Gravitational Field Of Multiple Masses

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Two particles of mass m and 4m are fixed at a separation d on a frictionless table. At what distance from the lighter mass, along the line joining them, is the net gravitational field zero?

RankGuru · Accepted Answer

NCERT Class 11, Chapter 8 (Gravitation) establishes that the gravitational fields produced by several masses add as vectors at any point. The null point on the line joining two masses lies where the two field contributions point in opposite directions and are equal in magnitude, so they cancel. Let the point be at distance $x$ from the lighter mass $m$, so it is $(d - x)$ from the heavier mass $4m$. Setting the field magnitudes equal, $\dfrac{Gm}{x^2} = \dfrac{G(4m)}{(d-x)^2}$. Cancelling $Gm$ and taking square roots gives $\dfrac{1}{x} = \dfrac{2}{d-x}$, so $d - x = 2x$, which yields $x = \dfrac{d}{3}$. The null point therefore lies closer to the lighter mass, as expected. The option $d/2$ would be correct only for two equal masses. The option $2d/3$ places the point closer to the heavier mass, which is backwards. The option $d/5$ results from mishandling the square root step. As a plausibility check, the field must vanish nearer the smaller mass so that its weaker pull can balance the stronger pull of the heavier mass, and $\dfrac{d}{3} < \dfrac{d}{2}$ confirms this is on the correct side.

Gravitational Field Of Multiple Masses

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