# Expanding of formula

1. Aug 13, 2015

### Vrbic

Hello,
I have problem with understanding of expanding of formula in this video (at 30min 05s), after few second he neglects term (r/R)^2).

For me there is missing factor 2 at second term (in "j" component). Similarly as here at 24min 10s.
If somebody knows what's the matter there...please let me know.
Thank you.

Last edited by a moderator: Aug 13, 2015
2. Aug 13, 2015

### RUber

It appears that he is using the assumption that r<<R to simplify the math.
When he has $\frac{GM}{R^2(1-\frac{r}{R})^2}$, he assumes a Taylor series.
To see this, rewrite the problem as $\left(\frac{GM}{R^2}\right) \frac{1}{(1- x)^2}$ and take the Taylor expansion about x=0.
You get $\left(\frac{GM}{R^2}\right)\left[ 1 -\epsilon\frac{2(1-\epsilon )}{(1-\epsilon)^3 } + \mathcal{O}(\epsilon^2)\right]$
Keeping only the $\mathcal{O}(\epsilon)$ terms, you are left with $\left(\frac{GM}{R^2}\right)\left[ 1 -2\epsilon\right]$ where epsilon is $\frac{r}{R}$.

Similarly, when he is dealing with the transverse alignment, he states the assumption that he is ignoring terms on the order of r^2/R^2.
So he assumes that $R_A\approx. R$, that is, the distance to A is approximately the same as the distance to B. And then he linearizes the problem to say that the slope is one that would be a straight line approximation that shifts r over the change of R. Essentially, the paths intersect at the center of the massive object.

This makes sense, when thinking of realistic problems, if objects in space are separated by miles, but are orbiting 100s of miles away from Earth, the r/R term is on the order of 10^-2, but its square is significantly smaller. Once the r<<R assumption no longer holds, more terms need to be accounted for to ensure accuracy.