The change in the gravitational potential energy when a body of mass m is raised to a height nR above the surface of the earth is ( here R is the radius of the earth )

$(\frac{n}{n + 1})$ mgR

$(\frac{n}{n - 1})$ mgR

nmgR

$\frac{mgR}{n}$

Solution

$(\frac{n}{n + 1})$ mgR

Gravitational potential energy of mass m at any point at a distance r from the centre of earth is

U = $- \frac{G Mm}{r}$

At the surface of earth r = R,

∴ U_s = $- \frac{G Mm}{R}$

= - mgR $(∵ g = \frac{GM}{R^{2}})$

At the height h = nR from the surface of earth

r = R + h

= R + nR

r = R ( 1 + n )

∴ U_h = $- \frac{G Mm}{R (1 + n)}$

= $- \frac{mg R}{(1 + n)}$

Change in gravitational potential energy is

ΔU = U_h $-$ U_s

= $- \frac{m gR}{1 + n} - (- m gR)$

= $- \frac{m gR}{1 + n} (1 - \frac{1}{1 + n})$

ΔU = m gR $(\frac{n}{1 + n})$

Some More Questions From Gravitation Chapter

For Daily Free Study Material Join wiredfaculty