Gravitation - Gravitational Potential Energy

Calculate the change in gravitational potential energy when a body of mass $m$ is raised from the surface of the Earth to a height $h$ equal to the radius of the Earth $R_e$.

The initial potential energy at the surface is $U_i = -\frac{GM_em}{R_e}$. The final potential energy at height $h = R_e$ (total distance $r = 2R_e$) is $U_f = -\frac{GM_em}{2R_e}$. The change in potential energy is $\Delta U = U_f - U_i = -\frac{GM_em}{2R_e} - (-\frac{GM_em}{R_e}) = \frac{GM_em}{2R_e}$. Since $g = \frac{GM_e}{R_e^2}$, we can substitute $GM_e = gR_e^2$, giving $\Delta U = \frac{m(gR_e^2)}{2R_e} = \frac{1}{2}mgR_e$.

This demonstrates that for large heights, the simple $mgh$ formula is inaccurate. Using $mgh$ would yield $mgR_e$, which is double the actual value required to reach that altitude.

Find the gravitational potential at a point on the surface of the Earth. (Given: $M_e = 6 \times 10^{24} \text{ kg}$, $R_e = 6.4 \times 10^6 \text{ m}$, $G = 6.67 \times 10^{-11} \text{ N m}^2\text{kg}^{-2}$)

Using the formula for gravitational potential $V = -\frac{GM_e}{R_e}$: $V = -\frac{6.67 \times 10^{-11} \times 6 \times 10^{24}}{6.4 \times 10^6}$ $V \approx -6.25 \times 10^7 \text{ J/kg}$

Gravitational potential represents the work done per unit mass. The negative value indicates that work must be done against the gravitational field to move the mass to infinity.