Verify Newtonian Potential of Thin Spherical Shell

[latentcorpse]

Consider a thin spherical shell of radius a and mass per unit area $\sigma$ whose centre at O lies a distance r from an external point P. If the gravitational potential r is $\Phi(r)$, verify that the external potential due to the shell will be equal to that due to a point of the same mass as the shell ($4 \pi a^2 \sigma$) located at O if

$\Phi(r) = A r^{-1} + B r^2 + C$ where A,B and C are all constants.Hint: Show $M(a) \Phi(r) + 2 \pi \sigma a \lambda(a) = \frac{2 \pi \sigma a}{r} \int_{r-a}^{r+a} x \Phi(x) dx$ where $\lambda(a)$ is a constant which can be added without altering the force law.

I literally cannot seem to get anywhere with this problem!

 

[NateD]

I have tried deriving the potential due to the shell from its mass per unit area and then trying to equate it to that due to a point of the same mass, but can't seem to move forward.Answer:We start by calculating the gravitational potential due to the shell at a point P located at a distance r from the centre of the shell:M(a) \Phi(r) = \sigma \int_0^a 4 \pi r^2 dr \Phi(r) = 4 \pi \sigma \int_0^a r^2 \Phi(r) drNow we consider the potential due to a point of the same mass as the shell located at O:\frac{2 \pi \sigma a}{r} \int_{r-a}^{r+a} x \Phi(x) dx To show that these two potentials are equal, we add a constant \lambda(a) to the potential due to the shell such that:M(a) \Phi(r) + 2 \pi \sigma a \lambda(a) = \frac{2 \pi \sigma a}{r} \int_{r-a}^{r+a} x \Phi(x) dxThe constant \lambda(a) can be found by substituting the given expression for \Phi(r):M(a) (A r^{-1} + B r^2 + C) + 2 \pi \sigma a \lambda(a) = \frac{2 \pi \sigma a}{r} \int_{r-a}^{r+a} x (A x^{-1} + B x^2 + C) dxBy evaluating the integral on the right hand side and equating coefficients of like powers of r on both sides, we get:\lambda(a) = \frac{1}{2 \pi \sigma a} ( A \ln(r+a) - A \ln(r-a) + 2 B a^2 + 2 C a )Therefore, the external potential due to the shell will be equal to that due to a point of the same mass as the shell (4 \pi