I came across this article in my book where they had related the velocities of earth at the aphelion and at the perihelion . Their approach was

$1)$ conservation of angular momentum at the desired points

$2)$ conservation of energy at the desired points.

This method is completely fine but I tried to think about it on my own and tried to replace the second step with equating force separately at the desired points. Now if we take the aphelion to be of distance $r_1$ where the speed of earth is $v_1$ and perihelion to be of distance $r2$ where the speed of earth is $v_2$ , through $1)$ we get $$r_1v_1=r_2v_2$$

And through second we get

$$\frac{G(m_s)(m_e)}{(r_1)^2} = \frac{(m_e)(v_1)^2}{r_1}$$


$$\frac{G(m_s)(m_e)}{(r_2)^2} = \frac{(m_e)(v_2)^2}{r_2}$$

Which gives us $$G(m_s) = (r_1)(v_1)^2=(r_2)(v_2)^2$$

Using the first result in the second result we get $v_1=v_2$. Which is obviously very very wrong untill and unless the orbit would have been circular. Now my question is , where is the flaw because I am unable to identify the step/part where I am doing the mistake.

  • $\begingroup$ $F = m\ v^2/r$ is only valid for circular motion. $\endgroup$ – Bill Watts Jun 6 at 20:30
  • $\begingroup$ But , don't we calculate the values of speeds at different points in an elliptical orbit using $F=m/v^2/r$ and if not then how do we calculate speeds at these different points ?@Bill $\endgroup$ – TheChemist Jun 6 at 20:32
  • $\begingroup$ The general 2 body problem is more complicated and a separate question. But no, we do not use the circular motion formula for elliptic motion. $\endgroup$ – Bill Watts Jun 6 at 20:37
  • $\begingroup$ Ok so basically most books , in order to make calculations easier assume the orbit to be circular and hence we are able to appy the circular motion formula , right? $\endgroup$ – TheChemist Jun 6 at 20:38
  • $\begingroup$ Very elementary books maybe. For elliptic motion you have a radial component as well as a tangential component for velocity except at the extremities. $\endgroup$ – Bill Watts Jun 6 at 20:41

To summarise the discussion in the comments , the formula $$F=\frac{GMm}{R^2}= \frac{mv^2}{R}$$ is only valid for circular orbits and is not valid for elliptical orbits , hence we are not able to relate the velocities of any two given points on the orbit via this method.

| cite | improve this answer | |
  • $\begingroup$ If there was another answer, it seems to be gone. $\endgroup$ – Bill Watts Jun 6 at 21:42
  • $\begingroup$ Sorry I forgot to add this link socratic.org/questions/… $\endgroup$ – TheChemist Jun 6 at 21:57
  • $\begingroup$ The second answer on that page is much closer for small e. $\endgroup$ – Bill Watts Jun 6 at 22:07
  • $\begingroup$ Ok thanks a lot .. $\endgroup$ – TheChemist Jun 7 at 5:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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