Question

Three identical point masses, each of mass 1 kg lie in the x-y plane at points (0,0,0), (0,0.2m) and (0.2m, 0). The net gravitational force on the mass at the origin is

• A. $1.67\times {10}^{-9}(\hat{i}+\hat{j})N$
• B. $3.34\times {10}^{-10}(\hat{i}+\hat{j})N$
• C. $1.67\times {10}^{-19}(\hat{i}+\hat{j})N$
• D. $3.34\times {10}^{-10}(\hat{i}+\hat{j})N$

Answer 1

The correct option is A

$1.67\times {10}^{-9}(\hat{i}+\hat{j})N$

Let particle A lies at origin, particle B and C on y and x-axis respectively

$\overrightarrow{F_{AC}}=\frac{G{m}_A{m}_B}{r_{AR}^2}\hat{i}=\frac{6.67\times {10}^{-11}\times 1\times 1}{(0.2{)}^2}\hat{i}=1.67\times {10}^{-9}\hat{i}NSimilarly\overrightarrow{F_{AC}}=1.67\times {10}^{-9}\hat{j}N\therefore NetforceonparticleA\overrightarrow{F}=\overrightarrow{F_{AC}}+\overrightarrow{F_{AB}}=1.67\times {10}^{-9}(\hat{i}+\hat{j})N$