Question

From a solid sphere of mass M and radius R, a spherical portion of radius \[\dfrac{R}{2}\] is removed, as shown in the figure. Taking gravitational potential \[V=0\] at \[r=\infty \], the potential at the centre of the cavity thus formed is:
(G = gravitational constant)

Answer 1

Hint: Gravitational potential at a point is defined as the amount of work done to bring a small test mass from zero potential to the point. At the infinite distance, the potential will be zero. Gravitational potential can be found out from the gravitational field.
Formula used: \[E=\dfrac{GMr}{{{R}^{3}}},r < R\], where M is the mass of the sphere, R is the radius of the sphere, G is the gravitational constant and r is the distance to the point.
To find the potential from the gravitational field \[V=\int\limits_{\infty }^{r}{\vec{E}.d\vec{r}}\] can be used.
\[\text{Mass}=\text{density}\times \text{volume}\], can be used to find the mass of the removed part of the sphere.
The potential at the centre of the sphere will be, \[V=-\dfrac{3}{2}\dfrac{GM}{R}\]

Complete step-by-step answer:

Consider a sphere of radius R. Assume a spherical cavity of the radius \[\dfrac{R}{2}\] in the given sphere. P is the centre of the point of that cavity.
The gravitational potential energy at the outside of the surface will be,
\[V=\dfrac{-GM}{r}\]………………….(1), where G is the gravitational constant, M is the mass of the sphere and r is the distance from the sphere. If r becomes infinity, then the potential will become zero. It is already mentioned in the question.
On the surface of the sphere, the potential will be,
\[V=\dfrac{-GM}{R}\]………………….(2), Where G is the gravitational constant, M is the mass of the sphere and R is the radius of the sphere.
Here, we have to find the potential inside the cavity. The gravitational field inside of a solid sphere will be,
\[E=\dfrac{GMr}{{{R}^{3}}},r < R\]……………………(3)
We can find the gravitational potential from this.
\[V=\int\limits_{\infty }^{r}{\vec{E}.d\vec{r}}\], ……………………(4)
The gravitational potential at a point, in this situation, can be defined as bringing a unit mass from infinity to that point. So here, the mass has to bring to the surface of the sphere first then to the desired point. So we can split this integration into two parts.
\[V=\left[ \int\limits_{\infty }^{R}{\vec{E}.d\vec{r}+\int\limits_{R}^{r}{\vec{E}.d\vec{r}}} \right]\]…………………….(5)
We can assign the corresponding gravitational field equations into this equation.
\[V=\left[ \int\limits_{\infty }^{R}{\dfrac{GM}{{{R}^{2}}}dR+\int\limits_{R}^{r}{\dfrac{GMr}{{{R}^{3}}}dr}} \right]\]
After integration, this equation will be,
\[V=\left[ -\dfrac{GM}{R}+\dfrac{GM({{r}^{2}}-{{R}^{2}})}{2{{R}^{3}}} \right]\]
By using L.C.M
\[V=\left[ -\dfrac{GM}{R}\left[ \dfrac{1-({{r}^{2}}-{{R}^{2}})}{2{{R}^{2}}} \right] \right]\]
\[V=\left[ -\dfrac{GM}{2{{R}^{3}}}\left[ 2{{R}^{2}}-({{r}^{2}}-{{R}^{2}}) \right] \right]\]
\[V=\left[ -\dfrac{GM}{2{{R}^{3}}}\left[ 3{{R}^{2}}-{{r}^{2}} \right] \right]\]………………………(6)
Here we can take the r as \[\dfrac{R}{2}\]since we are trying to find the potential at the centre of the cavity. So the potential at the point P due to the solid sphere.
\[{{V}_{s}}=-\dfrac{GM}{2{{R}^{3}}}\left[ 3{{R}^{2}}-{{\left( \dfrac{R}{2} \right)}^{2}} \right]\]
\[{{V}_{s}}=-\dfrac{GM}{2{{R}^{3}}}\left[ \dfrac{11{{R}^{2}}}{4} \right]\]
\[{{V}_{s}}=-\dfrac{11GM}{8R}\]………………………(7)
Next, we can find out the mass of the removed part from the solid sphere.
\[\text{Mass}=\text{density}\times \text{volume}\]
The density of the removed part will be the same as the density of the solid sphere. Since the sphere is uniformly distributed.
So the mass of the removed part will be,
\[m=\dfrac{M}{\dfrac{4}{3}\pi {{R}^{3}}}\times \dfrac{4}{3}\pi {{\left[ \dfrac{R}{2} \right]}^{3}}\]
\[m=\dfrac{M}{8}\]…………………..(9)
Now we have to find the potential at the point P due to the removed sphere of radius \[\dfrac{R}{2}\]. In this sphere, the point P is the centre of that sphere. So the r will be zero in equation 6.
\[V=\left[ -\dfrac{GM}{2{{R}^{3}}}\left[ 3{{R}^{2}} \right] \right]=-\dfrac{3}{2}\dfrac{GM}{R}\]
So we can assign corresponding values to this equation.
\[{{V}_{r}}=-\dfrac{3Gm}{2\dfrac{R}{2}}\], here \[m=\dfrac{M}{8}\]
\[{{V}_{r}}=-\dfrac{3GM}{8R}\]………………(10)
Now two potentials are acting on the point P. So the net potential acting on the point will be,
\[{{V}_{p}}=-\dfrac{11GM}{8R}-\left[ -\dfrac{3GM}{8R} \right]\]
\[{{V}_{p}}=-\dfrac{GM}{R}\], this is the gravitational potential acting at the point P.

Note: The potential will be always negative since we are doing work to push the mass away from the planet to infinity. Here we are considering the same sphere and some portions are removed from the sphere. That’s why we are finding the difference between \[{{V}_{s}}\] and \[{{V}_{r}}\]. Otherwise, the total potential acting at the point will be the sum of two potentials.