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:

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

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. :( – StuartHa 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

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

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 $T=\sqrt{4p^2*r^3/GM}$.

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Here is the time to learn latex! – peterh Mar 18 '15 at 6:18

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)

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Welcome to MSE. Please have a look at our guide to typesetting math here. – A.P. Mar 31 '15 at 17:55