# Deriving Big G

1. Dec 6, 2006

### Poop-Loops

We did the Cavendish experiment and I'm writing up the lab now. We need to go through the theory of it and how we got to our equation that we'll use. Since the class is a mish mash of different labs, this was given to us, but not entirely. I completely understand what's going on up until a point.

Basically, I'm stuck at the last two equations. How he says "and then we substitute" blah blah. It just seems like he took some really big steps that I can't see. Like ending up with "b" and "d" where it wasn't even used before.

I suck at deriving stuff, since this is the first class that I've ever had to do that before. I'm just not seeing how he got there. And I checked Google to see if someone else has done this before, but they also just take it in chunks, which is weird. Thanks in advance.

EDIT: Well actually, I know where he had to put in "b" and "d", and I figured out where "b" comes in, but have no idea how he managed to stick "d" in there. =/

2. Dec 6, 2006

### nrqed

There is a mistake in the formula given there!!

Instead of $$kT^2 = {\tau \over \theta} T^2 = 4 \pi T^2$$

it should be

$$kT^2 = {\tau \over \theta} T^2 = 4 \pi^2 I$$

so that $k = { 4 \pi^2 I \over T^2}$.

Now, if you use the I they give at the very beginning and the expression for theta and use that in $2 F_G d = k \theta$ you should get the final expression for G.

Hope this helps

patrick

Last edited: Dec 6, 2006
3. Dec 6, 2006

### Poop-Loops

Yes, that makes sense now. Thanks a lot. My prof tends to make errors like that, so it's not surprising.

4. Dec 6, 2006

### Poop-Loops

I'm assuming the last equation for G also has L squared, and not just L, then? Everything else matches what I derived.

5. Dec 7, 2006

### AlephZero

Oops - Nonsense deleted...

Last edited: Dec 7, 2006
6. Dec 7, 2006

### nrqed

No, the last equation is correct. I don't know how you coul dget L squared. L comes inserting the expression for the angle theta which contains just one power of L.

Patrick

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