Gravitational Potential Energy

Question

A 500 kg payload is carried by rocket from the Earth's surface to a great height where it is effectively free of Earth's gravity. Taking g as 9.8 m/s squared and Earth's radius as 6400 km, how much work is done against gravity?

RankGuru · Accepted Answer

NCERT Class 11, Chapter 8 (Gravitation) defines the gravitational potential energy of a mass $m$ at the Earth's surface as $U = -\dfrac{GMm}{R}$, with the convention that potential energy is zero at \infty where gravity no longer acts. Moving the payload to a great height where gravity is effectively negligible is equivalent to raising it to \infty, so the work done against gravity equals the change in potential energy, $W = U_\infty - U_{surface} = 0 - \left(-\dfrac{GMm}{R}ight) = \dfrac{GMm}{R}$. Using the surface relation $g = \dfrac{GM}{R^2}$, this conveniently becomes $W = mgR$. Substituting the data: $W = 500 	imes 9.8 	imes 6.4 	imes 10^{6} = 3.136 	imes 10^{10}\,	ext{J} \approx 3.14 	imes 10^{10}\,	ext{J}$. The option $3.14 	imes 10^{7}\,	ext{J}$ uses the radius in kilometres rather than metres. The option $6.27 	imes 10^{10}\,	ext{J}$ doubles the value, as if computing $mgR$ times two by mistakenly adding a factor for escape. The option $1.57 	imes 10^{10}\,	ext{J}$ halves the correct result. As a plausibility check, lifting half a tonne completely out of Earth's gravity well should demand a very large energy on the order of $10^{10}\,	ext{J}$, which matches the computed value.

Gravitational Potential Energy

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