Gravitational Field Due to Uniform Ring

Gravitational Field Due to Uniform Ring

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Last updated on September 6th, 2019 at 08:05 am

In this article, we will rigorously derive expressions for the Gravitational Field due to a Point Mass and the Gravitational Field due to a Uniform Ring at a Point on its Axis. As already discussed, the calculation of gravitational potential employs the idea that the potential energy is equal to the work done by a mass m under the influence of a source mass M. We then divide this work done by the mass of that object to get the potential. Similarly, here we will calculate the force due to a source mass M on a test mass m and the gravitational field can be calculated as the force per unit mass.

To know more about the gravitational field and the potential, we recommend starting your journey from the Laws of Gravitation.

Gravitational Field Due to a Point Mass

Simplest of the problem is the calculation of the gravitational field due to a point mass. For that, we shall consider a point mass M at point A. We shall call this “source mass” and will calculate the force due to this source mass on another point mass m, placed at point P. m is known as a test mass. By definition, the field is the force felt by a unit mass at point P. Thus if \vec{E} be the field due to source mass M, then this field exerts a force \vec{F} on the test mass m.

gravitational field due to point mass

    \[  \vec{F}=m\times\vec{E} \]

Again from the law of gravitation, we can define the force as

    \[  \vec{F}=-\frac{GMm}{r^2}\hat{r} \]

Here we have used the “-” sign to denote the direction of force along \vec{PA}. Thus equating these two forces, we find

    \[  \boxed{\vec{E}=-\frac{GM}{r^2}\hat{r}} \]

We will use this expression in the following section to calculate the field due to uniform ring.

Example: Let us consider the earth to be a point mass. Force of attraction due to earth on a nearby object is known as the weight of that object and is given by m\vec{g}. Thus the gravitational field due to earth on a nearby object is

    \[\vec{E}=\frac{m\vec{g}}{m}=\vec{g}\]

Gravitational Field Due to a Point Mass as a function of distance from the point mass

In the diagram above, we have shown the gravitational field due to a point mass as a function of distance from the point mass.

Gravitational Field due to a Uniform Ring at a Point on its Axis

Here we will calculate the gravitational field at an axial point P due to a uniform ring of mass M. The center of the ring is at point O. From the figure shown below, the circumference of the ring makes an angle \theta with the line OP drawn from the center of the ring. We will consider a very small element of the circumference around point A and having a mass of dM. Now to calculate the field at point P due to the mass element dM, we will use the expression of the field derived in the previous section.

    \[  dE'=\frac{GdM}{z^2} \]

gravitational field due to uniform ring

The direction of this field is along \vec{PA}. We can divide this field into two components, one horizontal and another vertical. However due to the axial symmetry of the total system, the vertical component will vanish. The horizontal component of the gravitational field due to the tiny mass element dM is

    \[  dE=dE'cos(\theta)=\frac{GdM}{z^2}cos(\theta) \]

To calculate the field at point P due to the whole ring, we have to integrate dE over the ring.

    \[  E=\int{\frac{GdM}{z^2}cos(\theta)} \]

    \[  =\frac{Gcos(\theta)}{z^2}\int{dM} \]

    \[  =\frac{GMcos(\theta)}{z^2} \]

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

Or

    \[  \boxed{\vec{E}=-\frac{GMr}{(a^2+r^2)^{\frac{3}{2}}}\hat{r}} \]

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

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