Practicing Success

Target Exam

CUET

Subject

Physics

Chapter

Gravitation

Question:

There are two fixed heavy masses of magnitude M of high density at a distance 2d apart. On the axis, a small mass m moves in a circle of radius R in the yz plane between the heavy masses. Then the velocity of the small particle.

Options:

$\left\{\frac{2 G M}{\left(R^2+d^2\right)^{3 / 2}}\right\}^{1 / 2}$

$\left\{\frac{3 G M}{\left(R^2+d^2\right)^{3 / 2}}\right\}^{1 / 2}$

$\left\{\frac{4 G M}{\left(R^2+d^2\right)^{3 / 2}}\right\}^{1 / 2}$

$\left\{\frac{GM}{2\left(R^2+d^2\right)^{3 / 2}}\right\}^{1 / 2}$

Correct Answer:

$\left\{\frac{2 G M}{\left(R^2+d^2\right)^{3 / 2}}\right\}^{1 / 2}$

Explanation:

Force of attraction between M and m is F = $\frac{GMm}{R^2 + d^2}$

By symmetry Fx components will cancel.

∴ The net force which provide the centripetal force.

$2 F_y = 2 ~. \frac{G M m}{\left(R^2+d^2\right)} \times \frac{R}{\left(R^2+d^2\right)^{1 / 2}}$

$=2 R \frac{G M m}{\left(R^2+d^2\right)^{3 / 2}}$

$\Rightarrow \frac{m v^2}{R}=2 R \frac{G M m}{\left(R^2+d^2\right)^{3 / 2}} \Rightarrow  v=\left\{\frac{2 G M}{\left(R^2+d^2\right)^{3 / 2}}\right\}^{1 / 2}$