Acceleration due to gravity at the centre of a hemisphere

In summary: G\rho \sin\theta \cos\theta dr d\theta d\phi####g = \int_0^{2\pi} \int_0^{\pi/2} \int_0^R G\rho \sin\theta \cos\theta r^2dr d\theta d\phi####= \int_0^{2\pi} d\phi \int_0^{\pi/2} G\rho \sin\theta \cos\theta \left[ \frac{r^3}{3} \right]_0^R d\theta####= \frac{\pi}{2} \int_0^{2\pi} G\r
  • #1
AdityaDev
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Homework Statement



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

Homework Equations



##F=\frac{GMm}{r^2}##

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.
 
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  • #2
Why not use spherical coordinates [itex](r,\theta,\varphi)[/itex] instead? It's a lot easier to get the integration limits in that case.
 
  • #3
i will try. but what about the expression for dV in cylindrical coordinates and the upper limit for z?
 
  • #4
Well since the equation of a hemisphere is [itex]0<x^2+y^2+z^2<R^2[/itex] with [itex]z>0[/itex], if you use the fact that [itex]x^2+y^2=r^2[/itex], you get [itex]0<z<\sqrt{R^2-r^2}[/itex] 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.
 
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  • #5
why is it dz.dr.rdθ? is it the volume of a small cylinder?
 
  • #6
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
then its volume should be 2πr.dr.dz.
 
  • #8
Cylindrical coordinates are fine. But, consider integrating over r first and then z. The upper limit of r will depend on z.
 
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  • #9
AdityaDev said:
##dg=\frac{Gdm}{r^2+z^2}cos\phi##
##dg=\frac{G\rho dV}{{(r^2+z^2)}^{3/2}}##

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.
 
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  • #10
AdityaDev said:
then its volume should be 2πr.dr.dz.
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 [itex]dV=rdrd\theta dz[/itex]. Basically the r term comes from the Jacobian, which is required in order for the volume element be invariant when performing a coordinate transform [itex](x,y,z)\rightarrow(r,\theta,z)[/itex], so [itex]dV=dxdydz\rightarrow dV'=|J|drd\theta dz=rdrd\theta dz[/itex].
 
  • #11
TSny said:
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.
sorry. its a typo. i left out the term 'z' in the numerator.
 
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  • #12
kontejnjer said:
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 [itex]dV=rdrd\theta dz[/itex].
Do you have picture of this? its difficult to visualize.
 
  • #13
is this the shape?
sector.gif
 
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  • #14
@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.
 
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  • #15
I also got the correct answer.
##g=\pi G\rho R##
 
  • #16
AdityaDev said:
I also got the correct answer.
##g=\pi G\rho R##

Can you help me? I got
##g=\pi G\rho R/3##

my calculus
##dg = \frac {G \rho cos\theta dV}{R^2}##

for spherical coordinate $$dV = r^2drsin\theta d\theta d\phi$$

so:
##dg=\frac {G\rho }{R^2} r^2dr sin\theta cos\theta d\theta d\phi##

##g=\frac {G\rho}{R^2} \int_0^R r^2 \, dr \int_0^\frac {\pi}{2} sin\theta cos\theta \, d\theta \int_0^{2\pi} \, d\phi##

##g = \frac {G\rho}{R^2} \cdot \frac{R^3}{3} \cdot \frac {1}{2} \cdot 2\pi##

##g = \frac {G\rho R\pi}{3}##

##g=\frac {\pi G\rho R}{3}##
 
  • #17
Yalanhar said:
Can you help me? I got
##g=\pi G\rho R/3##

my calculus
##dg = \frac {G \rho \cos\theta dV}{R^2}##
That should be ## r^2##, not ##R^2##.
for spherical coordinates
##dV = r^2dr\sin\theta d\theta d\phi##

so:
##dg=\frac {G\rho }{R^2} r^2dr \sin\theta \cos\theta d\theta d\phi##
That gives you:

##dg=\dfrac {G\rho }{r^2} r^2dr \sin\theta \cos\theta d\theta d\phi##
 

Related to Acceleration due to gravity at the centre of a hemisphere

1. What is the acceleration due to gravity at the centre of a hemisphere?

The acceleration due to gravity at the centre of a hemisphere is approximately half of the acceleration due to gravity at the surface of the hemisphere. This is because the gravitational force decreases as you move towards the centre of a hemisphere, becoming zero at the very centre.

2. Does the acceleration due to gravity at the centre of a hemisphere change with the size of the hemisphere?

No, the acceleration due to gravity at the centre of a hemisphere does not change with the size of the hemisphere. It only depends on the mass and radius of the hemisphere.

3. How does the acceleration due to gravity at the centre of a hemisphere compare to the acceleration due to gravity at the centre of a full sphere?

The acceleration due to gravity at the centre of a hemisphere is half of the acceleration due to gravity at the centre of a full sphere. This is because the mass of a hemisphere is half of the mass of a full sphere, and the acceleration due to gravity is directly proportional to the mass of an object.

4. Does the acceleration due to gravity at the centre of a hemisphere vary with the density of the hemisphere?

No, the acceleration due to gravity at the centre of a hemisphere does not vary with the density of the hemisphere. It only depends on the mass and radius of the hemisphere.

5. How is the acceleration due to gravity at the centre of a hemisphere calculated?

The acceleration due to gravity at the centre of a hemisphere can be calculated using the formula g = (2/3) * (G * M)/R, where g is the acceleration due to gravity, G is the gravitational constant, M is the mass of the hemisphere, and R is the radius of the hemisphere.

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