Question

Four particles having masses m,2 m, 3 m and 4 m are placed at the four corners of a square of edge a.Find the gravitational force acting on a particle of mass m placed at the center.

Answer 1

The calculate the gravitational force on 'm' at unline due to other mouse.
$$ \xrightarrow { FOD } = \dfrac{G\times m \times m \times 2m}{(a/r^2)^2} = \frac{8Gm^2}{a^2}$$
$$ \xrightarrow { FOI } = \dfrac{G\times m \times 2m}{(a/r^2)^2} = \dfrac{6Gm^2}{a^2} $$
$$ \xrightarrow { FOA } = \dfrac{G\times m \times m}{(a/r^2)^2} = \dfrac{2Gm^2}{a^2} $$
$$ Resultant \xrightarrow { Fof } = \sqrt { 64\left( \dfrac { Gm^2 }{ a^ 2 } \right) +36\left( \dfrac { Gm^2 }{ a^2 } \right) =10\dfrac { Gm^2 }{ a^2 } }$$
$$ \xrightarrow { Fof } =\sqrt { 64\left( \dfrac { Gm^{ 2 } }{ a^ 2 } \right) +36\left( \dfrac { Gm^{ 2 } }{ a^{ 2 } } \right) =2\sqrt { 5 } \dfrac { Gm^{ 2 } }{ a^{ 2 } } } $$
The net resultant force will be
$$ F=\sqrt { 100\left( \dfrac { Gm^{ 2 } }{ a^ 2 } \right) +20\left( \dfrac { Gm^{ 2 } }{ a^{ 2 } } \right) -2\left( \dfrac { Gm^ 2 }{ a^ 2 } \right) \times 20\sqrt { 5 } } $$
$$ \dfrac{Gm^2}{a^2} \sqrt{40.4} = 4\sqrt{2} \dfrac{Gm^2}{a^2} $$