Satellite orbits — linking gravity to circular motion — Physics

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Negative and fractional indices (Mathematics) · coming soon

A machine that never stops falling

The International Space Station has no engine pushing it along its orbit. It has no string tying it to the Earth. It flies about 400 km above the surface, where gravity is still about 89% as strong as on the ground. So why does it not fall down?

The honest answer surprises most people. It does fall. It has been falling, all the time, since it was launched. It simply never gets any closer to the ground.

This lecture explains how that is possible. It joins the two ideas of the last few lectures. Circular motion needs a resultant force towards the centre. Gravity, as a field, gives every kilogram a pull towards the centre. Put them together and you can find the speed of any orbit, the time it takes, and the one height where a satellite seems to hang still in the sky.

Falling around the Earth

Newton imagined a cannon on a very high mountain, above the air, firing a ball sideways. A slow ball falls to the ground nearby. A faster ball lands farther away. Now ask: could a ball be so fast that the ground curves away beneath it as quickly as it falls?

Work it out for one second. In one second, starting from moving sideways, a ball falls $\frac{1}{2} g t^{2} = \frac{1}{2} \times 9.8 \times 1^{2} = 4.9$ m. Now look at the Earth. Its surface is curved. If you travel in a straight line along the ground, the surface drops away below you. Over a distance of about 7.9 km, it drops by 4.9 m.

In one second, a ball moving sideways at 7.9 km/s would travel 7.9 km in a straight line. Gravity pulls it down 4.9 m in that second. Over 7.9 km, the Earth's surface curves away by the same 4.9 m. So the ball is no closer to the ground than before.

So a ball moving sideways at about 7.9 km/s falls 4.9 m every second, and the ground falls away 4.9 m every second. It keeps falling, and it never lands. That is it. That is the whole idea of an orbit. An orbit is free fall with enough sideways speed to miss the Earth.

Real satellites fly above most of the air, because at 7.9 km/s air resistance would slow them quickly. Try the cannon yourself below. It is placed 400 km up, at the height of the Space Station, and fires sideways.

Simulation — Newton's cannon, 400 km above the Earth

Choose a launch speed and press Launch. Find the speed that gives a circle. Then try faster speeds. Can you make the ball escape?

Launch speed: 7.30 km/s

Paths of a ball fired sideways from 400 km above the Earth, worked out from Newton's law of gravitation. Speeds are measured relative to the Earth's centre, and air is ignored.

Below 7.67 km/s, the path curves into the Earth. At 7.67 km/s, it is a circle. Faster, it becomes an ellipse that swings out and comes back. At about 10.85 km/s or more, it escapes and never returns.

Launch speed	What happens	Time for one orbit
7.3 km/s	falls back into the Earth	—
7.67 km/s	a circle, 400 km up	≈ 92 min
9.0 km/s	an ellipse, up to ≈ 8 600 km high	≈ 188 min
11.0 km/s	escapes	—

What did you see? At 7.3 km/s, the ball curved down and hit the Earth. At 7.67 km/s, it went all the way round in a circle and came back to the cannon after about 92 minutes. At 9 km/s, it swung far out and came back: an ellipse. At 11 km/s, it never came back. The circle is not the only orbit. It is the special case where the speed is exactly right for that height.

The field supplies the centripetal acceleration

Now make the idea exact. The last lecture showed that the gravitational field at distance $r$ from the centre of a planet of mass $M$ is

g = \frac{GM}{r ^{2}}

For a satellite, gravity is the only force. So the field strength is the satellite's whole acceleration, and it points to the centre. A circle of radius $r$ at speed $v$ needs a centripetal acceleration $\frac{v ^{2}}{r}$ . A circular orbit is possible when the field supplies exactly that:

\frac{GM}{r ^{2}} = \frac{v ^{2}}{r}

Multiply both sides by $r$ . Then $v^{2} = \frac{GM}{r}$ .

Speed of a circular orbit

For a circular orbit of radius $r$ around a body of mass $M$ :

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

Here $r$ is measured from the centre of the body. The mass of the satellite does not appear.

Two things follow. First, the satellite's own mass has gone, as it did in the circular motion lectures. A space station and a spanner dropped by an astronaut orbit together at the same speed. Second, $r$ is underneath. A higher orbit is a slower orbit. This surprises people, because a higher orbit also has a longer path. Both effects make the time for one orbit longer.

Check it against the field you already know. At the Space Station, the last lecture found $g \approx 8.69$ N/kg at $r = 6.77 \times 1 0^{6}$ m. Then $v = g r = 8.69 \times 6.77 \times 1 0^{6} \approx 7670$ m/s, about 7.67 km/s. That is the speed the simulation needed for a circle.

