Total energy of elliptical orbit

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The total energy of an elliptical orbit is expressed as E_{tot} = -GMm/(2a), where 'a' is the semi-major axis. This formulation is justified by recognizing that the total energy remains constant at any point in the orbit, similar to circular orbits. The discussion highlights the derivation of orbital speed in elliptical orbits, showing that at the semi-major axis, the speed aligns with that of a circular orbit. The thread also touches on the relationship between total energy and specific distances in the orbit, leading to questions about deriving energy equations without circular logic. The conversation concludes with a request for equations to solve a specific problem regarding a comet's speed at a given distance from the Sun.
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Why is the total energy of an elliptical orbit given by:
E_{tot}=\frac{-GMm}{2a}
Where a=semi major axis.
I agree for a circular orbit I can do the following:
F_c=F_g
ma_c=\frac{GMm}{r^2}
\frac{v^2}{r}=\frac{GM}{r^2}
v^2=\frac{GM}{r}
Since the total energy also equal to the kinetic plus potential energy we have:
E_{tot}=\frac{1}{2}mv^2-\frac{GMm}{r}=\frac{1}{2}m\frac{GM}{r}-\frac{GMm}{r}=\frac{-GMm}{2r}
Ok this is a similar form for circular orbit. But how can we just put a in instead of r for elliptical orbit? What is the justification?
 
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Consider the fact that the total energy of the orbiting object is constant for any point of its orbit. Then take the visa-vis equation for finding the orbital speed at any point of an elliptical orbit:

\sqrt{GM \left (\frac{2}{r}-\frac{1}{a} \right )}

Note that when r=a, you get

v=\sqrt{\frac{GM}{a}}

or

v^2=\frac{GM}{a}

Thus the total energy of a elliptical orbit with a semi-major axis of 'a' is the same as a that for a circular orbit with a radius of 'a'.
 
Well actually I was trying to derive the vis-visa equation through the total energy. So I must have an alternate way to explain it or i'll be circular logic.
 
A comet is in an elliptical orbit around the Sun. Its closest approach to the Sun is a distance of 5e10 m (inside the orbit of Mercury), at which point its speed is 9e4 m/s. Its farthest distance from the Sun is far beyond the orbit of Pluto. What is its speed when it is 6e12 m from the Sun?

please show the equations needed, if you feel generous work the problem out as well.
 
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