Gravitational Potential and Field Intensity Due to a Spherical Shell and a Solid Sphere

Gravitational Potential and Field Intensity due to a Spherical Shell:

Consider spherical shell of mass M and radius R. We need to find the gravitational potential and field intensity at a point p at a distance r from the centre of the shell. For this let us divide the shell into large no. of elementary rings by a plane perpendicular to OP as shown in a figure.

Consider one of such rings of radius y. Then from figure,

\(y=Rsin\theta\)

Surface area of the ring = circumference × width \begin{align*}(2\pi y).AB\end{align*}\begin{align*}2\pi Rsin\theta d\theta \end{align*}

\begin{align*}(2\pi y).AB\end{align*}\begin{align*}2\pi Rsin\theta d\theta \end{align*}

Mass per unit surface area of the shell = \begin{align*}\frac{M}{4\pi R^{2}}\end{align*}

Mass of the ring of radius y is

\begin{align*}Mass=\frac{M}{4\pi R^{2}}2\pi R^{2}sin\theta Rd\theta \end{align*}\begin{align*}=M/2sin\theta d\theta \end{align*}

Since the point P is equidistant (x) from every point on the ring, the gravitational potential at point P due to the ring is \begin{align*} \end{align*}\begin{align*}dV=-\frac{G\frac{M}{2}sin\theta}{x}->Eqn I \end{align*}

From figure,

\begin{align*}x^{2}=y^{2}+CP^{2} \end{align*}\begin{align*}=R^{2}sin^{2}\theta + (OP-OC)^{2} \end{align*}\begin{align*}=R^{2}sin^{2}\theta + (r-Rcos\theta)^{2} \end{align*}\begin{align*}X^{2}=R^{2}-2rRcos\theta +r^{2} \end{align*}

Diff. on both sides,

\begin{align*}2x\frac{\mathrm{d}x }{\mathrm{d} \theta}= 2Rrsin\theta \end{align*}\begin{align*}or, x=Rrsin\theta\frac{\mathrm{d} \theta}{\mathrm{d} x} \end{align*}

With this value eqn I becomes

\begin{align*}dV=-\frac{1}{2}\frac{GM}{Rr}{\mathrm{d}x } \end{align*}

Therefore, the gravitational potential a point P due to the spherical shell is

\begin{align*}V=\int dV=-\frac{GM}{2Rr}{\mathrm{d}x }-> Eqn-II \end{align*}

Case I:

when the point p lies outside the shelI

Gravitational field intensity \begin{align*}E=-\frac{dV}{dr} \end{align*}\begin{align*}=-\frac{GM}{r^{2}} \end{align*}
In this case, limit of integration runs from r-R at E to r+R at F as shown in figure, \begin{align*}\therefore V= -\frac{GM}{2Rr}\int_{r-R}^{r+R}{\mathrm{d}x } \end{align*}\begin{align*}=-\frac{Gmm}{2Rr}[x]_{r-R}^{r+R} \end{align*}\begin{align*}=-\frac{GMm}{2Rr}[r+R-r+R] \end{align*}\begin{align*}=-\frac{GM}{2rR}2R \end{align*}\begin{align*}V=-\frac{GM}{r} \end{align*}

Case II:

When point P lies on the surface of the shell,

In this case, limit of integration runs from O at E to 2R at F.

\begin{align*}\therefore V=-\frac{GM}{2Rr}\int_{0}^{2R}dx=-\frac{GM}{R}\end{align*}

But on the surface of the shell, r=R

\begin{align*}\therefore V =-\frac{GM}{R}\end{align*}

Also, Gravitational field intensity at the surface,\begin{align*}E=-\frac{dV}{dr}=-\frac{GM}{r^{2}}\end{align*}

Again, on the surface, r=R

\begin{align*} E=-\frac{GM}{R^{2}} \end{align*}

Case III:

When the point P lies inside the shell,

In this case, limit of integration runs from R-r at E to R+r at F

\begin{align*}\therefore V=-\frac{GM}{2Rr}\int_{R-r}^{R+r}dx \end{align*} \begin{align*}=-\frac{GM}{2rR}(R+r-R+r) \end{align*} \begin{align*} =-\frac{GM}{R} \end{align*}

This shows that the gravitational potential at any point inside the spherical shell is constant and equal to that on the surface of the shell.

Gravitation field inside the shell

\begin{align*}E=-\frac{dV}{dr}=0 \end{align*}i.e. no gravitational field inside the shell.

Gravitational Potential and Field Intensity Due to a Solid Sphere

Consider a Solid sphere of radius R and mass M.

