Hi @JohnRennie. I wanted to ask something.

@Abcd Morning :-)

@JohnRennie Finding the force due to the ring (and then using newton's third law) and considering the sphere as a point mass helps

But then, directly finding the force on the ring due to the sphere which can be treated as a point mass doesn't help. (without using Newton's third law. )

Isn't it just a matter of calculating the force on an element of the ring then integrating round the ring?

yes, but how? i wasn't getting the right answer using that.

Give me a moment and I'll draw a diagram ...

No

I have a diagram.

Let me send

Here @JohnRennie ^

let's treat the sphere as a point of mass M placed at its centre, using shell theorem.

Ok

Ha. Just beat me to it :-)

yeah :).

If we consider an element of the ring subtending an angle $d\theta$ then its mass is $md\theta/2\pi$, so the force is $dF = \frac{GM md\theta}{4a^2 2\pi}$

@JohnRennie where does it subtend $d\theta$?

That's the top view of the ring so we are looking down on the ring and the sphere is below it.

We consider a small element of the ring subtending an angle $d\theta$ measured from the centre of the ring.

ok

now integration gives: $F = \dfrac{GMm}{4a^2}$?

No, because force is a vector. You can't just sum the magnitudes of the $dF$ s

right.

we need $\hat{j}$ components of force, from symmetry.

If we go back to the side view, there will be a component of the force that acts horizontally inwards towards the centre of the ring. When we integrate this component will sum to zero.

yes

So we just integrate the vertical component, which I guess is what you mean by the $j$ component.

yes

Oh, then we'll have to use another angle.

Then, after that cosine rule.

Wait, i/m slightly uncertain.

It's just going to add a factor of $1/\sqrt{3}$

What about $\theta$ @JohnRennie? Were's it.

Theta is the angle measured from the centre of the ring to the edge of the ring. It integrates from 0 to $2\pi$ to go round the ring.

Ok.

$dF_v = \frac{GM md\theta}{4a^2 2\pi} \cos\phi$

yes

And nothing in that equation is a function of $\theta$ so integration gives $$F_v = \frac{GM m}{4a^2 2\pi} \cos\phi \int_0^{2\pi} d\theta $$

yes, and $\cos \phi = \dfrac{\sqrt 3}{2}$

Done @JohnRennie. Thank you :).