Orbits of planets and satellites (A-level only)

AQA·Physics·3.7.2.4

Circular orbital motion

A planet or satellite in a circular orbit is continually pulled towards the centre of the body it orbits. Gravity changes the direction of its velocity, keeping it on a circular path. The gravitational force therefore supplies the centripetal force needed for the orbit; it is not an additional force.

For an orbiting body of mass $$m$$, let $$r$$ be its orbital radius, measured from the centre of the spherical central body. Let the central body have mass $$M$$, and let $$G$$ be the gravitational constant. If the orbital speed is $$v$$, equating the gravitational force to the required centripetal force gives:

GMmr2=mv2r\frac{GMm}{r^2}=\frac{mv^2}{r}

Cancelling $$m$$ and rearranging gives the orbital speed:

v=GMrv=\sqrt{\frac{GM}{r}}

The orbiting body’s mass cancels, so the circular orbital speed depends on the central mass and the orbital radius, not on the mass of the orbiting body. Around the same central mass, a larger orbital radius gives a lower speed.

Circular orbital period

The orbital period $$T$$ is the time taken to complete one orbit. The distance travelled in one circular orbit is its circumference, $$2\pi r$$. Dividing this distance by the orbital speed and substituting the expression for $$v$$ gives:

T=2πrv=2πr3GMT=\frac{2\pi r}{v}=2\pi\sqrt{\frac{r^3}{GM}}

Squaring this expression gives the relationship between period and radius:

T2=4π2r3GMT^2=\frac{4\pi^2r^3}{GM}

For circular orbits around the same central mass, $$G$$ and $$M$$ are constant, so $$T^2\propto r^3$$. A larger orbit has a longer period: the body travels a greater distance at a lower orbital speed.

Energy of an orbiting satellite

A satellite’s total mechanical energy is the sum of its kinetic energy and its gravitational potential energy. To assign a value to gravitational potential energy, we take it to be zero when the satellite and the body it orbits are infinitely far apart.

For a satellite of mass $$m$$ at a distance $$r$$ from the centre of a body of mass $$M$$, its gravitational potential energy is:

U=−GMmrU=-\frac{GMm}{r}

Here $$G$$ is the gravitational constant. The potential energy is negative because energy must be supplied to separate the satellite from the body to an infinite distance.

Energy in a circular orbit

In a circular orbit, gravity provides the force that keeps the satellite moving in a circle. Equating the gravitational force to the required centripetal force gives $$v^2=GM/r$$, where $$v$$ is the satellite’s orbital speed. Substituting this into $$K=\frac{1}{2}mv^2$$ gives its kinetic energy:

K=GMm2rK=\frac{GMm}{2r}

The kinetic energy is positive and has half the magnitude of the negative potential energy. Adding the two gives the total mechanical energy:

E=K+U=−GMm2rE=K+U=-\frac{GMm}{2r}

The negative total energy indicates that the satellite is gravitationally bound: energy must be added for it to reach infinite separation with zero speed.

Moving to a larger circular orbit

A larger orbital radius makes $$U$$ less negative, so the satellite’s gravitational potential energy increases. At the same time, $$K=GMm/(2r)$$ decreases: the satellite moves more slowly in the larger circular orbit.

The increase in potential energy is greater than the decrease in kinetic energy. The total energy therefore increases—it becomes less negative—so energy must be supplied to move the satellite to the larger circular orbit.

Escape velocity

An object launched away from a gravitating body slows down as gravity pulls it back. Escape velocity is the minimum launch speed needed for the object to reach an infinite distance with its speed tending to zero, without further propulsion.

As the object moves away, its kinetic energy decreases while its gravitational potential energy increases. Taking gravitational potential energy to be zero at infinity, the minimum escape condition is that the object's initial total energy is zero. It then has just enough kinetic energy to reach infinity with none remaining.

Escape velocity equation

For an object of mass $$m$$ launched from a distance $$r$$ from the centre of a spherical body of mass $$M$$, the minimum energy condition is:

12mvesc2−GMmr=0\frac{1}{2}mv_{\mathrm{esc}}^2-\frac{GMm}{r}=0

The first term is the object's initial kinetic energy; the second is its gravitational potential energy. Here, $$G$$ is the gravitational constant. Rearranging gives:

vesc=2GMrv_{\mathrm{esc}}=\sqrt{\frac{2GM}{r}}

The object's mass $$m$$ cancels, so the escape velocity is independent of the escaping object's mass. It depends on the mass of the gravitating body and the launch distance from its centre. The equation applies at or outside the surface of the spherical body and assumes that no energy is lost, for example through air resistance.

Synchronous and geostationary orbits

A satellite is in a synchronous orbit when it takes the same time to complete one orbit as the body it orbits takes to complete one rotation. The time for one orbit is called the satellite’s orbital period.

A synchronous orbit around Earth has a period of approximately one sidereal day: 23 hours 56 minutes. This is the time Earth takes to rotate once relative to the distant stars.

Conditions for a geostationary orbit

A geostationary orbit is a synchronous orbit in which a satellite remains above the same point on Earth’s equator. For this to happen, the orbit must be:

  • Circular, so the satellite moves at a constant orbital speed.
  • In the equatorial plane, so the satellite stays above the equator.
  • In the same direction as Earth’s rotation, so the satellite travels around Earth in step with the surface below.

Because the satellite completes each orbit in the time Earth takes to rotate once, it stays above the same equatorial point as Earth turns.

The orbital radius is approximately $$4.22\times10^7,\mathrm{m}$$, measured from Earth’s centre. This corresponds to an altitude of approximately $$3.58\times10^7,\mathrm{m}$$ above Earth’s surface.

Having the correct period alone does not make an orbit geostationary. For example, a synchronous orbit tilted relative to the equator would carry the satellite north and south of the equator, rather than keeping it above one point.

Low and geostationary satellite orbit applications

The usefulness of a satellite orbit depends partly on how far the satellite is from Earth and whether it stays above the same place. These factors affect the detail it can observe, the time signals take to travel and how continuously it can cover a location.

Low Earth orbit satellites

Low Earth orbit satellites are relatively close to Earth. Their closeness allows them to collect detailed observations of the surface, so they are used for Earth observation. They are also used for communications: signals have a relatively short distance to travel between the ground and the satellite, giving shorter signal travel times.

A low Earth orbit satellite moves across the sky relative to an observer on the ground. It does not remain in range of one location continuously. To provide continuous communication coverage at that location, multiple satellites may be needed so that another can take over as one moves out of range.

Geostationary satellites

A geostationary satellite orbits above the equator in the same direction and with the same rotation period as Earth. It therefore remains above the same equatorial location. From the ground, it appears stationary and can provide continuous coverage of a large region. This makes it useful for communications and weather observation.

A geostationary satellite is farther from Earth than a low Earth orbit satellite, so signals take longer to travel to and from it. Its coverage is also limited at high latitudes: from locations far north or south, the satellite appears low in the sky or is below the horizon.

Orbits of planets and satellites (A-level only) | AQA A-level Physics (7408) Revision Notes | KnowledgeSet