A satellite is in orbit at an altitude of 62.5km, with a speed along the earth-relative position vector of 3.98km/s, and a speed perpendicular to the earth-relative position vector of 7.763km/s. What is the eccentricity of its orbit?

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    $\begingroup$ Given those constraints, (and without further acceleration) it is either on an escape vector from Earth, or in a sub-orbital hop to another point on Earth. So since it is 'not in orbit' I'd argue that the question is meaningless. $\endgroup$ Feb 3 '16 at 9:36
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    $\begingroup$ homework problem? $\endgroup$ Feb 3 '16 at 10:04
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    $\begingroup$ @AndrewThompson - The magnitude of the velocity vector is 8.72 km/s, less than the 11.1 km/s escape velocity at that altitude. The eccentricity is less than one. This looks like homework, so Carl, it's best if you do this by yourself. Hint: Use the eccentricity vector. $\endgroup$ Feb 3 '16 at 13:03
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    $\begingroup$ 62.5 km?! It won't be in orbit for very long. Also you didn't say if the 3.98 km/s is up or down. (You don't need that sign for the eccentricity.) If it's down, then it really won't be in orbit for very long. $\endgroup$
    – Mark Adler
    Feb 3 '16 at 17:48
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    $\begingroup$ Ended up using the vis-viva equation along with locating the semimajor axis distance to finally find eccentricity. Thanks for the help. $\endgroup$ Feb 5 '16 at 0:05

You should write everything in vector form and then it is just a matter of applying a couple of formulas.

1) Use the given position and velocity values to write the position and velocity vectors, $\vec{r}$ and $\vec{v}$

2) Compute $\vec{h} = \vec{r} \times \vec{v}$ (where $\times$ is the cross product)

3) Compute the eccentricity $\vec{e} = \dfrac{1}{\mu}(\vec{v} \times \vec{h})-\dfrac{\vec{r}}{|\vec{r}|}$


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