Why here is gravitational force not $$\frac{MGm}{r^2}$$?

Question

anonymous · Answer

http://www.scholarpedia.org/article/Celestial_mechanics#Newton.E2.80.99s_Celestial_Mechanics They claim that it's$$\frac{mG(M+m)}{r^2}$$

anonymous · Answer

That is, assuming $$U=-\nabla F$$

anonymous · Answer

I've looked at the article and I could not find where the author  says the force depends on the sum of the masses but he does say the potential energy of M and m is$$G(M+m)m/r$$ which I do not understand.   I also found other discrepancies in the article compared to standard treatises on mechanics so I do no put much faith in it.  

Maybe someone else can  look at it but I think he is off base.

The internet is not necessarily the best source of scientific info either because it may not be edited or the author may have made a mistake or may not be schooled well enough in the discipline.  So beware, seek other sources to corroborate  the info.

anonymous · Answer

I went back to the article and I think I know what he did.  Although I didn't see where he stated it but I think he is expressing his equations in the center of mass coordinate system. In the CM coor.$$F _{CM}=ma = -G(M+m)m/r ^{2}$$  Without verifying the accuracy of any other of his statements or equations my criticism was unjustified and I retract my previous comments on his approach.

However I still maintain that info on the internet must be scrutinized carefully for accuracy , validity of approach and assumptions and even typos as you do not know how well it was edited.  Some discourses are not particularly meant for learning fundamentals.

anonymous · Answer

http://www.physicsforums.com/showthread.php?t=677089