Gravitational Potential Energy

NCERT Class 11, Chapter 8 (Gravitation) defines the gravitational potential energy of a mass $m$ at the Earth's surface as $U=-\frac{GMm}{R}$, 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, $W=U_{\infty }-U_{surface}=0-\left(-\frac{GMm}{R}\right)=\frac{GMm}{R}$. Using the surface relation $g=\frac{GM}{R^2}$, this conveniently becomes $W=mgR$. Substituting the data: $W=500\times 9.8\times 6.4\times 10^6=3.136\times 10^{10}\text{J}\approx 3.14\times 10^{10}\text{J}$. The option $3.14\times 10^7\text{J}$ uses the radius in kilometres rather than metres. The option $6.27\times 10^{10}\text{J}$ doubles the value, as if computing $mgR$ times two by mistakenly adding a factor for escape. The option $1.57\times 10^{10}\text{J}$ 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 $10^{10}\text{J}$, which matches the computed value.