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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


1 Answer 1


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.

  • $\begingroup$ 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???? $\endgroup$ Apr 26, 2019 at 10:11
  • $\begingroup$ @Snmohith Raju. That's correct. You have $$ \lim_{k\rightarrow 0^+} E(k) = \frac{GM}{R^2}, \qquad \lim_{k\rightarrow 0^-} E(k) = 0, \qquad E(0) = \frac{GM}{2R^2}$$ $\endgroup$ Apr 26, 2019 at 12:28

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