Case I:When the point P lies outside the sphere:

Let us divide the Solid sphere into large no. of concentric spherical shells having masses \(m_1, m_2, ..., m_n\). Let r be the distance of point P from the centre of the spherical shells. Then the gravitational potential at point P due to such shells is given by \begin{align*}V_1=-\frac{Gm_1}{r}, V2=-\frac{Gm_2}{r}, ..., V_n=-\frac{Gm_n}{r} \end{align*} The total gravitational potential at point P due to the whole mass is \begin{align*}V=V_1+V_2+...+V_n \end{align*}\begin{align*}=\frac{G(m_1+m_2+...+m_n)}{r} \end{align*}\begin{align*}=\frac{GM}{r} \end{align*} where \(M=m_1 + m_2 +...+ m_n\) therefore, the field intensity at point P is\begin{align*}E=-\frac{dV}{dr}=-\frac{GM}{r^{2}} \end{align*}

Case II:When the point P lies on the surface of the sphere

In this case also the potential and field intensity due to the solid sphere resembles well with that to the spherical shell.

Gravitational potential on the surface

\begin{align*}V=-\frac{GM}{R} \end{align*}

Field intensity,\begin{align*}E=-\frac{GM}{R^{2}} \end{align*}

Case III:When the point P lies inside the surface of the sphere.

Let us divide the sphere into concentric spherical shells. Let us take anyone of such shells having radius x and thickness dx.

Surface area of the shell\begin{align*}=4\pi x^{2} \end{align*}Volume of the shell\begin{align*}=4\pi x^{2}dx \end{align*} Mass of the shell\begin{align*}=4\pi x^{2}dx\rho \end{align*}

Subcase I:If the point P lies outside the shell of radius x.

The gravitational potential at point P due to the shell is \begin{align*}dV_1=-\frac{Gmass}{r} \end{align*}r=distance of point P from the centre of the shell\begin{align*}i.e. dV_1=-G\frac{4\pi x^{2}dx\rho }{r} \end{align*}

Therefore, the potential at point P due to such shells, for which the point P lies outside is given by\begin{align*}V_1=\int dV_1 \end{align*}\begin{align*}=-G\frac{4\pi dx\rho }{r}\int_{0}{r}x^{2}dx \end{align*}\begin{align*}=-\frac{-4\pi \rho G r^{3}}{3r} \end{align*}\begin{align*}=\frac{-4\pi \rho Gr^{2}}{3} \end{align*}

Subcase II:When the point P lies inside the Shell of radius x

The potential at point P due to this shell is \begin{align*}dV_2=-\frac{Gmass}{x} \end{align*}\begin{align*}-G\frac{4\pi x^{2}dx\rho }{x} \end{align*}\begin{align*}=-4\pi \rho Gxdx \end{align*}Thus the potential at point P due to the shells for the point P lies inside is given by\begin{align*}V_2=\int dV_2=-4\pi \rho G\int_{r}^{R}xdx \end{align*}\begin{align*}=-\frac{4\pi \rho G(R^{2}-r^{2})}{2} \end{align*}\begin{align*}-2\pi \rho G(R^{2}-r^{2}) \end{align*}

Total potential at point P is given by \begin{align*}V=V_1+V_2 \end{align*}\begin{align*}=-\frac{4\pi }{3}\rho Gr^{2}-2\pi \rho G(R^{2}-r^{2}) \end{align*}\begin{align*}=-2\pi \rho G(\frac{2r^{2}}{2}+R^{2}-r{2}) \end{align*}\begin{align*}=-\frac{2\pi \rho G}{3}(3R^{2}-r^{2})->I \end{align*}\begin{align*}\frac{2R^{3}}{2R^{3}}(\frac{2\pi \rho G}{3}(3R^{2}-r^{2}))\end{align*}\begin{align*}V=-\frac{GM(3R^{2}-r^{2})}{2R^{3}} ->II\end{align*}

Field intensity is given by \begin{align*}E=-\frac{dV}{dr} \end{align*}\begin{align*}=-{-\frac{2\pi \rho G}{3}(-2r)} \end{align*}\begin{align*}-\frac{4\pi \rho Gr}{3}(from-I) \end{align*}\begin{align*}i.e.E\prec r \end{align*}Also in terms of the sphere\begin{align*}E=-\frac{dV}{dr}from(2) \end{align*}\begin{align*}E=-\frac{GM}{R^{3}}r \end{align*}\begin{align*}i.e.E\prec r \end{align*}

References

Adhikari, Pitri Bhakta. A Textbook of Physics Volume-I. Kathmandu: Sukunda Pustak Bhawan, 2015.

Feynman, Richard P. The Feynman Lectures on Physics Volume 1. Noida: Dorling Kindersley (India) Pvt. Ltd., 2014.

Mathur, D S. Mechanics. New Delhi: S. Chand & Company Pvt. Ltd., 2015.

Young, Hugh D, Roger A Freedman and A Lewis Ford. University Physics. Noida: Dorling Kindersley (India) Pvt. Ltd., 2014.