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Orbital speed

Gravity and orbits · level 6, lesson 3 of 6

A question to start

A relay satellite will orbit Mars 300 km up. How fast must it go to stay in a circle, and how long will each orbit take? Would a satellite at the same height round the Earth go faster or slower?

The idea

Anything moving in a circle of radius \(r\) at speed \(v\) is accelerating towards the centre, by

\[a = \frac{v^2}{r}.\]

(The Forces and motion topic derives this; a stone on a string needs the string's pull to keep turning.) For an orbit, gravity provides the pull. Set \(GM/r^2\), gravity's acceleration at distance \(r\), equal to \(v^2/r\):

\[\frac{GM}{r^2} = \frac{v^2}{r} \qquad\Rightarrow\qquad v = \sqrt{\frac{GM}{r}}.\]

The satellite's own mass has cancelled again: a space station and a bolt beside it orbit together.

The period is one lap divided by the speed:

\[T = \frac{2\pi r}{v}.\]
rvGM/r²

Low orbits, just above the air:

Mars, 300 km up Earth, 400 km up
\(r\) 3,690 km 6,771 km
\(GM\) 42,828 km³/s² 398,600 km³/s²
Speed \(v\) 3.41 km/s 7.67 km/s
Period \(T\) 1 h 53 min 1 h 32 min

To stay in low orbit round Mars takes less than half the speed of the space station. That is the speed a ship must reach to get into orbit after leaving the surface of Mars, about 3.4 to 3.5 km/s (plus what it loses on the way up).

Farther out, \(v\) shrinks as \(1/\sqrt r\): four times as far, half the speed. And the path gets longer, so the period grows even faster (lesson 4).

Worked examples

1. The relay satellite. \(v = \sqrt{42828 / 3689.5} = \sqrt{11.61} = 3.41\) km/s. \(T = 2\pi \times 3689.5 / 3.41 = 6{,}800\) s \(= 1\) h 53 min.

2. Phobos. \(v = \sqrt{42828 / 9376} = \sqrt{4.568} = 2.14\) km/s. \(T = 2\pi \times 9376 / 2.14 = 27{,}500\) s \(= 7.6\) h. ✓ (The real figure is 7 h 39 min.)

3. The Earth round the Sun. \(GM_\odot = 1.327 \times 10^{11}\) km³/s², \(r = 1.496 \times 10^8\) km. \(v = \sqrt{887} = 29.8\) km/s. ✓

Try it

  1. Find the speed of a satellite in a circular orbit 1,000 km above Mars.
  2. Find its period.
  3. Find the orbital speed of Mars round the Sun (\(r = 2.279 \times 10^8\) km).
  4. Find the orbital speed of Deimos (\(r = 23{,}463\) km).
  5. A satellite moves to an orbit four times as far from the centre. What happens to its speed?
Answers
  1. \(r = 4389.5\) km. \(v = \sqrt{42828/4389.5} = \sqrt{9.757} = 3.12\) km/s.
  2. \(T = 2\pi \times 4389.5 / 3.12 = 8{,}840\) s, about 2 h 27 min.
  3. \(\sqrt{1.327\times10^{11} / 2.279\times10^8} = \sqrt{582} = 24.1\) km/s.
  4. \(\sqrt{42828/23463} = \sqrt{1.825} = 1.35\) km/s.
  5. It halves.

Common mistake

Using the height instead of the radius. In \(v = \sqrt{GM/r}\), \(r\) is from the planet's centre: radius plus height.

Where it leads