From speed to period: Kepler's third law

The time for one orbit is called the period, $T$ . It links speed and radius in a second way, because one orbit is a distance $2 π r$ . Step through the argument below. Each line comes with the reason it is allowed.

Step through — from gravity to T² ∝ r³

Press Next to add one step. Read the reason before you go on.

Each step gives one line and the reason it is allowed.

1. Gravity is the only force: F = GMm/r². It is the whole resultant force on the satellite.
2. A circle needs a resultant force mv²/r towards the centre.
3. So GMm/r² = mv²/r. Gravity supplies exactly what the circle needs.
4. Divide by m and multiply by r: v² = GM/r. The satellite's mass cancels.
5. One orbit is a distance 2πr in a time T, so v = 2πr/T, and v² = 4π²r²/T².
6. Set the two expressions for v² equal: 4π²r²/T² = GM/r.
7. Multiply by r and by T², and divide by GM: T² = (4π²/GM) × r³.

Result: for everything orbiting the same body, T² is proportional to r³. This is Kepler's third law.

Period of a circular orbit

$T^{2} = \frac{4 π ^{2}}{GM} r^{3}$ , so $T = 2 π \frac{r ^{3}}{GM}$

For everything orbiting the same body, $T^{2}$ is proportional to $r^{3}$ .

Johannes Kepler found this pattern in the planets' motion in 1618, and published it in 1619 in his book Harmonices Mundi (The Harmony of the World). He had no idea why it was true. He found it by searching through years of measurements. About seventy years later, Newton's law of gravitation explained it, in exactly the steps you have just followed.

You can test it with real planets. Measure distances in units of the Earth's distance from the Sun, and periods in Earth years. Then for the Earth, $T^{2} = 1$ and $r^{3} = 1$ . If Kepler is right, $T^{2} \div r^{3}$ should be 1 for every planet.

Planet	Distance from Sun (Earth = 1)	Period (Earth years)	T² ÷ r³
Mercury	0.387	0.241	1.00
Venus	0.723	0.615	1.00
Earth	1	1	1
Mars	1.524	1.881	1.00
Jupiter	5.204	11.86	1.00
Saturn	9.572	29.43	0.99

The values are worked out from NASA's planetary fact sheet. Saturn is about 25 times as far from the Sun as Mercury, yet both obey the same rule to within about 1%. The small differences come from rounding and from the fact that real orbits are slightly oval.

Stop and think

Two spacecraft are in the same circular orbit. One is 50 km behind the other and wants to catch up. Should it fire its engine forwards, to speed up?

Show the answer

No, and this is one of the strangest facts in spaceflight. Speeding up pushes the spacecraft into an ellipse that swings out to a higher orbit. Higher orbits are slower and longer, so it falls further behind. To catch up, it must slow down a little. Then it drops into a slightly lower orbit, which is shorter and faster, and it gains on the other craft. Later it speeds up again to return to the first orbit. You saw the first half of this in the simulation: faster than 7.67 km/s sent the ball outwards, not along the circle.

Worked examples

In these examples, $G = 6.67 \times 1 0^{- 11}$ N m²/kg², the Earth's mass is $5.97 \times 1 0^{24}$ kg, and its radius is $6.37 \times 1 0^{6}$ m. So $GM \approx 3.98 \times 1 0^{14}$ N m²/kg.

Worked Example 1 — The Space Station

The Space Station orbits 400 km above the surface. Find its speed and its period.

First the radius: $r = 6.37 \times 1 0^{6} + 0.40 \times 1 0^{6} = 6.77 \times 1 0^{6}$ m.

v = \frac{3.98 \times 1 0 ^{14}}{6.77 \times 1 0 ^{6}} \approx 7670 m/s

Then $T = \frac{2 π r}{v} = \frac{2 π \times 6.77 \times 1 0 ^{6}}{7670} \approx 5550$ s, which is about 92 minutes.

Check against reality: the European Space Agency gives a height of 400 km and 92 minutes for one orbit ✓. In one day the station goes round about $1440 \div 92 \approx 16$ times, so the astronauts see about 16 sunrises a day.

Worked Example 2 — Where must a geostationary satellite be?

A geostationary satellite stays above one point on the equator. It must go round once in the time the Earth turns once. The Earth turns once, relative to the stars, in one sidereal day: 23 h 56 min 4 s, which is 86 164 s. (The 24-hour day is a little longer, because the Earth also moves along its orbit round the Sun.)

