Mass Density Gradient

There's a super clever trick to solve this problem which I've picked up reading Feynman Lectures on Electromagnetism.

The most challenging part of this problem is to find a gravitational field produced by hollow sphere of mass surface density proportional to one of the coordinates - in this case $x$ . This distribution has no physical meaning since it implies that there's negative mass, which is absurd. This negative mass should be interpreted mathematically, abstractly, as if it repels objects instead of attracting them (think of electricity as analogy).

For start, consider hollow sphere with surface density $\sigma = \sigma_{0}\cos{\theta}$ , where $\sigma_{0}$ denotes maximum surface density and $\theta$ is the angle between the position vector and positive direction of some coordinate axes, let it be $x$ . It can be shown that such a distribution is equivalent to superposition of two solid spheres of uniform mass density with first consisting of "attracting mass", center of which is displaced by infinitesimal along positive direction of $x$ axis and second consisting of "repelling mass", center of which is displaced by infinitesimal along negative direction of $x$ axis.

Potential energy of a test mass at distance $r$ from center of the solid sphere is: $U = -\frac{\rho V}{r},$ leaving off constant $Gm$ which is here unnecessary. Above we argued that potential energy produced by our specific distribution is a superposition of two solid spheres: $\begin{aligned} U &= U_{+}+U_{-} \\ &= -\frac{\rho V}{r-d/2}+\left ( -\frac{-\rho V}{r+d/2} \right ) \\&= -\rho V\frac{d}{r^{2}} = -\sigma_{0}S\frac{a}{3r^{2}} \end{aligned}$ where $d/2$ is an infinitesimal displacement, $S$ is the surface area of the sphere, and $a$ is the radius of the sphere. Force experienced by a test mass is $F = -\nabla{U}$ or in our case: $F = -\sigma_{0}S\frac{2a}{3r^{3}}$

In general $x = a\cos{\theta}$ , and since in this problem $a = 1$ , and in second scenario $\sigma_{0} = -\alpha$ , force is: $F = \alpha S\frac{2}{3r^{3}}$

Total force in the second scenario is: $F_{2} = -\alpha S\frac{1}{r^{2}} + \alpha S\frac{2}{3r^{3}}$

Finally, we have: $\begin{aligned} F_{1} &= F_{2} \\ -S\frac{1}{r^{2}} &= -\alpha S\frac{1}{r^{2}} + \alpha S\frac{2}{3r^{3}} \\ 1 &=\alpha \left ( 1-\frac{2}{3r} \right ) \\ \alpha &= \dfrac{3r}{3r-2}\end{aligned}$ which makes final answer: $\boxed{\alpha = 1.5}$ .

I would like to elaborate this technique in detail and actually offer a proof why it works, but currently I don't have much time. I may write more about it in a week or two if there's anyone interested.

At distances greater than $1$ from the centre, the gravitational force in Scenario 1 is that same as that of a point particle of mass $4\pi \sigma_1 = 4\pi$ situated at the origin. Thus the force on the test particle (of mass $m$ ) in Scenario 1 is $\pi Gm$ , acting towards the origin.

In Scenario 2, the force on the test particle will also act towards the origin, and will have magnitude $\int_0^\pi \frac{Gm \times 2\pi \sin\theta \times \alpha(1-\cos\theta)}{(2 - \cos\theta)^2 + \sin^2\theta} \times \frac{2 - \cos\theta}{\sqrt{(2 - \cos\theta)^2 + \sin^2\theta}}\,d\theta \; = \; 2\pi G\alpha m\int_{-1}^1 \frac{(1-u)(2-u)}{(5 - 4u)^{\frac32}}\,du \; = \; \tfrac23Gm\pi \alpha$ and hence, since these two attractions are equal, we deduce that $\alpha = \boxed{\tfrac32}$ .