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I'm combining three equations from my physics text book:

  • Newton's law of gravitation: $F = -\frac{GMm}{r^2}$

  • The centripetal force equation: $F = \frac{mv^2}{r}$

  • The equation for the speed of an object traveling in a circle: $v = \frac{2 \pi r}{T}$

I wanted to create an equation to find the Time period, $T$ and ended up with: $T = \frac{2 \pi r^2}{GM}$ Which is wrong...


EDIT

I've worked it out again, this is my working:

I put Newton's law of gravitation and the centripetal force equation equal to each other:

$\frac{GMm}{r^2} = \frac{mv^2}{r}$

Divide both sides by $r$:

$\frac{GMm}{r} = mv^2$

Sub in $v = \frac{2 \pi r}{T}$ for $v$:

$\frac{GMm}{r} = m(\frac{2 \pi r}{T})^2$

Divide both sides by $m$:

$\frac{GM}{r} = (\frac{2 \pi r}{T})^2$

Root both sides:

$\sqrt{\frac{GM}{r}} = \frac{2 \pi r}{T}$

Flip both sides and divide by $2 \pi r$:

$T = \frac{2 \pi r}{\sqrt{\frac{GM}{r}}}$


EDIT 2 Which I can simplify:

Multiply both sides by $\sqrt{\frac{GM}{r}}$:

$T \times \sqrt{\frac{GM}{r}} = 2 \pi r$

Square both sides:

$T^2 \times \frac{GM}{r} = (2 \pi r)^2$

Divide both sides by $\frac{GM}{r}$:

$T^2 = \frac{(2 \pi r)^2}{\frac{GM}{r}}$

Clean it up:

$T^2 = \frac{(2 \pi r)^2 \times r}{GM}$

Take out $r$ to get the final answer:

$T^2 = \frac{(2 \pi)^2}{GM}r^3$

If you take out the constant you get Kepler's law (as Ross Millikan said):

$T^2 \propto r^3$

Is this correct? It's going in my A-Level Physics notes and I don't want to be learning the wrong stuff when it comes to the exam. This is correct now, thanks guys!

If anybody's interested, I've open sourced the notes here

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Your equations are right... you just plugged in wrong. :( –  Stuart Dec 10 '12 at 19:31
    
Actually, questions of this kind belong to the physics forum: physics.stackexchange.com –  Dominik Dec 10 '12 at 20:18
    
Ahh, I didn't know about that... Can somebody with lots of reputation move it please? –  Todd Davies Dec 10 '12 at 20:21

1 Answer 1

up vote 5 down vote accepted

It is not correct. Kepler's third lawstates: The square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit. Your solution has the square, not the $\frac 32$ power of the axis. You are using the correct input, so if you show your work we may find the problem. $v$ should be proportional to $\frac 1{\sqrt r}$

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Okay, hang on, I'll edit the question! –  Todd Davies Dec 10 '12 at 19:27
    
I've added all my working, does that look right? –  Todd Davies Dec 10 '12 at 19:36
    
I've accepted your answer, as you lead me to the right answer, as described in the question. Thanks mate! –  Todd Davies Dec 10 '12 at 20:12

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