Rearrange $T^{2} = \frac{4 π ^{2}}{GM} r^{3}$ to make $r$ the subject: $r^{3} = \frac{GM T ^{2}}{4 π ^{2}}$ .

r^{3} = \frac{3.98 \times 1 0 ^{14} \times ( 86 164 ) ^{2}}{4 π ^{2}} \approx 7.48 \times 1 0^{22} m^{3}

Take the cube root: $r \approx 4.21 \times 1 0^{7}$ m, about 42 140 km from the centre. The height is $42140 - 6370 \approx 35770$ km. The measured radius of the geostationary orbit is 42 164 km. The small difference comes from rounding $G$ and the Earth's mass.

The speed is $\frac{2 π r}{T} = \frac{2 π \times 4.21 \times 1 0 ^{7}}{86 164} \approx 3070$ m/s. A satellite racing at about 3 km/s appears to hang still, because the ground below it turns at the same rate. There is only one such radius above the Earth, because $T$ fixes $r$ .

Worked Example 3 — GPS, from a ratio

GPS satellites go round the Earth twice a day: once every half sidereal day. How high are they?

There is no need to use $G$ again. Compare with the geostationary orbit. Since $T^{2} \propto r^{3}$ , $r \propto T^{2/3}$ . Halving the period multiplies the radius by $(\frac{1}{2})^{2/3} \approx 0.630$ :

r \approx 0.630 \times 42140 \approx 26550 km

The height is $26550 - 6370 \approx 20180$ km. The US government's GPS website gives "approximately 20 200 km" ✓. A ratio like this is often the fastest route: when two objects orbit the same body, compare them directly.

Worked Example 4 — Weighing the Earth with the Moon

The Moon orbits at $r \approx 3.84 \times 1 0^{8}$ m with a period of 27.3 days. Treat the orbit as a circle and find the Earth's mass.

Rearrange $T^{2} = \frac{4 π ^{2}}{GM} r^{3}$ to make $M$ the subject: $M = \frac{4 π ^{2} r ^{3}}{G T ^{2}}$ . The period in seconds is $27.3 \times 86400 \approx 2.36 \times 1 0^{6}$ s.

Work out the top and the bottom separately. Top: $4 π^{2} \times (3.84 \times 1 0^{8})^{3} \approx 2.24 \times 1 0^{27}$ . Bottom: $6.67 \times 1 0^{- 11} \times (2.36 \times 1 0^{6})^{2} \approx 371$ . Divide the top by the bottom:

M \approx 6.0 \times 1 0^{24} kg

The accepted value is $5.97 \times 1 0^{24}$ kg ✓. This is how astronomers find the mass of any body that has something orbiting it. The same calculation with the Earth's orbit gives the Sun's mass, about $2.0 \times 1 0^{30}$ kg.

Three classic mistakes

Trap 1 — Using the height for r

Every formula in this lecture uses the distance from the centre of the Earth. For the Space Station, $r = 6770$ km, not 400 km. Using 400 km would give a speed of about 31.5 km/s, which is more than enough to escape the Earth.

Trap 2 — Leaving T in hours or days

$G$ is in SI units, so $T$ must be in seconds and $r$ in metres. Put $T = 24$ into the geostationary calculation and the radius comes out about 180 km, which is inside the Earth. Convert first. For geostationary orbit, use one sidereal day, 86 164 s.

Trap 3 — Thinking a satellite needs a forward force

A free-body diagram of a satellite has one arrow: gravity, towards the centre. There is no forward force and no outward force. The satellite keeps moving sideways because nothing stops it, as Newton's first law says. Gravity only bends that motion into a circle.

Try it yourself

Work each one on paper first. Open a hint only when you are stuck. Use $GM \approx 3.98 \times 1 0^{14}$ N m²/kg for the Earth and $R = 6370$ km.

A weather satellite orbits 1000 km above the surface. Find its speed and its period in minutes.

Hint 1
Add the height to the Earth's radius first.

Hint 2
$r = 7.37 \times 1 0^{6}$ m. Use $v = GM / r$ , then $T = 2 π r / v$ .

Full answer
$v = 3.98 \times 1 0^{14} \div 7.37 \times 1 0^{6} \approx 7350$ m/s, about 7.35 km/s. $T = 2 π \times 7.37 \times 1 0^{6} \div 7350 \approx 6300$ s, about 105 minutes. Check: it is higher than the Space Station, so it should be slower (7.35 < 7.67 km/s ✓) and take longer (105 > 92 min ✓).
Satellite B orbits the Earth at four times the radius of satellite A. How do their periods and speeds compare?

