Question

Four particles having masses, m, 2m, 3m, and 4m 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 centre.

Answer 1

Given:

Masses of four particles placed at courners of square:

m, 2m ,3m ,4m

Mass of particle placed at center of square: m

To find :

The gravitational force acting on central particle.

Solution :

Length of each side of square : a

Distance of each corner from center r :

$\bf \frac{a}{ \sqrt{2} }$

We know that formula of Gravitational force F :

$\\ \bf\frac{Gm_1 m_2 }{ {r}^{2} }$

(equation 1)

where G = Gravitational constant.

m's are the masses of two objects.

r = distance between two objects.

So, using equation 1.

Force on center(m) by mass m:

$\\ \bf F_1 = 2\frac{Gm \cdot m}{ {a}^{2} } = 2\frac{G {m}^{2} }{ {a}^{2} }$

Force on center (m) by mass 2m:

$\\ \bf F_2= 2\frac{Gm \cdot 2m}{ {a}^{2} } = 4\frac{G {m}^{2} }{ {a}^{2} }$

Force on center (m) by mass 3m:

$\\ \bf F_3 = 2\frac{Gm \cdot 3m}{ {a}^{2} } = 6\frac{G {m}^{2} }{ {a}^{2} }$

Force on center (m) by mass 4m:

$\\ \bf F_4 = 2\frac{Gm \cdot 4m}{ {a}^{2} } = 8\frac{G {m}^{2} }{ {a}^{2} }$

Net force by particle of mass m and 3m (in the direction of 3m) :

F₁₃ = F₃− F₁

$\\ \bf F_{13} = 6\frac{G {m}^{2} }{ {a}^{2} } - 2\frac{G {m}^{2} }{ {a}^{2} } = 4\frac{G {m}^{2} }{ {a}^{2} }$

Net force by particle of mass 2m and 4m (in the direction of 4m) :

F₂₄ = F₄− F₂

$\\ \bf F_{24} = 8\frac{G {m}^{2} }{ {a}^{2} } - 4\frac{G {m}^{2} }{ {a}^{2} } = 4\frac{G {m}^{2} }{ {a}^{2} }$

By using Pythagoras theorem, Net force on system is

F :

$\bf \: F = \sqrt{(F_{13} ^{2} ) \: + (F_{24} ^{2} )}$

Putting required values from above :

$\bf \: F = \sqrt{( \frac{4G {m}^{2} }{ {a}^{2} } )^{2} \: + ( \frac{4G {m}^{2} }{ {a}^{2} } )^{2} \: }$

So, the force applied on the center of square having mass m =

$\\ \bf \: F = \frac{4\sqrt{2} G {m}^{2} }{ {a}^{2} } \: \: \: N$