Question

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 potential energy of the body is –

(2)$\frac { \mathrm { mgR } } { 4 }$

• A. ΔU = mg(3R)
• B. ΔU = mg(R)
• C. ΔU = -mg(3R)
• D. ΔU = -mg(R)

Answer 1

Answer: (B)

ΔU = mg(R)

Answer 2

Change in potential energy

DU = $\frac { \mathrm { mgh } } { 1 + \mathrm { h } / \mathrm { R } }$

given h = 3R

so DU = $\frac { 3 } { 4 } \mathrm { mgR }$