Hint 1
$T^{2} \propto r^{3}$ and $v^{2} \propto \frac{1}{r}$ .

Hint 2
$T$ is multiplied by $4^{3/2}$ , and $v$ by $\frac{1}{4}$ .

Full answer
$T_{B} = 4^{3/2} T_{A} = 8 T_{A}$ , so B takes 8 times as long. $v_{B} = v_{A} \div 4 = \frac{1}{2} v_{A}$ , so B moves at half the speed. Check: B's path is 4 times as long, at half the speed, so it takes $4 \times 2 = 8$ times as long ✓.
Explain why a ball fired sideways at 7.9 km/s just above the ground would orbit the Earth (if there were no air), using only the numbers 4.9 m and 7.9 km. Then check 7.9 km/s with $v = GM / r$ .

Hint 1
How far does any object fall in its first second? How far does the Earth's surface curve away over 7.9 km?

Hint 2
For the check, use $r = 6.37 \times 1 0^{6}$ m.

Full answer
In one second the ball falls 4.9 m. In the same second it travels 7.9 km sideways, and over 7.9 km the surface curves away by 4.9 m. So after one second it is just as high above the ground as before, and the same is true for every following second. Check: $3.98 \times 1 0^{14} \div 6.37 \times 1 0^{6} \approx 7900$ m/s ✓. The two methods agree because both describe the same thing: gravity's 9.8 m/s² equals the centripetal acceleration $v^{2} / r$ at the surface.
(a) The Earth orbits the Sun at $1.496 \times 1 0^{11}$ m once every 365.2 days. Find the mass of the Sun. (b) Jupiter is 5.20 times as far from the Sun as the Earth. Without using $G$ , predict its period in years.

Hint 1
(a) Use $M = \frac{4 π ^{2} r ^{3}}{G T ^{2}}$ with $T$ in seconds and $G = 6.67 \times 1 0^{- 11}$ .

Hint 2
(b) In Earth units, $T^{2} = r^{3}$ , so $T = 5.2 0^{3/2}$ .

Full answer
(a) $T = 365.2 \times 86400 \approx 3.16 \times 1 0^{7}$ s. $M = \frac{4 π ^{2} \times ( 1.496 \times 1 0 ^{11} ) ^{3}}{6.67 \times 1 0 ^{- 11} \times ( 3.16 \times 1 0 ^{7} ) ^{2}} \approx 2.0 \times 1 0^{30}$ kg. That is about 330 000 times the Earth's mass. (b) $T = 5.2 0^{1.5} \approx 11.9$ years. NASA's fact sheet gives 4331 days, which is $4331 \div 365.2 \approx 11.86$ years ✓. A rule found from the Moon and the Space Station predicts the year of a planet 778 million km away.

The bigger idea

Before Newton, the motion of the planets and the fall of a stone seemed to belong to two different worlds. The orbit shows they are one. The Space Station, the Moon and Jupiter are all falling, exactly like a dropped pen. They have so much sideways speed that they keep missing what they fall towards. One law, $F = \frac{GM m}{r ^{2}}$ , with $F = ma$ , explains all of them.

That law also creates a scarce resource. There is only one geostationary radius, so every satellite that must appear fixed in the sky shares one ring above the equator. At the end of 2011, about 400 active satellites were in it, according to a report to a United Nations space committee. The International Telecommunication Union records each one's position and radio frequencies, so that neighbours do not interfere. A result from a few lines of algebra is now managed by international agreement.

This lecture also completes the mechanics module: motion, forces, momentum, circular motion and gravitation. They are not six topics. They are one way of thinking: find the real forces, and let Newton's laws tell you what the motion must be.

Key concepts in this lesson

A satellite in orbit is in free fall, with enough sideways speed that the Earth's surface curves away as fast as it falls.
For a circular orbit, the gravitational field supplies exactly the centripetal acceleration, $\frac{GM}{r ^{2}} = \frac{v ^{2}}{r}$ , so $v = \frac{GM}{r}$ .
The satellite's mass cancels, and a higher orbit is slower and takes longer.
The period satisfies $T^{2} = \frac{4 π ^{2}}{GM} r^{3}$ , which is Kepler's third law, $T^{2} \propto r^{3}$ , for everything orbiting the same body.
A geostationary orbit has a period of one sidereal day, 86 164 s, which fixes one radius: about 42 164 km from the Earth's centre.
In every orbit formula, $r$ is measured from the centre of the body and $T$ is in seconds.