Quick questions on Gravitational potential energy and escape velocity explained: HSC Physics Module 5

Near Earth's surface, gravitational field strength $g$ is approximately constant, and $U = mgh$ works fine. At astronomical scales, $g$ falls off as $1/r^2$, so the potential energy must be obtained by integrating the gravitational force from infinity inward:

Imagine releasing a stationary object from far away. Gravity does positive work pulling it inward, increasing kinetic energy. By conservation of energy, potential energy must decrease. Since we set $U = 0$ at infinity, $U$ becomes negative as the object approaches the source.

Escape velocity $v_{\text{esc}}$ is the minimum speed needed at a distance $r$ for an object to reach infinity with zero remaining kinetic energy. By conservation of energy:

When a mass moves from $r_1$ to $r_2$:

$U = -G M m / r$, not $+G M m / r$. The sign matters for energy conservation.

This approximation only works for small altitude changes near a planet's surface where $g$ is roughly constant. For satellites and planetary orbits, use the radial formula.

Zero potential energy is at infinity, not at the planet's surface or centre.

Orbital velocity at radius $r$ is $v_{\text{orb}} = \sqrt{G M / r}$. Escape velocity is $\sqrt{2}$ times larger: $v_{\text{esc}} = \sqrt{2} \cdot v_{\text{orb}}$.

It does not. A pebble and a rocket need the same escape velocity (though the rocket needs more total energy because $E = \frac{1}{2} m v^2$).