Two bodies of masses - M - 1 - and - M - 2 - are placed a distance R apart- Then the position where the gravitational field due to them is zero- the gravitational potential is- G -frac -displaystyle sqrt - - M - 1 - - - displaystyle R - - G -frac - displaystyle sqrt - - M - 2 - - - displaystyle R - - left- sqrt - - M - 1 - - -sqrt - - M - 2 - - right- - 2 - displaystyle frac - G - R - - left- sqrt - - M - 1 - - -sqrt - - M - 2 - - right- - 2 - displaystyle frac - G - R -

Question

Toppr Admin · Accepted Answer

Let O be the point where gravitational field due to the masses is Zero.
Thus, $\frac{G M_{1} m}{x^{2}} - \frac{G M_{2} m}{(R - x)^{2}} = 0 ⟹ x = \frac{\sqrt{M_{1}} R}{\sqrt{M_{1}} + \sqrt{M_{2}}}$
Now gravitational potential at point O, $V_{o} = [- \frac{G M_{1}}{x^{2}}] + [- \frac{G M_{2}}{(R - x)^{2}}]$
Put the value of $x$ , $⟹ V_{o} = - (\sqrt{M_{1}} + \sqrt{M_{2}})^{2} \frac{G}{R}$

Two bodies of masses M1 and M2 are placed at a distance R apart. Then at the position where the gravitational field due to them is zero, the gravitational potential is−G√M1R−G√M2R−(√M1+√M2)2GR−(√M1−√M2)2GR

Let O be the point where gravitational field due to the masses is Zero.Thus, GM1mx2−GM2m(R−x)2=0⟹x=√M1R√M1+√M2 Now gravitational potential at point O, Vo=[−GM1x2]+[−GM2(R−x)2]Put the value of x, ⟹Vo=−(√M1+√M2)2GR