- #1
Yalanhar
- 39
- 2
- Homework Statement
- find the acceleration due to gravity at the centre of a solid hemisphere.
- Relevant Equations
- ##dg=\frac {GMdm}{r^2}##
Can someone help me? I am not sure where is my mistake
correct answer:
##g=\pi G\rho R##
my calculus
By symmetry, I only need to add the vertical component of gravity
##dg = \frac {GMdm}{d^2}\cdot cos\theta## (1) Where d=R
##\rho =\frac{dm}{dV} ##
##dm = \rho \cdot dV## (2)
(2) in (1)
##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}##
correct answer:
##g=\pi G\rho R##
my calculus
By symmetry, I only need to add the vertical component of gravity
##dg = \frac {GMdm}{d^2}\cdot cos\theta## (1) Where d=R
##\rho =\frac{dm}{dV} ##
##dm = \rho \cdot dV## (2)
(2) in (1)
##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}##