Newton's Law of Gravitation

  • #1
So I'm reading "Mathematical Physics" by Donald H.

Menzel, and I don't buy the following derivation from

section 2.12

The purpose of the derivation is derive the potential

energy at a point Po which is a distance Ro from the

center of a sphere of uniform density.

First they derive the amount an infinitesimal amount of

mass P(dv) located on the shell of the sphere contributes

to the potential energy at a point Po.


[tex]dV=-\frac{G \rho}{R}dxdydz[/tex]

The following expression is given to compute the

potential energy contribution of the infinitesimal piece

of mas dV at P(dV) on a point Po.

[tex]V=-G\rho \int_r^{r+dr}\int_{\theata

=0}^{\pi}\int_{\phi=0}^{2\pi}\frac{r^2sin^2( \theta ) dr

d \theata d \phi }{R}[/tex]
r is the distance from the center of the sphere to P(dV)
Ro is the distance from P to the center of the sphere
R is the distance from P(dV) to Po

A change in variables is derived by doing implicit

differentiation with r contant on the law of cosines:

[tex]R^2=R_o^2+r^2-2R_o r \ cos( \theta )[/tex]

which gives:

[tex]R \ dR = R_o r sin( \theta ) d \theta [/tex]

So far I agree but then they say that this implies:

[tex]V=-G\rho \int_r^{r+dr}\int_{R_o-r}^{R_o+r}\int_{0}^{2\pi}\frac{r}{R_o}drdR d \phi[/tex]

However, if I do the above subsitution I get an extra factor of [tex]sin( \theta )[/tex] left over.
 

Answers and Replies

  • #2
mjsd
Homework Helper
726
3
in the integral, i think it should just be
[tex]V= -G\rho \int \frac{1}{R} r^2 \sin\theta \, dr d\theta d\phi[/tex]
from change of variables
 
  • #3
in the integral, i think it should just be
[tex]V= -G\rho \int \frac{1}{R} r^2 \sin\theta \, dr d\theta d\phi[/tex]
from change of variables

That's what I started thinking last night after I posted. It should be easy enough to check.
 

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