Gravitational Field Due to Solid Sphere

Gravitational Field Due to Solid Sphere

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Last updated on August 29th, 2019 at 04:17 pm

Here we will discuss the Gravitational Field Due to Solid Sphere. We will discuss the scenario where the test point is outside as well as inside of the shell. In our previous article we have calculated Gravitational Field due to Spherical Shell and have shown that for an external test point, spherical shell behaves like a point mass system and the field for an internal point is always zero. Here we will use very similar treatment to calculate the field. So, if you are not familiar with the previous exercise, you can go back and start there.

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

Gravitational Field due to Solid Sphere at an External Point

Let us consider a uniform sphere of mass M. The center of mass is located at point O. We have to calculate the field due to this spherical shell at a test point P, as shown in the figure below. We denote the distance between O and P as r. Let us consider an elementary spherical shell (shown as a shaded region) with mass dM. The field at P due to this element will be

Gravitational Field due to Solid Sphere

    \[ dE=\frac{GdM}{r^2}\]

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and the direction of the field will be from P to O. Now we can calculate the field due to the sphere by integrating over all the shells making the whole sphere. Thus

    \[ E = \int{dE}=\int{\frac{G dM}{r^2}}=\frac{G}{r^2}\int{dM}\]

    \[ E= \frac{GM}{r^2}\]

Or,

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

Where  sign denotes the direction towards the center of sphere.

For an external test point, solid sphere behaves like a point mass placed at the center of the sphere.

Gravitational Field due to Solid Sphere at an Internal Point

Suppose the test point P is inside of the sphere and you have to calculate the field due to the sphere of radius a. The situation is shown in the figure below.

Gravitational Field due to Solid Sphere

The situation is a little tricky and we will further divide the sphere into two parts, first a sphere with radius r and second, a spherical shell with internal radius r and external radius a. Now from our earlier exercise, we can calculate the field due to the sphere of radius r and it can be immediately written as,

    \[ E'= \frac{GM'}{r^2}\]

However, here we have to take M' as the mass of the sphere with radius r. Thus, M'=M\frac{r^3}{a^3}, where M is the mass of the whole sphere. With this modification,

    \[ E'= \frac{GM}{a^3}r\]

Now, we have to calculate the field at point P due to spherical shell with internal radius r and external radius a. This situation is shown in the figure below.

Gravitational Field due to Solid Sphere

From our earlier exercise with Gravitational Field Due to Spherical Shell, we know that the field at an internal point is zero. Thus if we imagine our thick spherical shell is made of several constituent shells having different radius but the same center and integrate over it to find the field at P, we will end up with zero. Thus effectively this thick shell does not contribute to the field. Look how it is different from the scenario, Gravitational Potential due to Spherical Shell at an Internal Point. Thus the total field at an internal point is,

    \[ \boxed{\vec{E}= -\frac{GM}{a^3}\vec{r}}\]

Where  sign denotes the direction towards the center of the sphere.

Here we have plotted the Gravitational Field due to the Solid Sphere as a function of distance from the center of the sphere.

Gravitational Field Due to Solid Sphere as a function of distance from the center of the sphere

What is the Value of Gravitational Field at the Centre of the Sphere or Earth?

As evident from the graph shown above, the gravitational field at the center of the sphere or earth is zero. This is because of the fact that there is equal mass enclosing the center from every side which leads to the cancellation of an equal amount of force from every side. Alternative, you can think that at the very center, there is no mass enclosed (being a point, however, this point is enclosed from every side) and consequently the gravitational field is zero.

  Explore More About Gravitation

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