Gravitational Field due to Uniform Disc

Gravitational Field due to Uniform Disc

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Last updated on August 29th, 2019 at 04:18 pm

In our last article, we have discussed and calculated expressions for Gravitational Field due to Point Mass and Uniform Ring. Here we will use the final expression for the gravitational field due to a uniform ring as a ready reference. So if you are not familiar with that expression you can go back and start from there.

We will calculate the force due to the source mass having the shape of a disc on a unit mass to calculate the gravitational field due to a uniform disc at a point on its axis.

You can also start your journey with the Laws of Gravitation.

Gravitational Field due to Uniform Disc

Let us consider a uniform disc of mass M. The center of mass is located at point O. We have to calculate the field due to this uniform disc at a test point P, as shown in the figure below. The distance between O and P is r. We also have shown the direction of the resultant field which is along \vec{PO} in the figure below.

gravitational field due to uniform disc


Let us first draw a circle of radius x. We draw another concentric circle of radius x+dx. These two concentric circles form a disc of thickness dx. This is shown as a shaded region in the figure above. The area of this ring is  2\pi x dx and the corresponding mass will be 

    \[dM=\frac{M}{\pi a^2}(2\pi x dx)= \frac{2Mx dx}{a^2} \]

As discussed in our previous article the gravitational field due to this ring element is

    \[ dE=\frac{G\left(\frac{2Mxdx}{a^2}\right)r}{(r^2+x^2)^{3/2}} \]

    \[ =\frac{2GMr}{a^2}\frac{xdx}{(r^2+x^2)^{3/2}} \]

To calculate the magnitude of the gravitational field due to the whole disc, we have to integrate over the full disc. This can be done by varying x from 0 to a. Thus

    \[ E=\int_{0}^{a}{\frac{2GMr}{a^2}\frac{xdx}{(r^2+x^2)^{3/2}}} \]

    \[ =\frac{2GM}{a^2}\left[1-\frac{r}{\sqrt{r^2+a^2}}\right]\]

    \[ =\frac{2GM}{a^2}(1-Cos(\theta))\]


    \[ \vec{E}=-\frac{2GM}{a^2}(1-Cos(\theta))\hat{r}\]

where we have used “” sign to denote the direction of gravitational field along \vec{PO}. This is the final expression for the gravitational field due to uniform disc at a point on its axis. In the next article we will calculate the Gravitational Field due to Uniform Spherical Shell.

Electric Field due to Charged Disc

Here we have plotted gravitational filed due to a uniform disc at a point on its axis as a function of distance from the center of the disc.

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