If g is the acceleration due to gravity on the earth-s surface- the gain in the potential energy of an object of mass m raised from the surface of the earth to a height equal to the radius R of the earth iscfrac - 1 - 2 - mgR2mgRcfrac - 1 - 4 - mgRmgR

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Gain in potential energy -x2212-GMm2R-x2212-x2212-GMmR-GMmR-x2212-GMm2R-GMm2R-mgR22RAs-GM-gR2-12mgR

If $g$ is the acceleration due to gravity on the earth's surface, the gain in the potential energy of an object of mass $m$ raised from the surface of the earth to a height equal to the radius $R$ of the earth is
$\frac{1}{2} m g R$
$2 m g R$
$\frac{1}{4} m g R$
$m g R$

Gain in potential energy
$= (- \frac{G M m}{2 R}) - (- \frac{G M m}{R})$
$= \frac{G M m}{R} - \frac{G M m}{2 R}$
$= \frac{G M m}{2 R}$
$= \frac{m g R^{2}}{2 R} A s, G M = g R^{2}$
$= \frac{1}{2} m g R$

If g is the acceleration due to gravity on the earth's surface, the gain in the potential energy of an object of mass m raised from the surface of the earth to a height equal to the radius R of the earth is12mgR2mgR14mgRmgR

Gain in potential energy =(−GMm2R)−(−GMmR)=GMmR−GMm2R=GMm2R=mgR22RAs,GM=gR2=12mgR

If $g$ is the acceleration due to gravity on the earth's surface, the gain in the potential energy of an object of mass $m$ raised from the surface of the earth to a height equal to the radius $R$ of the earth is
$\frac{1}{2} m g R$
$2 m g R$
$\frac{1}{4} m g R$
$m g R$

Gain in potential energy
$= (- \frac{G M m}{2 R}) - (- \frac{G M m}{R})$
$= \frac{G M m}{R} - \frac{G M m}{2 R}$
$= \frac{G M m}{2 R}$
$= \frac{m g R^{2}}{2 R} A s, G M = g R^{2}$
$= \frac{1}{2} m g R$