Question

Three masses each of 1kg, are placed on the vertices of an equilateral triangle of sides
of length 0.4m. What is the total gravitational potential energy of the configuration?
PLEASE ANSWER URGENTLY. I HAVE MY PHYSICS EXAM TOMORROW
THANKS

Answer 1

Explanation:

Given Three masses each of 1 kg, are placed on the vertices of an equilateral triangle of sides  of length 0.4 m. What is the total gravitational potential energy of the configuration?

Given m = 1 kg and l = 0.4 m

Potential energy between two masses will be – Gm^2 / l

So there are 3 pairs AB, BC and CA

So net potential energy will be 3 x -Gm^2 / l

                                           = - 3 x 6.67 x 10^-11 x 1^2 / 0.4

                                          = - 5 x 10^-10 J

Answer 2

The gravitational potential energy of the system is:

              $-1.0422 \times 10^-^9 Joule$

                                               

Explanation:

• Three masses are placed on each vertex of an equilateral triangle.

• They all have equal magnitude of weight as 1 kg.

• The side length of triangle = 0.4 miter

We know the formula to find out the gravitational potential energy (U) as,

• $U = -G\left ( \dfrac{m_3m_2}{r_3_2}+\dfrac{m_3m_1}{r_3_1}+\dfrac{m_2m_1}{r_2_1} \right )$

We put the value of variables as;

$m_1 = m_2 = m_3 = 1 kg$

$r_3_2 = r_2_1 = r_3_1 = 0.4 \ miter$

Also, we know that $G = 6.67 \times 10^-^1^1 \ Nm^2/kg^2$

• $U = - 6.67 \times 10^-^1^1 \left ( \dfrac{1}{0.4}+\dfrac{1}{0.4}+\dfrac{1}{0.4} \right )$
• U =$-1.0422 \times 10^-^9 Joule$

This is the gravitational potential energy of the system.