A body of mass m kg. starts falling from a point 2R above the Earth's surface. Its kinetic energy when it has fallen to a point 'R' above the Earth's surface [R-Radius of Earth, M-Mass of Earth, G-Gravitational Constant]

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

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

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

Open in App

Solution

The correct option is B

Potential energy $U = \frac{- G M m}{r} = \frac{G M m}{R + h}$
$U_{i n i t i a l} = - \frac{G M m}{3 R} a n d U_{f i n a l} = - \frac{- G M m}{2 R}$
Loss in PE=gain in KE= $\frac{G M m}{2 R} - \frac{G M m}{3 R} = \frac{G M m}{6 R}$

Suggest Corrections

Similar questions

A body of mass m kg starts falling from a point 2R above the earth's surface. Its kinetic energy when it has fallen to a point 'R' above the earth's surface [R-Radius of earth]

A body of mass m kg. starts falling from a point 2R above the earth’s surface. Its kinetic energy when it has fallen to a point ‘R’ above the eart’s surface [R – Radius of earth, M – Mass of earth, G – Gravitational constant]

Q. A body of mass

m k g

starts falling from a distance

2 R

above the earth's surface. What is its kinetic energy when it reaches a distance

^{'} R^{'}

above the surface of the earth of radius

R

and mass

M

Q. Assuming the radius of the earth as

R

, the change in gravitational potential energy of a body of mass

m

, where it is taken from the earth's surface to a height

3 R

above its surface is

A body of mass m is placed on the earth's surface. It is taken from the earth's surface to a height h=3R. The change in gravitational energy of the body is

Join BYJU'S Learning Program

A body of mass m kg. starts falling from a point 2R above the Earth's surface. Its kinetic energy when it has fallen to a point 'R' above the Earth's surface [R-Radius of Earth, M-Mass of Earth, G-Gravitational Constant]

The correct option is B Potential energy U=−GMmr=GMmR+h Uinitial=−GMm3R and Ufinal=−−GMm2R Loss in PE=gain in KE=GMm2R−GMm3R=GMm6R

The correct option is B

Potential energy $U = \frac{- G M m}{r} = \frac{G M m}{R + h}$
$U_{i n i t i a l} = - \frac{G M m}{3 R} a n d U_{f i n a l} = - \frac{- G M m}{2 R}$
Loss in PE=gain in KE= $\frac{G M m}{2 R} - \frac{G M m}{3 R} = \frac{G M m}{6 R}$