Gravitational field of a hollow sphere

[Rikudo]

Homework Statement

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Relevant Equations

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Why the area of the thin rings are $2πasin\theta \, ds$? (a is the radius of the hollow sphere)

If we look from a little bit different way, the ring can be viewed as a thin trapezoid that has the same base length ( $2πa sin\theta$), and the legs are $ds$.
The angle between the leg and the lower base is $90-\theta$. Hence,we can conclude that the height of the trapezoid is $ds \, sin\theta$.

Since the base length is more or less the same, the area is just base x height.
This means, the area is :$2πa sin\theta \, ds \, sin \theta$.

 

[kuruman]

Why don't you substitute $ds=a~d\theta$, then integrate your area element over $\theta$ from $0$ to $\pi$. Do you get $4\pi a^2$ for the area of the shell?

 

[Rikudo]

Why don't you substitute $ds=a~d\theta$, then integrate your area element over $\theta$ from $0$ to $\pi$. Do you get $4\pi a^2$ for the area of the shell?

Yes.

But, $ds$ is not the height of the trapezoid. So, in my opinion, we are not supposed to multiply this with the $2πa sin\theta$ to get the area.

 

[haruspex]

the height of the trapezoid is dssinθ.

Sure, but the height of the trapezoid is not what you need. A wall length L and height H has area LH. If it then leans over at angle θ it still has area HL, not HL cos(θ).

 

[Rikudo]

A wall length L and height H has area LH. If it then leans over at angle θ it still has area HL, not HL cos(θ)

Umm... well, it is is true, but what is the relation of it with the area of ring?

but the height of the trapezoid is not what you need.

Why, though? We can't use $ds$ as its height since it is not perpendicular to the bases.

 

[Steve4Physics]

Why, though? We can't use $ds$ as its height since it is not perpendicular to the bases.

@Rikudo, I suspect your difficulty arises because you are not 'visualising' the infinitessimal 3D strip correctly.

The line element $ds$ is perpendicular to both the inner and outer edges of the strip. That means the (infinitessimal) strip has the same area as a rectangle measuring $2\pi a \sin \theta$ by $ds$.