# 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 $R$ and if ratio of the mass of the mass of the second object is $n$ times the mass of the first object then, if any object regardless of it's mass is placed at a distance of $\boxed{\frac{R}{\sqrt{n}+1}}$ from the first object the net gravitational force on it will be $0$. Proof:

Let there be $2$ objects named $A$ and $B$ with masses $m_A$ and $m_B$ respectively with a separation of $R$.

If $\frac{m_B}{m_A}=n$ then, $m_B=m_An$

Now suppose if any other object of mass $M$ is placed at a distance of $x$ from the first object (i.e. object $A$) then net force of gravity on it becomes $0$

=> Force on the third object by object $A$ $=$ Force on the third object by object $B$ $\frac{GMm_A}{x^2}=\frac{GMm_B}{(R-x)^2}$ $\frac{\cancel{GMm_A}}{x^2}=\frac{\cancel{GMm_A}n}{(R-x)^2}$ $\frac{1}{x^2}=\frac{n}{(R-x)^2}$ $nx^2=(R-x)^2$ $x\sqrt{n}=R-x$ $x(\sqrt{n}+1)=R$ $\boxed{x=\frac{R}{\sqrt{n}+1}}$ Note by Zakir Husain
1 year, 1 month ago

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This is the way i solved the question lol

- 5 months, 2 weeks ago