Gravitational potential of the body of mass m at a height h from surface of earth of radius R is (Take $g =$ acceleration due to gravity at earth's surface)
$- g (R + h)$
$- g (R - h)$
$g (R + h)$
$g (R - h)$

Question

Gravitational potential of the body of mass m a height h from surface of earth of radius R is -Take g - acceleration due to gravity earth-s surface-g -R - h-g -R - h-g -R - h-g -R - h-

Toppr Admin · Accepted Answer

If a point mass m is placed at a height h from surface of earth, the potential energy is
$U_{h} = - \frac{G M m}{(R + h)} = \frac{- g R^{2} m}{R (1 + \frac{h}{R})} = \frac{- g R^{2} m}{R} {(1 + \frac{h}{R})}^{- 1}$
$(∵ g = \frac{G M}{R^{2}})$

$U_{h} = \frac{- g R^{2} m (R - h)}{R^{2}}$ $= - g m (R - h)$

$\Rightarrow V = \frac{U_{h}}{m} = \frac{- g m (R - h)}{m} = - g (R - h)$

Gravitational potential of the body of mass m at a height h from surface of earth of radius R is (Take g= acceleration due to gravity at earth's surface)−g(R+h)−g(R−h)g(R+h)g(R−h)

If a point mass m is placed at a height h from surface of earth, the potential energy isUh=−GMm(R+h)=−gR2mR(1+hR)=−gR2mR(1+hR)−1(∵g=GMR2)Uh=−gR2m(R−h)R2=−gm(R−h)⇒V=Uhm=−gm(R−h)m=−g(R−h)

Gravitational potential of the body of mass m at a height h from surface of earth of radius R is (Take $g =$ acceleration due to gravity at earth's surface)
$- g (R + h)$
$- g (R - h)$
$g (R + h)$
$g (R - h)$