A body of mass $m$ rises to height $h = R / 5$ from the earth’s surface, where $R$ is earth’s radius. If $g$ is acceleration due to gravity at earth’s surface, the increase in potential energy is

m g h

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\frac{6}{7} m g h

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\frac{3}{6} m g h

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\frac{5}{6} m g h

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Solution

The correct option is D $\frac{5}{6} m g h$
Given: $h = \frac{R}{5}$

Potential Energy at surface of the earth, $(P E_{1}) = - \frac{G M_{m}}{R}$

Potential Energy at height: $h = \frac{R}{5}$

$(P E_{2}) = - \frac{G M m}{R + \frac{R}{5}} = - \frac{5 G M m}{6 R}$

Increase in Potential energy,

$= P E_{2} - P E_{1}$

$= - \frac{5 G M m}{6 R} + \frac{G M m}{R}$

$= \frac{G M m}{R}$

(Using, $g = \frac{G M}{R^{2}}$ and $h = \frac{R}{5})$

$\Rightarrow P E_{2} - P E_{1} = \frac{5}{6} m g h$

Hence, option $(B)$ is correct.

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A body of mass m rises to height h=R/5 from the earth’s surface, where R is earth’s radius. If g is acceleration due to gravity at earth’s surface, the increase in potential energy is

The correct option is D 56mghGiven: h=R5 Potential Energy at surface of the earth, (PE1)=−GMmR Potential Energy at height:h=R5 (PE2)=−GMmR+R5=−5GMm6R Increase in Potential energy, =PE2−PE1 =−5GMm6R+GMmR =GMmR (Using, g=GMR2 and h=R5) ⇒PE2−PE1=56mgh Hence, option (B) is correct.

A body of mass $m$ rises to height $h = R / 5$ from the earth’s surface, where $R$ is earth’s radius. If $g$ is acceleration due to gravity at earth’s surface, the increase in potential energy is