Question

A particle of mass m1 is kept as x=0 and another mass of me at x=d.When a third particle is kept at x=d/4,it experiences no net gravitational force due to the two particles.m1/me answer=9.

Answer 1

Please find below the solution to the asked query:

Given that M1 and M2 masses are at a distance "d" apart. Let "m" is the mass of the third particle kept on the line joining the two masses. Let the mass "m" is kept at a distance "x" from the mass M1. Then the mass "m" is at a distance d−x from mass M2. The situation is as shown in the figure.

So, the force on mass "m" due to mass M1 is,
FmM1 = GM1mx2
Where, G = universal gravitational constant.

The force on mass "m" due to mass M2 is,
FmM2 = GM2m(d−x)2
Therefore, the two forces are opposite to each other as shown in the figure.
So, If the particle of mass "m" should be at rest, then the two forces should be equal to each other in magnitude.
FmM1 = FmM2 ⇒ GM1mx2 = GM2m(d−x)2 ⇒ M1x2 = M2(d−x)2⇒ √M1 (d−x) = √M2 x ⇒  √M1 d = (√M1 + √M2)x⇒ x = (√M1(√M1 + √M2))d
 

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