Homework Help: Force due to gravity inside planet

1. Nov 27, 2012

Undoubtedly0

Here is a simple problem in classical gravitation.

Consider a spherical planet of radius R, and let the radial coordinate r originate from the plant's center. If the density of the planet is ρ from 0 ≤ r < R/2 and ρ/3 from R/2 < r < R, then my work tells me that the maximum force due to gravity inside the planet is at r = R/2, not at r = R as one might expect.

$$0\leq r\leq R/2,\qquad F_G = \frac{Gm\left(\rho \frac{4}{3}\pi r^3\right)}{r^2} = \frac{4\pi}{3}Gm\rho r \\ R/2 \leq r \leq R,\qquad F_G = \frac{Gm\left[ \rho\frac{4\pi}{3}\left(\frac{R}{2} \right)^3 + \frac{\rho}{3}\frac{4\pi}{3} \left( r^3 - \left(\frac{R}{2} \right)^3 \right) \right]}{r^2} = \frac{4\pi}{3}\frac{Gm\rho}{r^2}\left[ \left(\frac{R}{2} \right)^3 + \frac{1}{3}r^3 - \frac{1}{3}\left(\frac{R}{2} \right)^3 \right]$$

My work is above. Is this be correct, that the maximum force due to gravity would be at r = R/2? Thanks for your time!

2. Nov 27, 2012

Staff: Mentor

Neglecting 4pi/3 Gmρ as prefactor:

In the inner part: $F \propto r$
In the outer part: $F \propto (\frac{R}{2})^3/r^2 + 1/3 (r-\frac{R}{2}) = \frac{1}{r^2} ((\frac{R}{2})^3+\frac{1}{3}r^3-\frac{1}{3}(\frac{R}{2})r^2)$

I get a different third term for the force.

$\frac{dF}{dr}=-\frac{R^3}{4r^3} + \frac{1}{3}$ which is 0 at $4r^3=3R^3$, it has a minimum in the less dense region.

F(R/2) > F(R), so I can confirm your result.

3. Nov 27, 2012

Undoubtedly0

Thanks mfb. Regarding our difference, I think your third term might be mistaken. If we hollow out the planet from 0 ≤ r < R/2, then the force due to gravity in the domain R/2 ≤ r ≤ R is

$$F_G = \frac{Gm\frac{\rho}{3}}{r^2}\left[ \frac{4\pi}{3}r^3 - \frac{4\pi}{3}\left(\frac{R}{2}\right)^3\right]$$

By the shell theorem, it as if all the volume enclosed is contained in a point at the center. Do you agree?

4. Nov 27, 2012

Staff: Mentor

Oh, you are right.