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^{2}-GM/r=constant. To derive escape velocity we find for what v does E=0. But an orbit of this nature is certainly not circular! How can we apply the formula?

- Thread starter StephenPrivitera
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chroot

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Bound orbits are those with E < 0.

- Warren

- Warren

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As Warren pointed out, circular or eliptical orbits will have E<0

A parabolic orbit (the one followed by an object traveling exactly at escape velocity) E=0

For E > 0, you get a hyperbolic orbit.

One thing about circular orbits:

v= [squ](GM/r) at all points. if you substitute this for v in the formula you have it reduces to

E=-GMm/2r

Now, this turns out to be also true for eliptcal orbits if you substitute the semi-major axis(a) for r. (the semi-major axis is half of the longest dimension of the ellipse. It is also the Average length of the radius vector over the course of an orbit. )

This gives

E = -GMm/2a

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