# Acceleration due to gravity at centre of hemisphere

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1. May 16, 2015

1. The problem statement, all variables and given/known data

find the acceleration due to gravity at the centre of a solid hemisphere.
2. Relevant equations

$F=\frac{GMm}{r^2}$
3. The attempt at a solution

i decided to go for cylindrical coordinayes (which is way beyond my syllabus). I did some research though.
let me take a point P(r,θ,z) inside the sphere amd an elemental volume dV at P. This P exerts a force dF. But there is also a point Q(r,θ+180°,z) which cancells out the horizontal component of dF. let the line OP make an angle φ with the Z axis. (assuming the hemisphere lies on the xy plane with centre at O)
$dg=\frac{Gdm}{r^2+z^2}cos\phi$
$dg=\frac{G\rho dV}{{(r^2+z^2)}^{3/2}}$
now dV=dz.dr.rdθ
Is this expresion for dV true for all cases? how do you get that expression for dV?
(i found that expression on some video).

now i have to integrate the expression.
so $$g=\int_{\theta=0}^{2\pi}\int_{r=0}^R\int_{z=0}^?f(r,\theta,z)dz.dr.rd\theta$$
im finding it difficult to find upper limit for z.

Last edited: May 16, 2015
2. May 16, 2015

### kontejnjer

Why not use spherical coordinates $(r,\theta,\varphi)$ instead? It's a lot easier to get the integration limits in that case.

3. May 16, 2015

i will try. but what about the expression for dV in cylindrical coordinates and the upper limit for z?

4. May 16, 2015

### kontejnjer

Well since the equation of a hemisphere is $0<x^2+y^2+z^2<R^2$ with $z>0$, if you use the fact that $x^2+y^2=r^2$, you get $0<z<\sqrt{R^2-r^2}$ where r is a variable, and R is a constant (the radius). Also, the expression for dV is correct, and is only valid in cylindrical coordinates. For more general coordinates you need to calculate the Jacobian of transformation, here's a good start, see the "Examples" section.

5. May 16, 2015

why is it dz.dr.rdθ? is it the volume of a small cylinder?

6. May 16, 2015

### kontejnjer

Basically it's the volume of an infinitesimal cylindrical shell with mean radius r, thickness dr and height dz (think of a ring with a rectangular cross section).

7. May 16, 2015

then its volume should be 2πr.dr.dz.

8. May 16, 2015

### TSny

Cylindrical coordinates are fine. But, consider integrating over r first and then z. The upper limit of r will depend on z.

9. May 16, 2015

### TSny

I believe you left out something when substituting for $\cos \phi$. Not that your final expression for dg does not have the correct dimensions for acceleration.

10. May 16, 2015

### kontejnjer

Ah yes, my apologies. Think of a chunk of a ring with the ring being cut along the diameter, from two infinitesimally close diameters. Then the volume element is $dV=rdrd\theta dz$. Basically the r term comes from the Jacobian, which is required in order for the volume element be invariant when performing a coordinate transform $(x,y,z)\rightarrow(r,\theta,z)$, so $dV=dxdydz\rightarrow dV'=|J|drd\theta dz=rdrd\theta dz$.

11. May 17, 2015

sorry. its a typo. i left out the term 'z' in the numerator.

Last edited: May 17, 2015
12. May 17, 2015

Do you have picture of this? its difficult to visualize.

13. May 17, 2015

is this the shape?

14. May 17, 2015

@TSny and @kontejnjer so the limits for θ are 0 to 2π
for r its 0 to $\sqrt{R^2-z^2}$
for z its 0 to R.
So first by integrating with respect to θ, the cunk becomes a full ring. Then integrating with respect to r makes the ring a disc which is situated at z and then integrating with respect to z will make it a hemisphere since it keeps adding one disc over another with changing radius.

15. May 17, 2015

$g=\pi G\rho R$