Wednesday, November 16, 2011

The Gaussian Gravitational Constant

In Theoria Motus (1809) Gauss gave a formula which he stated was a constant for all bodies orbiting the Sun. This quantity, k, has come to be known as the Gaussian gravitational constant. The formula Gauss gives is,

ΔA/Δt, the areal velocity, is the area ΔA swept out by the body's radius in time Δt, and is a constant for a single body according to Kepler's 2nd law. p is the parameter in the formula for the elliptical orbit of a body and m is the body's mass given in solar masses. This is based on a two-body solution for an orbit about the Sun and gravitational perturbations by other bodies are ignored. In the two-body problem one can replace the mass of a body with its reduced mass. M presumably represents the mass of the Sun.

The Moon and the Earth are gravitationally bound so their motion is coupled and we would have to use the sum of their masses in the formula to get it work properly. But does M truly represents the mass of the Sun if we have to treat gravitationally bound together or is this only an approximation? The other planets are gravitationally bound to the Sun so in treating this as a two-body problem we are splitting the solar system into two parts, the Earth-Moon system and the rest of the Solar System with the total mass being M'(m1 + m2) where m1 is the mass fraction of the other bodies in the Solar System and m2 is the mass fraction of the Earth-Moon system. Instead of 1+m on the left we would have m1 + m2 = 1 and a mass term would not be present in the formula. Even though the fractions were different for each body and the "Sun" the sum of the two masses would have the same value, M'. Gauss' formula would then be approximately correct since the mass of Jupiter is about one thousandth that of the Sun. The formula that Gauss gave may work better for single orbits but over long periods of time one may notice a discrepancy with the perturbations of the orbits being a complication.

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