hey guys. maybe not exactly a gre problem, but certainly something that should be understood for the GRE.
Im trying to find the gravitation potential per unit mass (U) inside a spherical shell. I thought this would be easy but cant reproduce the answer in Marion,Thornton 5ed,p188. I wanted to try a different approach than they did.
So i imagine bringing in a point mass from infinity into a sphere with outer radius (a) and inner (b) and uniform mass distribution; U(inf) = 0
$$U$$ = $$U_a$$ + $$\Delta U(a>b)$$
$$U_a = \frac{GM}{a}$$
$$U_a= \frac{G}{a}\frac{4\pi}{3}\rho(a^3b^3)$$
where i expressed M in terms of rho. Thats was easy. Now for the potential difference across the shell.
$$U_{(a>b)} = \int_a^b (F\cdot dr)$$
$$= \int_a^b (\frac{GM}{r^2})dr$$
again subbing mass in terms of density and radius...
$$= G\int_a^b \frac{1}{r^2}\left[\frac{4\pi}{3}\rho(r^3b^3)\right]dr$$
After integrating...
$$=\frac{4}{3}\pi G \rho [b+b^2a\frac{b^3}{a}]$$
adding U_a
$$=\frac{4}{3}\pi G \rho [b+b^2a\frac{b^3}{a}] \frac{G}{a}\frac{4\pi}{3}\rho(a^3b^3)$$
Some simplifying
$$=\frac{4}{3}\pi G \rho [b+b^2a\frac{b^3}{a} a^2+\frac{b^3}{a}]$$
$$=\frac{4}{3}\pi G \rho [b+b^2a a^2]$$
whereas the answer in the book...
$$U(R<b) = 2\pi \rho G(a^2b^2)$$
love a good latex workout, feeling the burn. anyway, its probably something stupid im messing up on, but please help. Im so frustrated at this.
and if there are other ways you would solve this, please suggest them
gravitational potential

 Posts: 7
 Joined: Thu May 13, 2010 8:29 pm
Re: gravitational potential
I think you messed up on the integration. a should be squared, and the linear b term should not be there
Re: gravitational potential
yes, youre correct. Thanks. I must have worked through that integration a dozen times and kept coming out with something different. Not sure why i missed it. Anyway, I am posting the corrected solution below for anyone interested.
So i imagine bringing in a point mass from infinity into a sphere with outer radius (a) and inner (b) and uniform mass distribution; U(inf) = 0
$$U$$ = $$U_a$$ + $$\Delta U(a>b)$$
$$U_a = \frac{GM}{a}$$
$$U_a= \frac{G}{a}\frac{4\pi}{3}\rho(a^3b^3)$$
where I expressed M in terms of rho. Thats was easy. Now for the potential difference across the shell.
$$U_{(a>b)} = \int_a^b (F\cdot dr)$$
$$= \int_a^b (\frac{GM}{r^2})dr$$
again subbing mass in terms of density and radius...
$$= G\int_a^b \frac{1}{r^2}\left[\frac{4\pi}{3}\rho(r^3b^3)\right]dr$$
After integrating...
$$=\frac{4}{3}\pi G \rho [\frac{3}{2}b \frac{a^2}{2}\frac{b^3}{a}]$$
adding U_a
$$=\frac{4}{3}\pi G \rho [\frac{3}{2}b^2 \frac{a^2}{2}\frac{b^3}{a}] \frac{G}{a}\frac{4\pi}{3}\rho(a^3b^3)$$
Some simplifying
$$=\frac{4}{3}\pi G \rho [\frac{3}{2}b^2 \frac{a^2}{2}\frac{b^3}{a}  a^2 + \frac{b^3}{a}]$$
$$=2\pi G \rho [b^2  a^2]$$
whereas the answer in the book...
