# Gravitational Potential due to spherical shell

Tags:
1. Oct 25, 2016

### Elvis 123456789

1. The problem statement, all variables and given/known data
What is the gravitational potential both inside and outside a spherical shell of inner radius b and outer radius a?

2. Relevant equations
φ = ∫g⋅da = -4πGMencl
g = d∅/dr in the r hat direction

3. The attempt at a solution
I can get as far as getting the gravitational field for the three parts of the shell but im not really sure how to determine the limits of integration in order to get the potentials

for (R > a) g*4πR2= -4πG*(4/3*π(a3-b3)*ρ)
then ∅ = ∫[G*(4/3*π(a3-b3)*ρ)]/R2 dr

for (b< R < a) g*4πR2= -4πG*(4/3*π(R3-b3)*ρ)
g = 4/3*πρG*(b3/R2 - R)
∅ = ∫-[4/3*πρG*(b3/R2 - R)] dr

for (R < b) g = 0 because there is no enclosed mass
and ∅ = constant the constant being determined from the integration limits

Last edited: Oct 25, 2016
2. Oct 25, 2016

### kuruman

Make 3 drawings, one for each of the three regions. Note that the left side of the equation is always 4πr2g where r is the radius of the sphere that encloses whatever mass. What about the right side? What is an expression for Mencl in each of the 3 regions? That should give you a clue about the limits of integration.

3. Oct 25, 2016

### Elvis 123456789

The expressions for the enclosed mass in each region is in my original post. So im still not sure how to determine the limits of integration

4. Oct 25, 2016

### kuruman

OK, I now see what you mean by limits of integration. You got Mencl correctly. The problem is with the first equation, φ = ∫g⋅da. I would use r instead of a because a is defined here. The relation between g and φ in spherical symmetry is $g = -\frac{\partial \phi}{\partial r}$ so that $\phi(r) = - \int_{ref}^r g(r) ~dr$. Usually, the lower limit of integration (the reference of potential) is infinity. I hope this clarifies what you need to do.