# Acceleration due to gravity?

I was looking into orbitals and found something I haven't been able to understand.

http://www.math.ubc.ca/~cass/courses/m309-01a/hunter/satelliteOrbits.html

There is a part on the page which states the following: $$\vec a = \frac{G(M+m)\vec{r}}{r^3}\approx GM\vec{r} / r^3$$ since $m << M$

I'm not sure why this is the case and why we do not use $\vec a = \vec F/m$ here. The cubed radius is really throwing me off. Can anyone further explain this or point me to a reference with a proof? Thanks

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Very subtle: The $r$ in the numerator is written in boldface, $\textbf{r}$, which means that it's a vector, which has magnitude $r$ and a particular direction. To get a vector of length $1$ but with the same direction, just divide by $r$ again. That gives you the $r^3$ in the denominator. – Lagerbaer May 20 '13 at 3:40
Welcome to Physics SE! Please add yor level of knowlegde about Kepler's laws. – Stefan Bischof May 20 '13 at 5:17
Greetings Tyrick. Have a look at this link to learn how to typeset equations on the SE sites. – user26872 May 20 '13 at 6:01
Thanks, Lagerbaer. I simply didn't see the vector r, in the numerator. ^^ – Tyrick May 20 '13 at 15:34

## 1 Answer

We have $${\bf F} = \mu {\bf a}$$ where $\mu = M m/(M+m)$ is the reduced mass. Thus, $$\frac{G M m}{r^3}{\bf r} = \frac{M m}{M+m}{\bf a}$$ or $$\begin{eqnarray*} {\bf a} &=& \frac{G(M+m)}{r^3}{\bf r} \\ &=& \frac{G M}{r^3}{\bf r}\left(1+\frac{m}{M}\right) \\ &\simeq& \frac{G M}{r^3}{\bf r}. \end{eqnarray*}$$ Notice that $m/M \simeq 10^{-19}$. The error accrued by neglecting this term is very small indeed.

As mentioned by @Lagerbaer, the ratio you find troubling involves the vector ${\bf r}$ divided by $r^3$. This can be written as $$\frac{{\bf r}}{r^3} = \frac{\hat r}{r^2}$$ where $\hat r$ is the unit vector in the ${\bf r}$ direction. The magnitude of the gravitational force will be the familiar one since $|{\bf r}/r^3| = 1/r^2$.

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## protected by Qmechanic♦Sep 3 '14 at 9:01

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