# Gravitational field at point just outside the sphere using integration

consider a point P at a distance k*R from a hollow sphere

The Gravitational field at point P can obtained by the summation of gravitational fields due to small rings which make up the ring.

the gravitational field at a point y distance away from a ring of mass M is given by: $$\frac{GMy}{(R^2+y^2)^{3/2}}$$

now considering the sphere to made up of infinitesimally small rings we get

$$dE=\frac{G\,dM\,R\,(1+k+\cos x)}{(R^2\sin^2x+(R(1+k+\cos x))^2)^{3/2}}$$

$$dM=\sigma\,2\pi\,\sin x\,R\,dx$$ (dm=mass of the ring) ($$\sigma$$ is mass per unit area of the sphere)

which simplifies to: $$dE=\frac{G\,\sigma\,2\pi\,\sin x\,dx\,(1+k+\cos x)}{(\sin^2 x+(1+k+\cos x)^2)^{3/2}}$$ $$E=\int dE=\int_{0}^{\pi} \frac{G\,\sigma\,2\pi\,\sin x\,dx\,(1+k+\cos x)}{(\sin^2 x+(1+k+\cos x)^2)^{3/2}}$$

$$E=\int dE=G\,\sigma\,2\pi\,\int_{0}^{\pi} \frac{\sin x\,dx\,(1+k+\cos x)}{(\sin^2 x+(1+k+\cos x)^2)^{3/2}}$$ $$E=\int dE=\frac{G\,M}{2R^2}\int_{0}^{\pi} \frac{\sin x\,dx\,(1+k+\cos x)}{(\sin^2 x+(1+k+\cos x)^2)^{3/2}}$$ $$E=\int dE=\frac{G\,M}{R^2}\int_{0}^{\pi} \frac{\sin x\,dx\,(1+k+\cos x)}{2(\sin^2 x+(1+k+\cos x)^2)^{3/2}}$$ $$E=\int dE=\frac{G\,M}{R^2}\,I$$ $$I=\int_{0}^{\pi} \frac{\sin x\,dx\,(1+k+\cos x)}{2(\sin^2 x+(1+k+\cos x)^2)^{3/2}}$$ FOR k=2

$$I=\frac{1}{9}$$ aka $$E=\frac{GM}{9R^2}$$

which is correct

but for k=0

$$I=\frac{1}{2}$$ aka $$E=\frac{GM}{2R^2}$$

which half of the actual result

what is the reason for the contradiction

my guess it has do with the fact the fact that k=0 lies exactly on the sphere

You're right that it comes from the point being exactly on the sphere. Inside the sphere the gravitational field is 0, and just ouside the sphere the field is $$GM/R^2$$. When you take the point exactly on the sphere, the integral gives you the average between these to values, thus $$1/2$$.
You can think about it imagining that the sphere is a shell with a small, but non-zero thickness. Then $$k=0$$ would mean that you're in a point half-way through the shell, effectively onyl half of the shell is still pulling you in, and the forces from the other half cancel out.
• so if we take the value of the limit of the field at point such that $k \rightarrow 0^+$ then the value of field will turn out to be $E=\frac{GM}{R^2}$ ie the expected result, RIGHT???? – Snmohith Raju Apr 26 '19 at 10:11
• @Snmohith Raju. That's correct. You have $$\lim_{k\rightarrow 0^+} E(k) = \frac{GM}{R^2}, \qquad \lim_{k\rightarrow 0^-} E(k) = 0, \qquad E(0) = \frac{GM}{2R^2}$$ – Adam Latosiński Apr 26 '19 at 12:28