Orbital motionEdexcel A-Level Physics: Revision notes

A body moving in a circle at constant speed is accelerating towards the centre, so by Newton's second law there must be a resultant force towards the centre: the centripetal force. For a satellite or planet in a circular orbit, this force is provided entirely by the gravitational attraction of the body it orbits.

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

Here $M$ is the mass of the body being orbited, $m$ is the orbiting mass, and $r$ is the orbital radius measured to the centre. No other force is needed; the orbiting body is in free fall, continuously falling towards the centre while moving sideways.

Cancel $m$ and one factor of $r$ in $GMm/r^2 = mv^2/r$:

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

The speed depends only on the mass of the body being orbited and the orbital radius, not on the mass of the satellite. A heavier satellite at the same radius moves at the same speed.

A larger orbit means a lower speed. Worked example: at $r = 6.97\times10^{6}$ m around the Earth, $v = \sqrt{\frac{6.67\times10^{-11}\times5.97\times10^{24}}{6.97\times10^{6}}} = 7.56\times10^{3}$ m s⁻¹.

The period is the time for one orbit: $T = 2\pi r/v = 2\pi/\omega$. Substitute $\omega = 2\pi/T$ into $GMm/r^2 = m\omega^2 r$:

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

So $T^2 \propto r^3$ for all bodies orbiting the same mass. If the orbital radius is multiplied by 4, $T^2$ is multiplied by 64, so $T$ is multiplied by 8.

A larger orbit therefore has a longer period for two reasons: a longer path and a lower speed.

Using the Moon: $M = \frac{4\pi^2r^3}{GT^2} = \frac{4\pi^2(3.84\times10^{8})^3}{6.67\times10^{-11}\times(2.36\times10^{6})^2} = 6.0\times10^{24}$ kg. Measuring the orbit of a satellite lets you find the mass of the body it orbits.

A geostationary satellite stays above the same point on the Earth's surface. To do this it must:

• have a period of 24 hours, the same as the Earth's rotation
• orbit above the equator (the orbital plane contains the equator)
• move in the same direction as the Earth's rotation (west to east)

Using $T = 8.64\times10^{4}$ s in $r^3 = GMT^2/4\pi^2$ gives $r = 4.22\times10^{7}$ m, a height of about $3.6\times10^{7}$ m above the surface. There is only one geostationary radius.

Uses include satellite television and communications, because a fixed dish can point at them. Their distance gives a longer signal delay and poor coverage near the poles. Satellites in lower orbits, such as navigation or Earth-observation satellites, have shorter periods and move relative to the ground.

Astronauts in orbit have weight, but no contact force acts on them because they and the spacecraft are in free fall together, accelerating towards the Earth at the local value of $g$. Their centripetal acceleration equals $g$ at that radius, and this free fall is why they appear weightless.

To analyse any orbit problem: identify the central mass, choose the right radius from the centre, then equate $GMm/r^2$ to the centripetal force in whichever form includes the quantity you know ($v$, $\omega$ or $T$).