Kepler's Third Law (T² = 4π²a³/GM, Geostationary Satellites, Mass of the Central Star)

Derive it with a circular orbit

Kepler's third law has a short derivation if a circular orbit is assumed. The answer comes out in the same form for an ellipse, so we begin with a circle.

What holds a planet in a circular orbit is gravitation. It is the only force pointing at the centre, so it is the centripetal force outright. The centripetal force could be written , so we may set .

Writing gravitation equals centripetal force is allowed because no other force points at the centre. No separate centripetal force has been added; gravitation is simply playing that part.

Tidy up this one equation and the rest is arithmetic. Divide both sides by and multiply by to get .

The planet's mass drops out

The has just vanished from both sides, and that is the important part.

It means that on an orbit of the same radius, a heavy planet and a light one go round at the same angular velocity. The two in the figure differ in mass and yet keep step on the same orbit. The reason is the same as for objects falling regardless of weight: the force pulling and the reluctance to be moved are both proportional to the same , and they cancel.

Turn into a period. Putting in and tidying gives . The square of the period is proportional to the cube of the radius of the orbit.

All that is left on the right is , the mass of the central star, and the size of the orbit. Nothing about the orbiting body appears anywhere. Whatever a satellite weighs, at the same altitude it goes round with the same period.

T² and a³ fall on a straight line

The equation derived for a circle holds for an ellipse as it stands. Replace the radius with the semi-major axis and it reads . This is Kepler's third law.

Plot across and up the side, and the bodies orbiting the same central star fall on a single straight line. The inner points crowd near the origin in the figure because enters cubed. A little way outward sends leaping.

The slope of the line is , which contains nothing about the planets. It is fixed by the mass of the central star alone. So every planet of the Solar System lies on one line, and the moons of Jupiter lie on another.

Read the other way round, this becomes an instrument for weighing heavenly bodies. and can be measured with a telescope, so putting them into gives the mass of the central star. We can know the mass of a body we can never touch because of this equation.

Geostationary satellites

The third law can also be read as fixing the size of an orbit once the period is chosen. The geostationary satellite is the example.

To appear motionless from the ground, a satellite must circle above the equator with the same period as the Earth's rotation and in the same sense. That fixes the period, so solving for fixes the altitude to a single value.

In the figure the marker on the ground and the satellite stay in line as they turn. Looking up from the ground, the satellite appears pinned to one point in the sky.

The crux of the problem is that the altitude cannot be chosen freely. Fly lower and the period shortens and the satellite drifts west; fly higher and the period lengthens and it falls behind to the east. For that period there is one altitude and no other.

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