Here is a question from ANTHE: an exam in India. The question is:
Two point masses M and 3M are placed at a L distance apart. Another point mass m is between on the line joining them so that the net gravitational force acting on it due to masses M and 3M is zero. The magnitude of gravitational force acting on m due to M will be? (Please answer in terms of variables!)
This question looked for me as the same lagrange point between sun and earth.
Hope I will get the answer soon, thank you in advance.

  • 3
    $\begingroup$ nope! Lagrange points are based on equilibrium in a rotating frame where in addition to real gravitational potential energy, you also include a fictitious rotational pseudopotential that accounts for the apparent behavior of real rotational effects as they appear in the rotating frame. You can not argue that your problem is a special case for Lagrange points where rotation goes to zero because there must be a finite rotation speed that keeps $M$ and $3M$ in orbit around their center of mass. So better avoid thinking about rotation for your physics problem. $\endgroup$ – uhoh Oct 20 '18 at 12:16
  • 1
    $\begingroup$ Welcome to Space by the way! Your question is definitely off-topic here. I'd recommend you go to Physics SE instead, but don't post this question! I think there is a chat room for people working out problems. Try Problem Solving Strategies. Neither the site nor the chat room will tell you the answer, but in chat you might get some recommendations how to proceed. I'm not sure, I've never tried it. $\endgroup$ – uhoh Oct 20 '18 at 12:19
  • $\begingroup$ Read this please. Homework questions without any effort are usually closed. People on SE sites do not welcome Can you do my work for me? questions. $\endgroup$ – user10509 Oct 20 '18 at 15:22
  • $\begingroup$ @JanDoggen Read this please Your link is to a completely different SE site. I voted to close this question because it is "about other space sciences (physics..." I think that you shouldn't generalize like "People on SE sites do not welcome..." Let's try hard to be welcoming! I think it's better to simply encourage the OP add more to the question, or to ask the usual "what have you tried so far?" $\endgroup$ – uhoh Oct 21 '18 at 16:33

This is not the same thing as a Lagrange point, but is similar and I definitely understand the confusion. Lagrange points must rotate along with the the mass M so are slightly closer to mass 3M such that its orbital speed (in terms of angular velocity) is the same as that of M. In the situation described, the gravitational forces are perfectly cancelled meaning that mass m is not orbiting at all this situation is actually unstable and can only exist instantaneously.

As for the actual answer the problem is fairly simple, simply apply the inverse square law! the mass (and thus gravitational force) of 3M is three times that of M so the mass m must be root three times further from it than it is to M. The distance between M & m is then L/(1 + Sqrt (3)). Under newtonian gravity this gives us a force of gMm (4+2Sqrt(3)) / L^2.


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