Deriving an equation for the orbital period of a satellite 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:
$\frac{GMm}{r^2} = \frac{mv^2}{r}$
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
 A: 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 $\frac 1{\sqrt r}$
A: Since centripetal force exert on the orbited body is equal to the gravitational force exert between the two bodies, then
$$F_c=F$$
$$mv^2/r=GMm/r^2$$
$$v^2/r=GM/r^2$$ [since m and m do cancel]
$$w^2r=GM/r^2$$ (since v=wr)
$$w^2=GM/r^3$$
$$4p^2/T^2=GM/r^3$$ [where is a radian pie]
$$T^2=r^34p^/GM$$
Therefore period $T=\sqrt{4p^2*r^3/GM}$.
A: F(g)=GMm/(R+H)^2  (gravitational force) F(centripetal force) =Mv^2/R+H (here v is orbital velocity) 
here centripetal force is provided  by gravitational force
therefore 
GMm/(R+H)^2=Mv^2/R+H
by working out we get
v^2=Gm/R+H
H is comparatively small than R (radius of earth) and is negligible 
therefore 
v= squareroot of GM/R
but g=GM/R^2 (g is acceleration due to gravity) 
or GM/R=Rg
atlast v =squareroot of Rg(where v is orbital velocity) 
