Orbits of planets and satellites (AQA A-Level Physics): Revision Notes

7.2.4 Orbits of planets and satellites

Kepler's Third Law

Kepler's third law states that the square of the orbital period ($T$) is directly proportional to the cube of the radius ($r$) of an orbit. Mathematically, this is expressed as:

$$T^2 \propto r^3$$

This relationship can be derived using the following steps:

1. Equating Gravitational and Centripetal Forces For an object orbiting a mass $M$, it experiences a gravitational force towards the centre of this mass, which acts as the centripetal force.

Thus, we can equate the gravitational and centripetal forces:

$$\frac{mv^2}{r} = \frac{GMm}{r^2}$$

where $m$ is the mass of the orbiting object, $G$ is the gravitational constant, and $M$ is the mass at the centre.

2. Solving for $v^2$ Rearrange this equation to make $v^2$ the subject:

$$v^2 = \frac{GM}{r}$$

3. Expressing Velocity in Terms of $r$ and $T$ The orbital velocity $v$ can also be expressed as the rate of change of displacement, or:

$$v = \frac{2 \pi r}{T}$$

Substitute this into the previous equation:

$$\left(\frac{2 \pi r}{T}\right)^2 = \frac{GM}{r}$$

4. Rearrange to Show $T^2 \propto r^3$ Simplify to obtain:

$$T^2 = \frac{4 \pi^2}{GM} \times r^3$$

Since $\frac{4 \pi^2}{GM}$ is a constant, this confirms $T^2 \propto r^3$.

Total Energy of an Orbiting Satellite

The total energy of a satellite in orbit is the sum of its kinetic and potential energies, which remains constant. If a satellite moves to a lower orbit (closer to the central mass), it loses gravitational potential energy and gains kinetic energy, increasing its speed. This balance keeps the total energy of the satellite constant.

$$\text{Total energy} = \text{Kinetic energy} + \text{Potential energy}$$

Escape Velocity

The escape velocity is the minimum speed required for an object to leave the gravitational influence of a mass. At this velocity, the kinetic energy is equal to the magnitude of the gravitational potential energy.

Derivation:

$$\frac{1}{2}mv^2 = \frac{GMm}{r}$$

Solving for $v$:

$$v = \sqrt{\frac{2GM}{r}}$$

This result shows that escape velocity is independent of the object's mass.

Types of Orbits

1. Synchronous Orbit A synchronous orbit is one where the orbital period of the satellite matches the rotation period of the object it orbits. For instance, a satellite with a 24-hour orbit around Earth appears stationary relative to a fixed point on Earth.

2. Geostationary Orbit A geostationary satellite is a specific type of synchronous satellite that orbits directly above the equator with a period of 24 hours, so it remains above the same point on Earth. Geostationary satellites are widely used in TV broadcasting and telecommunications.

To calculate the orbital radius for a geostationary satellite:

$$T^2 = \frac{4\pi^2 r^3}{GM}$$

Rearranging and substituting values can yield the radius, typically about 42,000 km from Earth's centre (or 36,000 km above the surface).

3. Low-Orbit Satellites Low-Earth orbit satellites have smaller orbital radii than geostationary satellites, making them faster and useful for tasks like weather monitoring, Earth observation, and certain military applications. These satellites cover smaller areas due to their quick orbits, so multiple satellites are often used for continuous coverage.