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Standard XII

Physics

The gravitati...

Question

The gravitational potential energy of a body of mass 'm' at the earth's surface $- m g R_{e}$ . Its gravitational potential energy at a height $R_{e}$ from the earth's surface will be (Here $R_{e}$ is the radius of the earth)

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Solution

The correct option is D

$△ U = U_{2} - U_{1} = \frac{m g h}{1 + \frac{h}{R_{e}}} = \frac{m g R_{e}}{1 + \frac{R_{e}}{R_{e}}} = \frac{m g R_{e}}{2}$
$\Rightarrow U_{2} - (- m g R_{e}) = \frac{m g R_{e}}{2} \Rightarrow U_{2} = - \frac{1}{2} m g R_{e}$

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The gravitational potential energy of a body of mass 'm' at the earth's surface −mgRe. Its gravitational potential energy at a height Re from the earth's surface will be (Here Re is the radius of the earth)

The correct option is D △U=U2−U1=mgh1+hRe=mgRe1+ReRe=mgRe2 ⇒U2−(−mgRe)=mgRe2⇒U2=−12mgRe

The gravitational potential energy of a body of mass 'm' at the earth's surface $- m g R_{e}$ . Its gravitational potential energy at a height $R_{e}$ from the earth's surface will be (Here $R_{e}$ is the radius of the earth)

The correct option is D

$△ U = U_{2} - U_{1} = \frac{m g h}{1 + \frac{h}{R_{e}}} = \frac{m g R_{e}}{1 + \frac{R_{e}}{R_{e}}} = \frac{m g R_{e}}{2}$
$\Rightarrow U_{2} - (- m g R_{e}) = \frac{m g R_{e}}{2} \Rightarrow U_{2} = - \frac{1}{2} m g R_{e}$