Define gravitational potential energy. derive expression for gravitational potential energy in the gravitational field of earth at distance from center of the earth

The work done $dW$ by an external force to move a mass $m$ by an infinitesimal distance $dr$ against the gravitational force is given by:

$$dW = F_{ext} \cdot dr$$

The gravitational force $F_g$ between Earth (mass $M$) and an object (mass $m$) at a distance $r$ from Earth's center is given by Newton's Law of Universal Gravitation:

$$F_g = -\frac{GMm}{r^2}$$

The negative sign indicates that the force is attractive, directed towards the center of Earth. To move the object without acceleration, the external force $F_{ext}$ must be equal in magnitude and opposite in direction to the gravitational force:

$$F_{ext} = -F_g = \frac{GMm}{r^2}$$

The gravitational potential energy $U(r)$ at a distance $r$ from the center of Earth is the total work done to bring the mass $m$ from infinity (where potential energy is conventionally zero) to the distance $r$.

$$U(r) = \int_{\infty}^{r} dW = \int_{\infty}^{r} F_{ext} \cdot dr$$

Substitute the expression for $F_{ext}$:

$$U(r) = \int_{\infty}^{r} \frac{GMm}{r^2} dr$$

$$U(r) = GMm \int_{\infty}^{r} r^{-2} dr$$ $$U(r) = GMm \left[ \frac{r^{-1}}{-1} \right]_{\infty}^{r}$$ $$U(r) = GMm \left[ -\frac{1}{r} \right]_{\infty}^{r}$$ $$U(r) = GMm \left( -\frac{1}{r} - \left(-\frac{1}{\infty}\right) \right)$$

Since $\frac{1}{\infty} = 0$:

$$U(r) = GMm \left( -\frac{1}{r} - 0 \right)$$ $$U(r) = -\frac{GMm}{r}$$

The expression for gravitational potential energy at a distance $r$ from the center of Earth is: