A satellite of mass m is orbiting around the earth at a height equal to twice the radius of the earth (R). Its potential energy is given by :
$- 2 m g R$
$- 2 m g \frac{R}{3}$
$- m g \frac{R}{2}$
$- m g \frac{R}{3}$

Question

A satellite of mass m is orbiting around the earth at a height equal to twice the radius of the earth (R). Its potential energy is given by :
$- 2 m g R$
$- 2 m g \frac{R}{3}$
$- m g \frac{R}{2}$
$- m g \frac{R}{3}$

Toppr Admin · Accepted Answer

Gravitational potential energy at distance r from the center of the earth is given as-P-x2212-GMEmr-x2212-GMEm3R-x2212-GMEm-R-3-R2-x27F9-P-x2212-mgR3Option D is correct-

A satellite of mass m is orbiting around the earth at a height equal to twice the radius of the earth (R). Its potential energy is given by :−2mgR−2mgR3−mgR2−mgR3

Gravitational potential energy at distance r from the center of the earth is given as,P=−GMEmr=−GMEm3R=−GMEm(R/3)R2⟹P=−mgR3Option D is correct.

A satellite of mass m is orbiting around the earth at a height equal to twice the radius of the earth (R). Its potential energy is given by :
$- 2 m g R$
$- 2 m g \frac{R}{3}$
$- m g \frac{R}{2}$
$- m g \frac{R}{3}$

Gravitational potential energy at distance $r$ from the center of the earth is given as,
$P = - \frac{G M_{E} m}{r} = - \frac{G M_{E} m}{3 R} = - \frac{G M_{E} m (R / 3)}{R^{2}}$
$⟹ P = - m g \frac{R}{3}$
Option D is correct.