I'm combining three equations from my physics text book:
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Newton's law of gravitation: $F = -\frac{GMm}{r^2}$
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The centripetal force equation: $F = \frac{mv^2}{r}$
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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:
$\frac{GMm}{r^2} = \frac{mv^2}{r}$
Multiply 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
Best Answer
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 $\frac 1{\sqrt r}$