In our lecture about general mechanics, we have treated the simplified 3-body system consisting of the Sun, the Earth and the Moon. We assumed that the moon does not have any influence on the Sun or the Earth and wanted to find its equation of motion. Let us consider the frame of reference of the Earth as illustrated in this sketch:

Denote by $M$ the Sun's mass and by $m$ the Earth's mass. In the lecture notes, it is stated that gravitational and centrifugal force of the Sun and the Earth have to cancel each other which leads to

$$G \frac{mM}{R^2} = \frac{mM}{M+m}R\omega^2 \approx mR\omega^2.$$

I don't really see what is happening here. The LHS is just the gravitational force, which is clear. Which centrifugal force does this have to be equal to? If we look at the centrifugal force of the Earth, if I'm not mistaken, it should be zero, since in general it is expressed by $-m \omega \wedge (w \wedge y)$ and $y = 0$. However, if we look at the Sun's centrifugal force, then the mass should be $M$ and not $m$. What centrifugal force are we looking at exactly? Why are we plugging in $\frac{mM}{m+M}$ for the mass in the general expression?

  • $\begingroup$ See en.wikipedia.org/wiki/Reduced_mass - effectively it accounts for the fact that the center of the sun is not 'fixed'. $\endgroup$ – Kvothe Feb 7 '14 at 11:27
  • $\begingroup$ @Kvothe Can you post it as an answer? $\endgroup$ – Bernhard Feb 7 '14 at 11:40
  • $\begingroup$ @bernhard I was under the impression that one liners don't really constitute an answer, but sure ;) $\endgroup$ – Kvothe Feb 7 '14 at 11:45
  • $\begingroup$ @Kvothe Of course you have to add some information, and show that it reduces to $m$. $\endgroup$ – Bernhard Feb 7 '14 at 11:50
  • $\begingroup$ Are you sure that you have to deal with gravitation and centrifugal forces? You state that you want to find equations of motion for the Moon, neglecting its infuence on Earth and Sun. Therefore, I assume that the task is to describe Moon's path as the sum of two orbits (Earth around Sun plus Moon around Earth), and use time derivatives to calculate its velocity and acceleration, respectively. $\endgroup$ – Aziraphale Feb 7 '14 at 12:15

This is the reduced mass. This effectively accounts for the fact that the center of the sun is not 'fixed', but rotates around the center of mass (also called the barycenter) of the Earth and the sun together.

The expression $\frac{mM}{m+M}$ approximates to $m$ when $M>>m$.

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