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

123ryoma12 · Jun 8, 2015

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 · Jun 8, 2015

123ryoma12 said:
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 · Jun 8, 2015

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

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

123ryoma12

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InteGrand

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