Practicing Success

Target Exam

CUET

Subject

Physics

Chapter

Gravitation

Question:

Two identical hypothetical isolated planets A & B each of mass M have the coordinates as shown. The minimum speed of projection of a body from A to B is

Options:

$\sqrt{\frac{GM}{a}}$

$\sqrt{\frac{2GM}{3 a}}$

$\sqrt{\frac{GM}{3 a}}$

None of these

Correct Answer:

$\sqrt{\frac{2GM}{3 a}}$

Explanation:

When the body just passes the point P have the net gravitational field force/intensity is zero, it will be attached towards the planet B. That means the velocity of projection inorder to escape from A is that required to just reach the point P.

The potential difference between M & P is 

$V=\left|V_p-V_M\right|=\left\{-\frac{G M}{a}-\frac{G M}{2 a}\right\} - \left\{-\frac{G M}{(3 / 2) a}-\frac{G M}{(3 / 2) a}\right\}$

$\Rightarrow V=\left|-\frac{3 G M}{2 a}+\frac{4 G M}{3 a}\right|=\frac{G M}{6 a}$

∴    $\left|\Delta U_{g r}\right|=\left|K E_{M \rightarrow P}\right| \Rightarrow m V=\frac{1}{2} m v^2$

$\Rightarrow v=\sqrt{2 V}=\sqrt{\frac{G M}{3 a}}$