# Subbing in Orbital Velocity into Kepler's Law of Periods (1 Viewer)

#### 123ryoma12

##### Member
When rearranged with Gravitational force = Centripetal force you get
v= √(GM/r)
but when you sub in orbital velocity into kepler's law of periods and you rearrange you get
v= √(GMr)
Can someone point out what I'm doing wrong?

#### InteGrand

##### Well-Known Member
When rearranged with Gravitational force = Centripetal force you get
v= √(GM/r)
but when you sub in orbital velocity into kepler's law of periods and you rearrange you get
v= √(GMr)
Can someone point out what I'm doing wrong?
$You should be getting v=\sqrt{\frac{GM}{R}} in both cases. Here's how you get it from Kepler's law:$

$\frac{R^3}{T^2}=\frac{GM}{4\pi^2}\Rightarrow \frac{4\pi^2 R^2}{T^2}=\frac{GM}{R} (multiplying both sides by \frac{4\pi^2}{R})$

$Now notice that the L.H.S. is just v^2, since v=\frac{\text{distance}}{\text{time}}=\frac{2\pi R}{T} (circumference of circular orbit is 2\pi R, and the time taken is the period T). Hence we have v^2 = \frac{GM}{R}, i.e. v=\sqrt{\frac{GM}{R}}.$

#### 123ryoma12

##### Member
I found my mistake. I did v = 2(pi)r^2/T instead of 2(pi)r/t
Thanks for that ^_^