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This question already has an answer here:

Consider a planet of mass $m$ moving around the Sun of mass $M$ in a circular orbit of radius $r$. By equating the centripetal force to $\frac{mv^2}{r}$ to gravitational attraction $\frac{GMm}{r^2}$, one finds $v=\sqrt{\frac{GM}{r}}$ where $v$ is the planet's tangential velocity.

How does one calculate the tangential velocity if the planet's actual orbit (which is elliptical) is taken into account?

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marked as duplicate by sammy gerbil, stafusa, Jon Custer, peterh, Kyle Kanos Dec 19 '17 at 18:48

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Working out the equation for an elliptical orbit is surprisingly involved, at least compared to the circular orbit. However if we know the semi major axis of the orbit we can simply use conservation of energy. That is, the total energy is constant so when the planet moves nearer to the Sun the decrease in its potential energy is balanced out by an equal increase in the kinetic energy. Conservation of energy gives us the vis-viva equation:

$$ v^2 = GM\left(\frac{2}{r} - \frac{1}{a}\right) $$

For a circular orbit $a=r$ and the equation simplifies to the usual one for circular motion:

$$ v^2 = \frac{GM}{r} $$

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