Points to Ponder

When one object moves under the gravitational pull of another, two quantities stay constant: the total mechanical energy (kinetic plus potential) and the angular momentum. Linear momentum, however, is not conserved because an external gravitational force acts on the system.

Kepler’s second law (equal areas in equal times) is a direct consequence of angular momentum conservation. But this law isn’t unique to gravity — it holds true for any central force, where the force always points toward a fixed centre.

Kepler’s third law states $T^2 = K_S R^3$, where $K_S$ is the same constant for all planets orbiting the Sun in circular paths. The same relation applies to satellites orbiting the Earth, with a different constant.

An astronaut feels weightless inside a satellite not because gravity is weak up there. Both the astronaut and the satellite are in free fall toward the Earth, accelerating together at the same rate. That shared acceleration cancels the sensation of weight.

The gravitational potential energy for two particles separated by distance $r$ is $V = -\frac{G m_1 m_2}{r} + \text{constant}$. The constant is arbitrary; choosing it to be zero makes $V \to 0$ as $r \to \infty$, giving $V = -\frac{G m_1 m_2}{r}$. This choice does not affect the gravitational force.

Total mechanical energy = kinetic energy (always positive) + potential energy. With the zero of potential energy set at infinity, gravitational potential energy is negative. For a satellite in orbit, the total energy is negative, meaning it is bound to the Earth.

The familiar formula $mgh$ for gravitational potential energy near Earth’s surface is only an approximation. It comes from the difference in the exact expression $-\frac{GMm}{r}$ when $h$ is small compared to Earth’s radius. …