# Calculation of Gravitational Potential at the centre of the cube [closed]

I came across this problem in a test and have been able to come up with a solution however I am unsure if it is correct.

I started by building a cube of twice the initial dimensions to bring point P to the centre of a larger cube. This was to bring an element of symmetry to the figure. Since, each cube contributes potential V to the centre, the net gravitational potential will be 8V since there are 8 cubes. Now I wanted to find the Gravitational Potential at the centre of my original cube. So I considered an imaginary solid cube inside the larger one such that point P lies at the centre of both the cubes i.e the larger and smaller one. Now, I found the factor by which the gravitational potential had increased at the centre. Since, gravitational potential is directly proportional to the mass of the particle and inversely proportional to the distance, I considered the fact that on average, particles on the inner imaginary solid cube would be at a distance twice of that between particles on the outer larger sphere and point P. Since the mass of the larger cube is 8 times the mass of the smaller one, I concluded that the potential at point P due to a cube with same dimensions as the original cube would be 2V. Therefore the potential due to the remaining parts of the larger cube would be 6V. If we divide the remaining part of the cube providing a potential of 6V into 8 symmetrical parts, they resemble what the problem asks us to find. Thus, the answer came out to be 3V/4. However I recently received a hint that this was incorrect and have not been able to conclude why.

## closed as off-topic by Aaron Stevens, Buzz, rob♦Feb 5 at 6:31

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – Aaron Stevens, Buzz, rob
If this question can be reworded to fit the rules in the help center, please edit the question.

The Gravitational Potential at the corner of a cube of given density should be proportional to $$L^2$$ where $$L$$ is the length of an edge. The potential is the potential due to the cube minus the potential due to the removed cube, $$1-{(\frac{1}{2}})^2 = \frac{3}{4}$$ of the original cube, in agreement with your method. So $$\alpha = 3$$ and $$\beta=4$$, so $$\alpha + \beta = 7$$. But it could equally be $$\alpha = 6$$ and $$\beta=8$$, so $$\alpha + \beta = 14$$. That part doesn't make any sense to me.

• Thank You. Yes I understand what you mean but the question had answers only between 0 to 9 so you would have to consider 3/4 only – Aditya Sriram Feb 5 at 7:01

Each cube in the top layer has a matching cube in the bottom layer exactly opposite it:

Top layer:

t1   t2
t3   t4


Bottom Layer

b1    b2
b3    b4


t1 is pulling the center with the same force but in exactly the opposite direction as b4, so the net force is 0. The applies to t2/b3, t3/b2 and t4/b1. All of these have a net zero affect, so the net gravitational force at the center point is 0.

This exactly analogous to why the net gravity at the center of a sphere is 0.

• The question asks about gravitational potential, not gravitational force. – Michael Seifert Feb 4 at 17:27