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So gravitational potential is given by $V(r)={\frac{GM}{r}}$ and the tangential velocity of a satellite is the square root of $V$, i.e. $\sqrt V$. So how do these two relate, if at all?

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The relation is indeed correct. Let me show you why.

This type of problem is solved by equating forces. Other classical mechanics problems are by equating energies or, in rare cases, momentum.

The gravitational potential indeed is $$ V(r) = G \frac Mr \,.$$ This is not directly useful for this problem. Here you rather want to look at the gravitational acceleration. Is it given by the derivative with respect to $r$, that is $$ a(r) = - G \frac M r \,.$$ The gravitational force is $$ F_\text g(r) = - G \frac{mM}{r^2} \,,$$ where $m$ is the mass of the satellite and $M$ the mass of the earth. The connection of the last two equations if $F = ma$.

As the satellite is moving on a circular path, a centripetal force is needed in order to keep in on that circular math. This force is $$ F_\text c(r) = - m \frac{v^2}{r} \,. $$ The only stable orbit is one for which $F_\text c = F_\text g$ is given. Therefore we need to equate those forces. We get $$ - m \frac{v^2}{r} = - G \frac{mM}{r^2} \iff v^2 = G \frac{M}r \,. $$

So you indeed see that the square root of this potential is the tangential velocity. I did not think that it would, but now I am convinced. I hope you are too!

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It's called the virial theorem: in a bound system, the average of the potential energy of a $\frac{1}{r}$-potential V is related to the average of the kinetic energy T like $\frac{\langle V \rangle}{2} = - \langle T\rangle$.

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The tangential velocity is related to the radius of a circular orbit and the angular velocity of the orbiting object. Specifically: $$\overrightarrow{v}=\overrightarrow{\omega}\times\overrightarrow{r}\to v=\omega r$$ For the satellite, the centripetal acceleration is equal to $r\omega^2$, which is the same (according to the above relation) as the tangential velocity $v$ divided by the radius. The gravitational potential energy of mass 1 orbiting mass 2 is: $$U=\frac{Gm_1m_2}{r}=rF_g=rm_1a_c=r\frac{m_1v^2}{r}$$ this gives the relation that $$v^2=\frac{Gm_2}{r}$$

The fact that they're related stems from how the tangential velocity is determined by how fast you're going around a circle and how big that circle is. The reason this tangential velocity exists at all is because there's some constant centripetal force that causes an angular velocity.

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