# Clarifying doubt related to gravitation inside a shell

I proved that gravitational field due to a ring would be $$-GMz\over\sqrt{z^3+R^3}$$ for any particle lying on the axis perpendicular to its plane and passing through centre.

My teacher told me that gravitation field inside a shell is $$0$$ because a shell is combination of many small rings.
If we take this idea under consideration and take some point inside the shell then gravitation field at that point will be $$0$$.
Now if we join the point with centre of the shell and name it as $$z$$-axis. Further we cut shell into thin rings by making the slices perpendicular to $$z$$-axis.
Then one side of the point will have more rings (lets call left) than other side.
So the rings symmetrically big to those on left side will cancel each other. Leaving the resultant gravitational field on left side towards centre.

My view of thinking contradicts the statement made by my teacher. Well I have seen this result having being used in many times to state something. So it is likely that he is correct but still I want to clarify my doubt.

PS: I don't know of any software to make a picture ,so I can't attach a picture. Sorry for it.

Here's a picture taking the point inside shell to be $$C$$

• I think at least a sketch (hand drawn + photo) would be helpful here. The way I read your description the "symmetric" rings on the left and right of the point are not the same size + distance away so don't precisely cancel in the way you say they do. They would only do this if the point was actually at the centre. – jacob1729 Dec 15 '18 at 16:08
• @jacob1729- I got it. Got zero on integrating and understood the way too. Thanks. – Love Invariants Dec 15 '18 at 16:28
• You should write the integral as an answer to your own question – Dale Dec 15 '18 at 16:39
• Is the shell assumed to be spherically symmetric, cf. Newton's shell theorem? – Qmechanic Dec 15 '18 at 17:16
• @Qmechanic- Obviously. – Love Invariants Dec 15 '18 at 19:42