Gravitational Potential Energy of a Spherical Shell

We all have experienced this instinctively when a big weight is lifted above our head we feel it be a potentially dangerous situation. However, since the weight can be adequately secured, it is not necessarily hazardous. Something causes the force that keeps the weight from falling due to gravity to fail. To use proper physics terms, we are talking about the  weight ‘s gravitational potential ability that is converted into electricity.

The symbol often assigned to gravitational potential energy is Ug. Gravitational potential energy denotes the energy an object possesses as a result of being positioned at a given function in a gravitational field.

Gravitational potential energy of a spherical shell

We will learn through this conversation that Newton’s law of gravitation holds true for massive things with spherical mass distributions.

Let us consider the following points:

• A point mass m outside a spherical shell at a distance r from the centre of the shell
• Mass of the spherical shell is M
• Radius of shell is R
• A is the area

Here, we find the gravitational potential energy of the system of the spherical shell and the point mass.

Let us take a thin spherical shell (ring element) of mass dM and thickness dl=R on the spherical shell. Each particle in the ring is at a distance l from the point mass. The gravitational potential energy of the ring element and the point mass is:

dU=−GmdM / l

The area of the shell element is dA=2πR2sinθdθdθ. Also point to be noted is that the area of the shell element is the circumference 2πRsinθ times the thickness dl=Rd.

The ratio of the mass of the shell element to the total mass of the shell is now equal to the area of the shell element to the overall area of the shell, producing,

dM/ M=dA /A = 2πR2sinθdθ / 4πR2

or,

dM=1/2Msinθdθ

On substituting dM from the above equation we will get,

dU=−GmMsinθdθ / 2l

OA=Rcosθ and therefore,

l²=(r−Rcosθ)²+(Rcosθ)² or l²=r²−2rRcosθ+R²

ldl=rRsinθdθ

Let us substitute l from the above equation, we will get,

dU=−GmMdl / 2rR

Integrating the foregoing calculation yields the gravitational potential energy of the spherical shell system and the point mass.

r−R to r+R

U=−GmM2rRr+R-rr+R−dl=−GmMr

Thus,  it is clear that the gravitational potential energy of the system of the spherical shell and the point mass is as if the entire mass of the spherical shell were concentrated in the centre of the shell.

Since the gravitational force if F = -du/dr you can obtain the expression for the force as follows:

F=GmM * dr−1/dr =−GmM r²

If the aforementioned point mass is inside the spherical shell at a distance r from the center, the limits of integration while integrating change from R−r to R+r and the total gravitational potential energy of the system becomes:

U=−GmM/2rRr + ^r+R∫ _R-r−dl=−GmM / r

If there is R instead of r, then the gravitational potential energy when the point mass is inside the shell is constant, and the force on point mass is zero.

This final result is identical to what we would get if all of the mass was concentrated in the middle of the shell. This result will hold true for all shells. Since a sphere is constituted of such shells, it should be true for all spheres also. The phenomenon occurs even if the different shells have different mass densities, i.e. if the density is a function of the radius. From this we can derive that the gravitational pressure exerted by one planet on another planet is such that the mass of all the planets is concentrated in the centre.

Conclusion

We estimated g using the gap between the mass m and the Earth as the radius of the Earth while investigating Newton’s gravitational discoveries. To put it another way, we thought that all of the Earth’s mass is centred at its core. This hypothesis may appear logical when we are far away from the Earth (i.e., when the radius of the Earth is negligible), but it does not appear to be accurate while we are standing on the Earth’s surface. We can see, however, that this assumption holds true for any object outside of a gravitating sphere’s floor (to which the Earth is a good approximation). This is a significant outcome. It’s a combination of superposition, the inverse rectangular law, and spherical symmetry.