Energy required in very slowly moving a body of mass $m$ from a distance $2 R$ to $3 R$ (where $R$ is radius of earth) from centre of earth of mass $M$ is

\frac{G M m}{12 R}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{G M m}{3 R}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{G M m}{8 R}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{G M m}{6 R}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D $\frac{G M m}{6 R}$
Since the body moved from one point to other point very slowly, so kinetic energy of the body remains zero throughout the motion.

$∴ K E_{i} = K E_{f} = 0$

Mechanical energy at $2 R$ from centre of earth is given by

$E_{i} = P E_{i} + K E_{i} = - \frac{G M m}{2 R} + 0 = - \frac{G M m}{2 R}$

Similarly, mechanical energy at $3 R$ from centre of earth is

$E_{f} = P E_{i} + K E_{f} = - \frac{G M m}{3 R} + 0 = - \frac{G M m}{3 R}$

So, energy required,

$Δ E = E_{f} - E_{i} = - \frac{G M m}{3 R} - (- \frac{G M m}{2 R})$

$\Rightarrow Δ E = \frac{G M m}{6 R}$

Hence, option $(d)$ is correct.

Key Concept - The gravitational potential energy for $2$ point masses is given by $U = \frac{G m_{1} m_{2}}{R}$

Suggest Corrections

Similar questions

m

2 R

3 R

M =

R =

m

2 R

3 R

Join BYJU'S Learning Program

Explore more

Join BYJU'S Learning Program