Gravitational Potential Due to Solid Sphere

Gravitational Potential Due to Solid Sphere

The main interest of this article is to calculate Gravitational Potential due to Solid Sphere. However, if you are interested in a detailed study of Gravitation, you can start here. In our previous article, we have discussed how to calculate Gravitational Potential due to a Spherical shell. We will use that knowledge to solve this problem. Again, the main idea will be to calculate the amount of work done to move a unit mass system from a point of reference to the field of influence of the source mass M.

Gravitational Potential due to Solid Sphere

Let us consider a uniform sphere of mass M and radius a. The center of mass is located at point O. We have to calculate the potential due to this solid sphere at a test point P, as shown in the figure below. We denote the distance between O and P as r. By definition, the potential is the work done to move a unit mass from a point of reference, here infinity to the test point r.

gravitational potential due to solid sphere test point is outside of the sphere

We will first draw two concentric spherical shells of radius x and x+dx. The volume enclosed by this thick shell of interest is 4\pi x^2dx. We can easily calculate the mass due to this thick spherical shell as

    \[dM=\frac{M}{\frac{4}{3}\pi a^3}\times 4\pi x^2 dx\]

    \[=\frac{3M}{a^3}x^2dx\]

Now we can use the final expression for gravitational potential due to a spherical shell to calculate the gravitational potential due to a solid sphere at a point either outside or inside of the sphere.

Test Point P is Outside of the Sphere

The most trivial situation is that the test point P is outside of the sphere. We use our ready-made equation from earlier exercise to calculate the potential at this point.

    \[dV=-\frac{G dM}{r}\]

Thus the potential due to the solid sphere will be

    \[V=-\frac{G}{r}\int{dM}\]

    \[\boxed{V=-\frac{GM}{r}}\]

Again we note that the gravitational potential due to solid sphere at an external point is identical to the gravitational potential due to a point mass M situated at the center of the sphere.

Test Point P is Inside of the Sphere

Here the situation is more complex than the earlier one, however, we can divide the sphere into two parts. The first one which is a sphere of radius r and the second one is a thick spherical shell with the inner radius of r and outer radius of a. Let us assume that the mass of the inner sphere is M’, then

    \[M'=\frac{M}{\frac{4}{3}\pi a^3}\times \frac{4}{3} \pi r^3\]

    \[=\frac{Mr^3}{a^3}\]

gravitational potential due to solid sphere test point is inside pf the sphere

and the potential at a test point, P due to this sphere is

    \[V'=-\frac{GM'}{r}\]

    \[=-\frac{GMr^2}{a^3}\]

Now we have to calculate potential at point P due to the spherical shell of inner radius r and outer radius a. For that, we will divide the shell into several concentric shells. Let us consider one such shell with an inner radius of x and outer radius x+dx. The corresponding mass of the shell is

    \[dM''=\frac{M}{\frac{4}{3}\pi a^3}4\pi x^2 dx=\frac{3M}{a^3}x^2dx\]

The potential at an internal point P due to this shell is

    \[dV''=-\frac{GdM''}{x}=-\frac{3GM}{a^3}xdx\]

Thus the potential due to this part is

    \[V''=-\int_{r}^{a}{\frac{3GM}{a^3}xdx}\]

    \[=-\frac{-3GM}{2a^3}(a^2-r^2)\]

and the total gravitational potential due to solid sphere at an internal point is

    \[V=V'+V''\]

    \[\boxed{V=-\frac{GM}{2a^3}(3a^2-r^2)}\]

Gravitational Potential due to Cavity

In the following section, we will calculate the gravitational potential due to a spherical cavity. Let us consider a solid sphere of radius a. We create a spherical cavity as shown in the diagram below. The radius of the spherical cavity is \frac{a}{2}. Taking gravitational potential at \infty is zero, calculate the potential at the center of the cavity.

gravitational potential due to cavity

Let V, V_s and V_c are the potential due to the solid sphere, the sphere with a cavity and the cavity respectively. We have to calculate V_s at point P.

Point P is inside of the sphere thus the gravitational potential at point P due to the solid sphere is

    \[V(P)=-\frac{GM}{2a^3}\left(3a^2-\frac{a}{2}^2\right)}\]

    \[=-\frac{11GM}{8a}\]

Point P is at the center of the cavity thus the gravitational potential at point P due to the cavity is

    \[V_c(P)=-\frac{3GM'}{2a}\]

Here M' is the mass of the cavity and is \frac{M}{8}. Thus

    \[V_c(P)=-\frac{3GM}{8a}\]

Now, the total gravitational potential at point P is the sum of the potential due to the sphere with a cavity and the potential due to the cavity or V(P)=V_c(P)+V_s(P). Thus

    \[V_s(P)=V(P)-V_c(P)=-\frac{11GM}{8a}+\frac{3GM}{8a}=-\frac{GM}{a}\]

In the diagram below, we have plotted the gravitational potential due to a solid sphere as a function of distance from the center of the sphere. Please be careful about the y-axis in the plot.
 plot of Gravitational Potential Due to Solid Sphere as a function of distance

 

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