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You need to know a little physics here. The force on the satellite (mass m) is F = GMm/r^2 where M = the mass of the planet, G the gravitational constant and r the distance from the centre of the planet.

From Newton we know that force = mass X acceleration. When you study circular motion you will learn that the acceleration of an object moving in a circle is rw^2 where w = angular velocity (radians/ second). A radian is nearly 60 degrees (actually 180/pi) so there are 2.pi radians in a revolution.

There is therefore a relationship between w and T (time for a revolution) in that T = 2.pi/w.

So we can write m.r. 4.pi^2/T^2 = GMm/r^2

The mass of the satellite cancels out (just as well) and so by rearranging we get

r^3 . 4.pi^2 / GM = T^2 - which gives you Kepler's law.

Since we can measure r and t very easily we can work out GM very accurately as a product (much more accurately than we can measure G or M individually)

The maths is complicated by the fact that the satellite will probably be moving in a slightly elliptical orbit (in which case we replace r by the semi-major axis) and is also affected by variable density of the rocks of the planet (this is very noticeable in satellites going round the moon)

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Dortha Miller

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3y ago

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You need to know a little physics here. The force on the satellite (mass m) is F = GMm/r^2 where M = the mass of the planet, G the gravitational constant and r the distance from the centre of the planet.From Newton we know that force = mass X acceleration. When you study circular motion you will learn that the acceleration of an object moving in a circle is rw^2 where w = angular velocity (radians/ second). A radian is nearly 60 degrees (actually 180/pi) so there are 2.pi radians in a revolution.There is therefore a relationship between w and T (time for a revolution) in that T = 2.pi/w.So we can write m.r. 4.pi^2/T^2 = GMm/r^2The mass of the satellite cancels out (just as well) and so by rearranging we getr^3 . 4.pi^2 / GM = T^2 - which gives you Kepler's law.Since we can measure r and t very easily we can work out GM very accurately as a product (much more accurately than we can measure G or M individually)The maths is complicated by the fact that the satellite will probably be moving in a slightly elliptical orbit (in which case we replace r by the semi-major axis) and is also affected by variable density of the rocks of the planet (this is very noticeable in satellites going round the moon)


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