2
$\begingroup$

I want to have a general solution for calculating orbits. Wikipedia says how to put object on circle orbit. We must give it speed $v_1 = \sqrt{g(h_o +R)} $ Where $h_o$ - is the orbit height, and $R$ - Earth radius.

But how to calculate the speed that the object must have, if we want to place it in an elliptical orbit?

For example:

Having massive massive spherical object with mass $M$ and radius $R$.

which speed $V$ we must to give to this object to put it on elliptical orbit with eccentricity $\epsilon = E$ , and perihelium $P$?

Improve this question
$\endgroup$
2
  • $\begingroup$ Well. This could be a duplicate, but on my level of understanding physics and language it's too hard to make assumptions of such kind. Furthermore, given answer is very good in understanding. $\endgroup$
    – Vasiliy Stavenko
    Commented Jun 27, 2014 at 8:37
  • $\begingroup$ I don't see this as a duplicate. The proposed duplicate question addresses the concept of the shape of the orbit but doesn't address the concept of orbital velocity at all. The vis viva equation is the answer to Vasiliy's question. Since that's part of the accepted answer, reopening is not particularly important. But still, this question should not have been closed as a duplicate. $\endgroup$
    – David Hammen
    Commented Jul 1, 2014 at 15:24

1 Answer 1

Reset to default
3
$\begingroup$

You can solve this problem by using energy conservation. It holds

\begin{equation} -\frac{GMm}{P}+\frac{1}{2}mV^2 = - \frac{GMm}{2a}, \end{equation}

where $a$ is the semimajor axis and related to $P$ by $P=a(1-\epsilon)$. You can understand the RHS from Virial's theorem, for instance.

You may also want to check out the vis-viva equation.

psm.

Improve this answer
$\endgroup$
3
  • $\begingroup$ What does RHS stands for? $\endgroup$
    – Vasiliy Stavenko
    Commented Jun 27, 2014 at 8:27
  • $\begingroup$ Right hand side. :-) $\endgroup$
    – psm
    Commented Jun 27, 2014 at 8:29
  • $\begingroup$ The correct equation is: \begin{equation} -\frac{GMm}{P}+\frac{1}{2} \mu V^2 = - \frac{GMm}{2a}, \end{equation}. Where $\mu$ is the reduced mass. $\endgroup$
    – Nick
    Commented Feb 7, 2015 at 2:37

Not the answer you're looking for? Browse other questions tagged or ask your own question.