# What is the eccentricity of an orbit such that Vp = 2Va?

csmiller1993
Summary:: A question on a recent exam was, "At what eccentricity does an orbit experience a velocity at periapsis that is twice the velocity an apoapsis?" I don't know why the provided solution is correct.

On a recent exam, one of the questions was "At what eccentricity does an orbit experience a velocity at periapsis that is twice the velocity an apoapsis?"

In the exam solutions, the given answer is e = 3/5. with the following as the justification:

e + 1 = 4 - 4e
e = 3/5

I'm not sure where the professor got this relation. When I attempt to solve this problem I end up with e = 1/3. My work is attached. Am I missing something or is the professor wrong? I emailed her and she simply said to check my derivation. I've solved it with the vis viva equation as well and still came out to 1/3. Other students have also reported that they got 1/3.

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Homework Helper
Gold Member
I did it using ##e=\sqrt{1-\frac{b^2}{a^2}}## and got the same answer as you. You have the right to learn from your teacher how to do it correctly. I think it's time to show her your solution and say you couldn't find anything wrong with it. Then ask her what equation she used. It's entirely possible that she made a calculational mistake which she will find out once you ask. You will be doing yourself and your classmates a favor.

Homework Helper
Gold Member
I did it using ##e=\sqrt{1-\frac{b^2}{a^2}}## and got the same answer as you. I think it's time to show her your solution and couldn't find anything wrong then ask her what formula she used. It's entirely possible that she made a mistake. As her student, you have the right to learn from her how to do it correctly.

csmiller1993
I did it using ##e=\sqrt{1-\frac{b^2}{a^2}}## and got the same answer as you. I think it's time to show her your solution and couldn't find anything wrong then ask her what formula she used. It's entirely possible that she made a mistake. As her student, you have the right to learn from her how to do it correctly.
Thanks for the reply and for solving it through for yourself. I think I will talk to her because you're right, there's no reason other's should be losing points over this or be confused by it.

hutchphd
$$\frac{v_p^2}{v_a^2} = \left(\frac{1+e}{1-e}\right)^2$$