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Let us say a uniform ring of mass m and radius 3a is kept above a sphere of mass M and radius 3a at a distance of 4a such that line joining the centres of ring and sphere is perpendicular to the plane of the ring.

So my question is what will be gravitational force of attraction between them.

I tried solving it but I am facing little problem while integrating and also that should I take sphere as point mass or simply a sphere.

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closed as off-topic by John Rennie, Kyle Kanos, David Hammen, rob, Kyle Oman May 1 '15 at 21:06

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    $\begingroup$ You can treat the sphere as a single point (this follows from point 1 of the shell theorem). So you just need to integrate along the length of the ring. $\endgroup$ – lemon May 1 '15 at 16:57
  • $\begingroup$ @lemon I think itwouldbbbe good even as an answer. $\endgroup$ – peterh May 1 '15 at 16:58
  • $\begingroup$ @lemon If we can take sphere as point mass then we can say that it does not depend on radius mentioned. So I guess i must ignore 3a radius of sphere and integrate it with ring then I will get ? $\endgroup$ – Shashank May 1 '15 at 17:23
  • $\begingroup$ @ShashankSharma That is correct. $\endgroup$ – lemon May 1 '15 at 17:35
  • $\begingroup$ @lemon I hope answering my own question doesn't bring any bad effect on my question ? as I took help from yours or it would be great if you will answer I will accept it :) $\endgroup$ – Shashank May 1 '15 at 17:41
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As per @lemon mentioned :

You can treat the sphere as a single point (this follows from point 1 of the shell theorem). So you just need to integrate along the length of the ring.

So I just did like that only.

Taking sphere as point mass then point mass will experiences one gravitational force of attraction that is:

$$df=cos\theta$$

As vertical component will cancel out.

Integrating it we will get:

$$F=\frac{GMm}{r^2}.sin\theta$$

$$F=\frac{GMm}{5a^2}.\frac{3a}{5a}$$

$$F=\frac{3GMm}{125a^2}$$

Which is our required answer and that's how we can determine the gravitational force of attraction between ring and sphere.

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