sqrt{m+2}}}=\frac{\sqrt{m}R}{\sqrt m+\sqrt{9m}}=\frac{R}{4}\)
\(r_2=R-\frac{R}{4}=\frac{3R}{4}\)
Now gravitational potential at point P 
\(V_p=-\frac{GM}{\frac{R}{4}}-\frac{9GM}{\frac{3R}{4}}\)
\(=-\frac{16GM}{R}\)
Therefore, the correct option is (D) : \(-\frac{16Gm}{R}\).