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

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The discussion revolves around a question from an exam asking for the eccentricity of an orbit where the velocity at periapsis is twice that at apoapsis. The provided solution claims the eccentricity is e = 3/5, but several students, including the original poster, calculated e = 1/3 using the vis viva equation and other methods. There is uncertainty about the professor's derivation, with suggestions that she may have made a mistake in her calculations. Students are encouraged to present their findings to the professor to clarify the discrepancy and ensure accurate grading for all. The conversation highlights the importance of addressing potential errors in academic settings for the benefit of all students.
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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.

Thanks in advance!
 

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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.
 
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.
 
kuruman said:
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.
 
I would give twice full credit to the first student who pointed a mistake like this to me on a test. It didn't happen often but encouraged students to speak up if they perceived something wrong.
 
I also get ##e=1/3##. Think perhaps the professor missed the square on the right hand side of
$$\frac{v_p^2}{v_a^2} = \left(\frac{1+e}{1-e}\right)^2$$
 
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