$$U(R<b) = 2\pi \rho G(a^2b^2)$$
So i imagine bringing in a point mass from infinity into a sphere with outer radius (a) and inner (b) and uniform mass distribution; U(inf) = 0
$$U$$ = $$U_a$$ + $$\Delta U(a>b)$$
$$U_a = \frac{GM}{a}$$
$$U_a= \frac{G}{a}\frac{4\pi}{3}\rho(a^3b^3)$$
where I expressed M in terms of rho. Thats was easy. Now for the potential difference across the shell.
$$U_{(a>b)} = \int_a^b (F\cdot dr)$$
$$= \int_a^b (\frac{GM}{r^2})dr$$
again subbing mass in terms of density and radius...
$$= G\int_a^b \frac{1}{r^2}\left[\frac{4\pi}{3}\rho(r^3b^3)\right]dr$$
After integrating...
$$=\frac{4}{3}\pi G \rho [\frac{3}{2}b \frac{a^2}{2}\frac{b^3}{a}]$$
adding U_a
$$=\frac{4}{3}\pi G \rho [\frac{3}{2}b^2 \frac{a^2}{2}\frac{b^3}{a}] \frac{G}{a}\frac{4\pi}{3}\rho(a^3b^3)$$
Some simplifying
$$=\frac{4}{3}\pi G \rho [\frac{3}{2}b^2 \frac{a^2}{2}\frac{b^3}{a}  a^2 + \frac{b^3}{a}]$$
$$=2\pi G \rho [b^2  a^2]$$
whereas the answer in the book...
$$U(R<b) = 2\pi \rho G(a^2b^2)$$
Re: gravitational potential
Ok, here is a new problem. but since it is closely related im just threadrecycling here.
It actually comes from the 50 sample questions that is stickied.
QUESTION 26: Newtonian gravity
The mass density of a certain planet has spherical symmetry but varies in such a way that the mass inside every spherical surface with center at the center of the planet is proportional to the radius of the surface. If r is the distance from the center of the planet to a point mass inside the planet, the gravitational force on this mass is:
(A) not dependent on r
(B) proportional to r^2
(C) proportional to r
(D) proportional to 1/r
(E) proportional to 1/ (r^2)
So, first,
$$F =  \frac{GMm}{r^2}$$
but since M varies with r...
$$M = \int (kr)dv$$
$$M = k \int_0^r\int_0^{\pi}\int_0^{2\pi}(r)r^2 \sin(\theta)d\theta d\phi dr$$
After integration...
$$M=\pi k r^4$$
Plugging this into the force ....
$$F = Gm \frac{\pi k r^4}{r^2}$$
so F is proportional to r^2. BUT, the answer says that D is correct. I must be messing up big time to be so far off. please help.
It actually comes from the 50 sample questions that is stickied.
QUESTION 26: Newtonian gravity
The mass density of a certain planet has spherical symmetry but varies in such a way that the mass inside every spherical surface with center at the center of the planet is proportional to the radius of the surface. If r is the distance from the center of the planet to a point mass inside the planet, the gravitational force on this mass is:
(A) not dependent on r
(B) proportional to r^2
(C) proportional to r
(D) proportional to 1/r
(E) proportional to 1/ (r^2)
So, first,
$$F =  \frac{GMm}{r^2}$$
but since M varies with r...
$$M = \int (kr)dv$$
$$M = k \int_0^r\int_0^{\pi}\int_0^{2\pi}(r)r^2 \sin(\theta)d\theta d\phi dr$$
After integration...
$$M=\pi k r^4$$
Plugging this into the force ....
$$F = Gm \frac{\pi k r^4}{r^2}$$
so F is proportional to r^2. BUT, the answer says that D is correct. I must be messing up big time to be so far off. please help.

 Posts: 7
 Joined: Thu May 13, 2010 8:29 pm
Re: gravitational potential
It's a trick question, the mass (M) is proportional to the radius, not the mass density. So if M is linear with r, then the force is 1/r.
Re: gravitational potential
Ahhh, i see. I guess i should have read the question more carefully, but that was subtle (at least for me). Thanks