Orbital motionEdexcel International A Level Physics: Revision notes

An object moving in a circle at constant speed has an acceleration towards the centre, so by Newton's second law it needs a resultant force towards the centre, the centripetal force. For a satellite, planet or moon in a circular orbit this force is the gravitational attraction of the body it orbits. There is no outward 'centrifugal' force acting on the orbiting object, and no engine thrust is needed to keep a steady orbit.

Equating the gravitational force to the centripetal force for a satellite of mass $m$ and orbital radius $r$ around a mass $M$:

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

The mass of the satellite cancels, giving:

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

The speed depends only on the mass of the central body and the orbital radius, so a satellite in a higher orbit moves more slowly, and any satellite at the same radius has the same speed whatever its mass. Here $r$ is measured from the centre of the Earth, so $r = R_E + h$.

Worked example. The ISS at a height of 4.0 × 10⁵ m: $r = 6.37 \times 10^6 + 4.0 \times 10^5 = 6.77 \times 10^6$ m, so $v = \sqrt{6.67 \times 10^{-11} \times 5.97 \times 10^{24} / 6.77 \times 10^6} = 7.67 \times 10^3$ m s⁻¹.

Using $v = 2\pi r/T$ (or $\omega = 2\pi/T$) in $GMm/r^2 = m\omega^2 r$ gives:

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

So $T^2 \propto r^3$ for all bodies orbiting the same central mass. A higher orbit has a longer period because the satellite travels further, and more slowly. For the ISS, $T = 2\pi r/v = 5.5 \times 10^3$ s, about 92 minutes. For Mars around the Sun ($r = 2.28 \times 10^{11}$ m, $M = 1.99 \times 10^{30}$ kg) the formula gives $T = 5.94 \times 10^7$ s = 1.88 years.

To compare two orbits around the same body, use the ratio $T_1^2/T_2^2 = r_1^3/r_2^3$ and avoid calculating $GM$.

A geostationary satellite stays above the same point on the Earth's surface, so it must: orbit in the plane of the equator, move in the same direction as the Earth rotates, and have a period equal to the Earth's rotation period (one sidereal day, 8.62 × 10⁴ s). Using $r^3 = GMT^2/4\pi^2$ gives $r = 4.21 \times 10^7$ m, which is a height of 3.58 × 10⁷ m above the surface. They are used for communications and weather observation.

Satellites in low orbits (about 10² to 10³ km up) have short periods and are used for imaging and for the ISS. Navigation satellites such as GPS orbit at 2.66 × 10⁷ m with a period of about 12 hours, so each completes two orbits in one rotation of the Earth.

Astronauts in the ISS feel weightless because the station and the astronauts are all in free fall with the same acceleration, $g$ at that height ($g = GM/r^2$, about 8.7 N kg⁻¹ at the ISS). Gravity has not disappeared: it is providing the centripetal force. There is no contact force between them and the station, so they do not feel their weight.

In orbit questions, state that gravity provides the centripetal force, write $GMm/r^2 = mv^2/r$ (or $m\omega^2 r$), state that $m$ cancels, and be careful with units: radius in metres and time in seconds.