The radius of the earth is R and acceleration due to gravity at its surface is g- If a body of mass m is sent to a height R -6 from the earth-s surface- the potential energy increases byA- mgR-3B- mg R R -7C- mgR-8D- mg R

The correct option is B mgR7Initial potential energy (at earth's surface) , U_1 = GMmR= nbsp; gR^2 mR = mgR Final potential energy, U_2 = GMm(R+h) = ...

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Standard X

Physics

Potential Energy of an Object at a Height

The radius of...

Question

The radius of the earth is $R$ and acceleration due to gravity at its surface is $g$ . If a body of mass $m$ is sent to a height $\frac{R}{6}$ from the earth's surface, the potential energy increases by

\frac{m g R}{3}

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\frac{m g R}{7}

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\frac{m g R}{8}

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m g R

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Solution

The correct option is B $\frac{m g R}{7}$
Initial potential energy (at earth's surface) , $U_{1} = - \frac{G M m}{R} = \frac{- g R^{2} m}{R} = - m g R$

Final potential energy, $U_{2} = - \frac{G M m}{(R + h)} = \frac{- g R^{2} m}{(R + \frac{R}{6})} = - \frac{6}{7} m g R U_{2} - U_{1} = - \frac{6}{7} m g R - [- m g R] = \frac{m g R}{7}$

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The radius of the earth is R and acceleration due to gravity at its surface is g. If a body of mass m is sent to a height R6 from the earth's surface, the potential energy increases by

The correct option is B mgR7Initial potential energy (at earth's surface) , U1=−GMmR=−gR2mR=−mgR Final potential energy, U2=−GMm(R+h)=−gR2m(R+R6)=−67mgRU2−U1=−67mgR−[−mgR]=mgR7

The radius of the earth is $R$ and acceleration due to gravity at its surface is $g$ . If a body of mass $m$ is sent to a height $\frac{R}{6}$ from the earth's surface, the potential energy increases by