Gravitational Potential Due to Spherical Shell

Gravitational Potential Due to Spherical Shell

In the main Gravitation article, we have shown the expressions for the gravitational potential for different situations. In our last article, we also have discussed and calculated the Gravitational Potential due to Point Mass and Uniform Ring. Here we will discuss thoroughly about the Gravitational Potential Due to Spherical Shell. We will also discuss the situation when the test point is outside of the shell as well as when it is inside of the shell. Again, the main idea will be the calculation 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 Spherical Shell

Let us consider a uniform spherical shell of mass M. The center of mass is located at point O. We have to calculate the potential due to this spherical system 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 spherical shell

We will first draw a radius a=OA which makes an angle \theta with the line OP. We will now draw a circle by keeping the central line OA fixed. The radius of this circle (see figure) is aSin\theta. Thus the perimeter of this circle will be 2\pi a Sin\theta. Now consider another circle with radius aSin(\theta+d\theta). These two circles form a ring (shown as green) having a mass of dM with a width of ad\theta. Hence the area subtended by the ring is  (2\pi a Sin\theta)\times(ad\theta)= 2\pi a^2 Sin\theta d\theta and the mass enclosed is

    \[ dM=\frac{M}{4\pi a^2}(2\pi a^2 Sin\theta d\theta)= \frac{M}{2}Sin\theta d\theta \]

Considering the triangle OAP, we can write  z^2=a^2+r^2-2arCos\theta and taking differential on both side, we find

    \[ Sin\theta d\theta=\frac{zdz}{ar} \]

Thus we can write the expression of dM in terms of dz as  dM=\frac{M}{2ar}zdz.

From our previous exercise, we know that the potential at a central point P due to the ring is

    \[ dV=-\frac{GdM}{z}=-\frac{GM}{2ar}dz \]

We can now use this expression to calculate gravitational potential due to spherical shell at a test point which is either outside of the shell or inside of the shell.

Test Point P is Outside of the Shell

To calculate the potential at a point outside of the shell we have to integrate between \theta=0 to \theta=\pi. This corresponds to z=r-a to z=r+a. Thus,

    \[ V=-\frac{GM}{2ar}\int_{r-a}^{r+a}{dz}\\ =-\frac{GM}{2ar}[(r+a)-(r-a)] \]

    \[ \boxed{V=-\frac{GM}{r}} \]

This is surprising to note that the expression for the potential due to spherical shell at an external point is identical to the expression of potential due to a point mass M situated at the center O of the shell.

Test Point P is Inside of the Shell

As shown in the figure, now we have a situation where \theta=0 corresponds to z=a-r and \theta=\pi corresponds to z=a+r. Thus the potential due to the shell at an internal point can be calculated as,

gravitational potential due to spherical shell at an internal point

    \[ V=-\frac{GM}{2ar}\int_{a-r}^{a+r}{dz}\\ =-\frac{GM}{2ar}[(a+r)-(a-r)] \]

    \[ \boxed{V=-\frac{GM}{a}} \]

Thus inside a uniform spherical shell, the gravitational potential is fixed and has a value of -\frac{GM}{a}.

In the diagram below we have shown the gravitational potential due to a spherical shell as a function of distance from the center of the sphere.

Gravitational Potential Due to Spherical Shell as a function of distance

 

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