Practicing Success

Target Exam

CUET

Subject

Physics

Chapter

Gravitation

Question:

Two bodies of masses m and M are placed a distance d apart. The gravitational potential at the position where the gravitational field due to them is zero is V.

Options:

V = $-\frac{G}{d}(m + M)$

V = $-\frac{Gm}{d}$

V = $-\frac{GM}{d}$

V = $-\frac{G}{d}(m + M)^2$

Correct Answer:

V = $-\frac{G}{d}(m + M)^2$

Explanation:

Let gravitational field be zero at a point lying at distance x from M. Then,

$\frac{G M}{x^2}=\frac{G m}{(d-x)^2}$

$\Rightarrow \frac{d-x}{x}=\sqrt{\frac{m}{M}}$

$\Rightarrow \frac{d}{x}-1=\sqrt{\frac{m}{M}}$

$\Rightarrow x=\left(\frac{\sqrt{M}}{\sqrt{M}+\sqrt{m}}\right) d$              ......(1)

$\Rightarrow(d-x)=\left(\frac{\sqrt{M}}{\sqrt{M}+\sqrt{m}}\right) d$   ......(2)

Since, $V_p=-\frac{G m}{d-x}-\frac{G m}{x}$       ......(3)

Substituting (1) and (2) in (3), we get

$V_p=-\frac{G}{d}(\sqrt{m}+\sqrt{M})^2$