It's net gravitational is 0

After posting this problem I got an interesting result which can be stated as follows:

If two objects are separated by a distance of RR and if ratio of the mass of the mass of the second object is nn times the mass of the first object then, if any object regardless of it's mass is placed at a distance of Rn+1\boxed{\frac{R}{\sqrt{n}+1}} from the first object the net gravitational force on it will be 00.


Let there be 22 objects named AA and BB with masses mAm_A and mBm_B respectively with a separation of RR.

If mBmA=n\frac{m_B}{m_A}=n then, mB=mAnm_B=m_An

Now suppose if any other object of mass MM is placed at a distance of xx from the first object (i.e. object AA) then net force of gravity on it becomes 00

=> Force on the third object by object AA == Force on the third object by object BB GMmAx2=GMmB(Rx)2\frac{GMm_A}{x^2}=\frac{GMm_B}{(R-x)^2} GMmAx2=GMmAn(Rx)2\frac{\cancel{GMm_A}}{x^2}=\frac{\cancel{GMm_A}n}{(R-x)^2} 1x2=n(Rx)2\frac{1}{x^2}=\frac{n}{(R-x)^2} nx2=(Rx)2nx^2=(R-x)^2 xn=Rxx\sqrt{n}=R-x x(n+1)=Rx(\sqrt{n}+1)=R x=Rn+1\boxed{x=\frac{R}{\sqrt{n}+1}}

Note by Zakir Husain
1 year, 1 month ago

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This is the way i solved the question lol

Rohan Joshi - 5 months, 2 weeks ago

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