Question

As shown in figure, four masses each of mass 3√2 kg at the corners of a square of side 3 m.
Calculate the gravitational potential of system of these four particles. Also calculate the
gravitational potential at the centre of square. (G = 6.67 x 10-11 SI unit)
​

Answer 1

Answer:

-8G = -53.36 x $10^{-11}$ J/kg

Explanation:

A few things to note about Gravitational Potential:

• It is a scalar quantity. When you are finding G.P about a point, you should only add the values. It doesn't depend upon the direction or position of the particle.
• It is similar to gravitational potential energy. Gravitational potential is gravitational potential energy by mass of the 2nd particle

Gravitational Potential = -Gm/R

In this question:

About point 'O', net gravitational potential will be the sum of gravitational potential of `1 to 4.

Gravitational potential of the 4 particles are same as they are of the same mass, and are equidistant from the center.

Gravitational potential of one particle is equal to:

-Gm/R = -G($3\sqrt{2} / r$)

Using pythagoreas's theorem:

$(3/2)^2 + (3/2)^2 = x^2$

(9/2) = $x^2$

x = $3/\sqrt{2}$, where x is equal to r

So, -Gm/R = -G($3\sqrt{2} / r$) = -G(2) = -2G

Net Gravitational Potential = 4 x (Gravitational Potential of 1 particle)

= 4 x (-2G) = -8G

G = 6.67 x $10^{-11}$

= -8 x 6.67 x $10^{-11}$ = -53.36 x $10^{-11}$ J/kg

Now, some may wonder about the 'pythagoreas theorem' part. I found out the length of 'r' in the diagram by taking the right angle with 'r' has hypotenuse, and (length of the side)/2 as the 2 sides.

I hope it helps!