Physics · Ch 8 — Gravitation

Energy of an Orbiting Satellite

8.10

Energy of an Orbiting Satellite

For a satellite in a circular orbit around the Earth, the total mechanical energy is the sum of its kinetic energy (due to its motion) and its gravitational potential energy (due to its position in Earth's gravity). The key result is that this total energy is negative, which tells us the satellite is bound to the Earth — it cannot escape on its own.

Kinetic Energy in a Circular Orbit

We already know the orbital speed $v$ for a satellite at a distance $r = R_E + h$ from the Earth's centre (where $R_E$ is Earth's radius and $h$ is the altitude above the surface). From the condition that the gravitational force provides the necessary centripetal force:

\frac{G M_E m}{r^2} = \frac{m v^2}{r}

This gives $v^2 = \frac{G M_E}{r}$ . The kinetic energy $K$ is therefore:

K = \frac{1}{2} m v^2 = \frac{1}{2} m \left( \frac{G M_E}{r} \right) = \frac{G M_E m}{2r}

Since $r = R_E + h$ , we can write:

K = \frac{G M_E m}{2(R_E + h)}

This is equation (7.40) in the textbook. The kinetic energy is positive, as it always is for any moving object.

Potential Energy in a Circular Orbit

Gravitational potential energy is defined with the convention that it is zero at infinite separation. For two masses $m$ and $M_E$ separated by a distance $r$ , the potential energy is:

U = -\frac{G M_E m}{r}

For our satellite at distance $r = R_E + h$ from Earth's centre:

U = -\frac{G M_E m}{R_E + h}

This is equation (7.41). The potential energy is negative, reflecting the attractive nature of gravity. The closer the satellite is to Earth, the more negative (lower) its potential energy becomes.

Watch out

A common mistake is to forget the negative sign in the potential energy. The formula $U = -\frac{G M m}{r}$ is not optional — it is the definition. Without the negative sign, the total energy of a bound satellite would not be negative, and the entire analysis of bound orbits would break down.

Total Energy in a Circular Orbit

Adding the kinetic and potential energies:

E = K + U = \frac{G M_E m}{2(R_E + h)} - \frac{G M_E m}{R_E + h}

Factor out $\frac{G M_E m}{R_E + h}$ :

E = \frac{G M_E m}{R_E + h} \left( \frac{1}{2} - 1 \right) = -\frac{G M_E m}{2(R_E + h)}

This is equation (7.42). The total energy is negative. Notice that the magnitude of the potential energy is exactly twice the magnitude of the kinetic energy:

|U| = 2K

So the total energy is $E = -K = \frac{U}{2}$ .

E = -\frac{G M_E m}{2(R_E + h)}

Important

The total energy of a satellite in a circular orbit is negative and constant. This negative value is the signature of a bound system — the satellite cannot escape Earth's gravity unless it gains enough energy to make $E \ge 0$ .

Energy in an Elliptical Orbit

When the orbit is elliptical rather than circular, both the kinetic and potential energies vary continuously as the satellite moves along its path. At the perigee (closest point to Earth), the speed is highest, so kinetic energy is maximum and potential energy is minimum (most negative). At the apogee (farthest point), the speed is lowest, so kinetic energy is minimum and potential energy is maximum (least negative).

However, the total mechanical energy remains constant throughout the orbit. This is a direct consequence of gravity being a conservative force — no energy is lost or gained as the satellite moves. …

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