Question

A body of mass m is raised to a height 10R from the surface of the earth, where R is the radius of the earth. The increase in potential energy is (G = universal constant of gravitation, M = mass of the earth and g = acceleration due to gravity)
[MHT CET 2014]

Moderate

Solution

Potential energy at the earth surface $= \frac{- G M m}{R}$
Potential energy at a height h above the earth's surface
$= \frac{- G M m}{(R + h)}$ …(i)
$∴$ Change m' potential energy, $Δ PE = \frac{- G M m}{(R + h)} - (- \frac{G M m}{R})$
$= \frac{G_{↑} M m h}{R (R + h)}$ …(ii)
Substituting the value of h = 10R in Eq. (i), we get
$Δ PE = \frac{G M m \times 10 R}{R (R + 10 R)} = \frac{10 G M m}{11 R}$

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