- #1
ozone
- 122
- 0
Find the gravitational attraction of a solid hemisphere of radius a anddensity1 on a unit point mass placed at its pole.
My attempt:
Obviously I figured that spherical coordinates would work nicely for this problem.
I decided to invert the hemisphere thus placing the pole on the origin. Then I decided to split the upside down quarter disc into two integrals. The first integral was then for a cone shaped figure (r1 in the picture i made).
I set it up as follows
[itex] G \int_{\vartheta=0}^{2\pi} \int_{
\varphi=0}^{\pi/3}\int_{\rho=0}^{a/sin(\phi)} sin(\phi)cos(\phi)d\phi d\rho d\theta[/itex]
Plz let me know if i have made any mistakes with the above integrand (pi/3 is where a cone will intersect a circle of equal radius).
However I could not manage to think up the equation (in spherical coordinates) for the lower circular bound. I considered changing coordinate systems for the outer spherical part but setting up the equation for force of gravity might have been tricky. Any help will be greatly appreciated.
My attempt:
Obviously I figured that spherical coordinates would work nicely for this problem.
I decided to invert the hemisphere thus placing the pole on the origin. Then I decided to split the upside down quarter disc into two integrals. The first integral was then for a cone shaped figure (r1 in the picture i made).
I set it up as follows
[itex] G \int_{\vartheta=0}^{2\pi} \int_{
\varphi=0}^{\pi/3}\int_{\rho=0}^{a/sin(\phi)} sin(\phi)cos(\phi)d\phi d\rho d\theta[/itex]
Plz let me know if i have made any mistakes with the above integrand (pi/3 is where a cone will intersect a circle of equal radius).
However I could not manage to think up the equation (in spherical coordinates) for the lower circular bound. I considered changing coordinate systems for the outer spherical part but setting up the equation for force of gravity might have been tricky. Any help will be greatly appreciated.