And the rest is algebra: help me with this Maths question

Factual Questions
1

Ok, so this is kinda homework, but I’ve made an attempt and am stuck at the last step (or so I think).

The original equation is
T = 2Π * √( [7/5] * [(R-r)/g])
Re-arrange for g.

So this is what I did. (May be completely wrong, if so, please correct.)

T/2Π = √( [7/5] * [(R-r)/g] )
(T/2Π)² = [7/5] * [(R-r)/g]
(T/2Π)² * 5/7 = (R-r)/g
[(T/2Π)² * 5/7]/(R-r) = 1/g

So I guess what I’m asking is, what do I do with the inverse?

2

Simply take the reciprocal of both sides. For example,

1/x = 3

means

x = 1/3

Both sides have been “flipped” upside down.

3

Follow Cabbage’s method.
Formally what you do is multiply both sided by g giving:

{[(T/2Π)² * 5/7]/(R-r)}*g = 1

Then divide both sides by [(T/2Π)² * 5/7]/(R-r)]:

g = 1/[(T/2Π)² * 5/7]/(R-r)]

4

Don’t ask me what happened to the coding!

5

Yeh, I wasn’t quite sure.

Thanks, I’ll remember that from now on.

6

I tried doing the whole multiplying by g thing, but then when I got 1 on the RHS, I thought “that can’t be right”, and then got thouroughly(sp?) confused.

7

No that is correct, then you divide both sides by what was on the LHS before you multiplied by g and you end up with:

g = 1/LHS