Holed on

Principle : The field of a homogeneous sphere of radius $R$ at a point at distance $d \leq R$ from its center points inward and is proportional to $\rho d$ , where $\rho$ is the density.

First, consider the field of a solid sphere of radius $a + b$ , without the hole. For the blue point we have $d = a + b$ ; for the green point we have $d = a - b$ ; so we find that $\vec g_{1,bl} = a + b,\ \ \ \vec g_{1,gn} = a - b.$ We choose units so that the proportionality constant involving density, $G$ , and other constant factors is equal to 1. Moreover, the positive direction is chosen toward the center of the large sphere.

Next, consider the field of a solid sphere of radius $b$ that would fit in the hole. For both points we have $d = b$ , so we get $\vec g_{2,bl} = b,\ \ \ \vec g_{2,gn} = -b.$

Now for the given shape, which at the given points has fields $\vec g_0 = \vec g_{3,bl}$ and $\vec g' = \vec g_{3,gn}$ , respectively. By the superposition principle, if we add the field of this shape and the field of the small sphere together, we should find the field of the large solid sphere. Thus $\begin{cases} \vec g_{1,bl} = \vec g_{2,bl} + \vec g_{3,bl} \\ \vec g_{1,gn} = \vec g_{2,gn} + \vec g_{3,gn} \end{cases} \longrightarrow \begin{cases} a+b = b + g_0 \\ a-b = -b + g' \end{cases} \longrightarrow \begin{cases} g_0 = a \\ g' = a \end{cases}$ We see immediately that $\boxed{g' = g_0}$ .

Let the mass of the larger sphere be $M$ and the mass of the smaller sphere cut out to make the hole be $m$ .

Given that

$\rho = \dfrac{\text{Mass of large sphere}}{\text{Volume of large sphere}} = \dfrac{M}{\frac 43 \pi (a+b)^3}$

Therefore

$m = \rho \left( \dfrac 43 \pi b^3 \right) = \dfrac{M}{\frac 43 \pi (a+b)^3} \cdot \dfrac 43 \pi b^3 = M {\left( \dfrac{b}{a+b} \right)}^3$

At $\color{#3D99F6} \text{ blue point }$ the gravitational field lines due to both spheres is directed in the same direction. Since the direction is same so let us consider the sense of both gravitational fields to be in positive sense. Thus at the blue point, we have

$\begin{aligned} & E_{\text{large sphere}} = \dfrac{GM}{(a+b)^2} \\ & E_{\text{small sphere}} = \dfrac{Gm}{b^2} = \dfrac{GMb}{(a+b)^3} \end{aligned}$

As the small sphere is 'carved out', so the net field at blue point (by the principle of superposition) is

$E_{\text{net at blue}} = E_{\text{large sphere}} - E_{\text{small sphere}} = GM \left( \dfrac{1}{(a+b)^2} - \dfrac{b}{(a+b)^3} \right) = \dfrac{GMa}{(a+b)^3}$

At $\color{#20A900} \text{ green point }$ the gravitational field lines due to both spheres is directed in opposite directions. As the direction is opposite so let us consider the gravitational field of larger sphere to be positive while that of the smaller one to be negative. Thus at the green point, we have

$\begin{aligned} & E_{\text{large sphere}} = \dfrac{GM(a-b)}{(a+b)^3} \\ & E_{\text{small sphere}} = -\dfrac{Gm}{b^2} = -\dfrac{GMb}{(a+b)^3} \end{aligned}$

The net field at green point (by the principle of superposition) is

$E_{\text{net at green}} = E_{\text{large sphere}} - E_{\text{small sphere}} = GM \left( \dfrac{a-b}{(a+b)^3} + \dfrac{b}{(a+b)^3} \right) = \dfrac{GMa}{(a+b)^3}$

We find that

$E_{\text{net at blue}} = E_{\text{net at green}}$

Thus we get

$E_{\text{net at green}} = \boxed{g_0}$

Gravitational field at the surface of a solid sphere of mass M, radius R and density $\rho$ is given by $g = G\frac{M}{R^2} = G\frac{4/3 \pi R^3 \rho}{R^2} = G\frac{4 \pi R \rho}{3}$ We will use this general form repeatedly now.

The hole can be treated as having density $-\rho$ or a sphere that produces a 'repulsive' field! Moreover, since at the given points both the fields, have the same line of action, they can be added or subtracted algebraically.

At the blue point net field = $G\frac{4 \pi (a+b) \rho}{3} - G\frac{4 \pi b \rho}{3} = G\frac{4 \pi a \rho}{3} = g_0$

At the green point net field = $G\frac{4 \pi (a-b) \rho}{3} + G\frac{4 \pi b \rho}{3} = G\frac{4 \pi a \rho}{3} = g_0$

A simpler way would be to notice that the gravitational field must be constant( UNIFORM!) inside the cavity ( it is a pretty standard result, in both gravitation and electrostatics, so I skip the proof; which is easily obtained from simple vector laws), and is given as $4/3\pi G\rho\vec{r}$ , where $\vec{r}$ is the position vector of the center of cavity wrt the center of sphere. Clearly, it is independent of the position of any arbitrary point inside the cavity. Hence the result.

The idea of the solution is the actual field can be calculated by the difference between the field of a sphere without hole and field of a sphere with same density filling the hole.