Satellites and Orbits

Look up at a communications satellite and it seems to hang there, motionless, yet it is actually racing around the Earth at thousands of metres per second. It does not fall to the ground and it does not fly off into space. It is in a perfect balance: gravity pulls it towards the centre of the Earth just hard enough to keep bending its straight line motion into a circle. That is all an orbit really is, a constant fall that keeps missing the ground.

Gravity is the string

Picture swinging a ball on a string in a circle above your head. The string keeps tugging the ball inwards, towards your hand, and that inward pull is what stops the ball flying off in a straight line. For a satellite there is no string. Gravity does the tugging instead.

The inward force needed to hold something in a circle is called the centripetal force. For a satellite, gravity is exactly that force. So we set the gravitational force equal to the centripetal force:

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

The satellite mass $m$ sits on both sides, so it cancels. Rearranging for the speed gives a beautifully simple result:

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

The orbital speed depends only on the mass $M$ of the body being orbited and the radius $r$ of the orbit. It does not depend on the mass of the satellite at all.

The blue arrow is the velocity, which always points along the orbit, tangent to the circle. The red arrow is the gravitational force, which always points straight inward towards the planet. The force is at right angles to the velocity, so it never speeds the satellite up or slows it down. It only bends the path into a circle.

From speed to period

Once you know how fast a satellite moves, finding how long one lap takes is short. In one full orbit the satellite travels the circumference of its circle, a distance of $2\pi r$, at its steady speed $v$. The time for one lap is the period $T$:

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

Combine this with $v = \sqrt{GM/r}$ and a famous pattern drops out. Squaring and rearranging gives Kepler’s third law:

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

The right hand side contains only $G$, the central mass $M$ and constants. So $r^3/T^2$ is the same number for every satellite of a given central body. A low, fast satellite and a high, slow one obey the very same relationship.

See it for yourself

Drop a planet, a star and a moon onto the screen, give a body a sideways push, and watch gravity curve its path into an orbit. Speed it up too little and it spirals in; too much and it escapes. Find the sweet spot and it settles into a steady circle, exactly the balance the formulas describe.

How to actually solve one

Most satellite questions are a short, repeatable recipe.

1. If gravity holds the orbit, start from $\dfrac{GMm}{r^2} = \dfrac{mv^2}{r}$ and cancel the satellite mass $m$.
2. For orbital speed, use $v = \sqrt{\dfrac{GM}{r}}$. Remember to take the square root at the end.
3. For the period, use $T = \dfrac{2\pi r}{v}$, or jump straight to $\dfrac{r^3}{T^2} = \dfrac{GM}{4\pi^2}$.
4. For a geostationary satellite, set $T = 24$ hours first, then solve for the radius.

Keep your units in metres, kilograms and seconds, and the answers come out in metres per second and seconds.

Lock it in with active recall

Cover the answer and say each one out loud before you flip. Rate yourself honestly — the cards you find hard come back sooner, the ones you know are spaced further out.

See the recipe in action in the Worked Examples tab, then test yourself in Try It.