# Orbit related equations

1. Jun 9, 2008

### Davasaurous

1. The problem statement, all variables and given/known data
I'm just trying to rearrange a few equations for gravity, period, radius etc., and am a tad confused.

2. Relevant equations
(G*M)/R^2 = (4*pi^2*R)/T^2

Want to rearrange for T and R. :)

3. The attempt at a solution
I got T to a point of...

T^2 = (4*pi^2*R)*(R^2)/GM
I think that's right, but I'm sure it can be further simplified.

Any halp? :)

2. Jun 9, 2008

### rock.freak667

$$\frac{GM}{R^2} = \frac{4 \pi^2 R}{T^2}$$

$$\frac{GM}{R^3} = \frac{4 \pi^2}{T^2}$$

now re-arrange again.

3. Jun 9, 2008

### Davasaurous

Thanks! Just what I needed! Can't believe I forgot it actually, silly me.

ANYWAY, therefore...

$${T^2} = {4 \pi^2} \frac{R^3}{GM}$$

Yes??

and...

$${R^3} = {GM} \frac{4 \pi^2}{T^2}$$

and...

$${M} = \frac{4 \pi^2 R^3}{G T^2}$$

Just wondering if I could get these verified...

4. Jun 10, 2008

### gaugefield

The second equation is wrong...rest is fine!

5. Jun 10, 2008

### dynamicsolo

When in doubt, check the units. The gravitational constant G has the units N·(m^2)/(kg^2) = (m^3)/[kg·(sec^2)].

So the second equation couldn't be right, since the kg and the (sec^2) in the denominator of G have to be canceled out somehow in order to leave the (m^3) for R^3 on the left-hand side. The correct form must have the combination GM(T^2)...

6. Jun 10, 2008

### Davasaurous

So it'd be ..

R^3 = GMT^2? on 4pi^2

Oh, and I just rearranged the lorentz factor to subject v^2

v^2 = c^2(1-(1/lorentz)^2)

How's that?

Thanks guys

Last edited: Jun 10, 2008
7. Jun 10, 2008

### Kurdt

Staff Emeritus
Both look fine.