70.A body of mass'm' rises to a height R/5 from earths surface then the increase in potential energy is

Open in App

Solution

Gravitational potential energy at the earth's surface:

$P E_{1} = - \frac{G M m}{R}$

Gravitational potential energy at a height of R/5 from the earth's surface:

$P E_{2} = - \frac{G M m}{R + \frac{R}{5}} = - \frac{5 G M m}{6 R}$

Increase in potential energy:

$P E_{2} - P E_{1} = - \frac{5 G M m}{6 R} + \frac{G M m}{R} = - \frac{5 G M m}{6 R} + \frac{6 G M m}{6 R} = \frac{G M m}{6 R} = \frac{m g R}{6} [\sin c e \frac{G M}{R} = g R]$

Suggest Corrections

Similar questions

A body of mass m is taken from earth surface to the height h equal to radius of earth. The increase in potential energy will be

m

h = R / 5

R

g

m

R / 5

(R \to

m

R / 3

R

n R

(\frac{n}{n + a}) m g R

a =

Join BYJU'S Learning Program