Approach: We use the principle of superposition. A spherical asteroid with a cavity can be modeled as a complete sphere of density $+\rho$ and radius $R$, superimposed with a smaller sphere (the cavity) of density $-\rho$, radius $r$, and center at distance $d$ from the origin.

1. Calculating Field Strength

Let $M$ be the mass of the original asteroid and $m’$ be the mass of the removed material (cavity). Assuming uniform density $\rho$, the masses are proportional to their volumes:

$$ m’ = \frac{M}{\frac{4}{3}\pi R^3} \cdot \frac{4}{3}\pi r^3 = M \left( \frac{r}{R} \right)^3 $$

The gravitational field $g$ at any point on the surface is given by:

$$ \vec{g}_{net} = \vec{g}_{sphere} – \vec{g}_{cavity} $$

Let point A be the point on the equator closest to the cavity, and point B be the diametrically opposite point.

At Point A (Minimum Gravity):

Distance from center $O$ is $R$. Distance from cavity center $O’$ is $R – d$.

$$ g_A = \frac{GM}{R^2} – \frac{Gm’}{(R-d)^2} $$

We are given that $g_A$ is smaller than $g_0$ by a fraction $\eta_1$. Thus, the contribution of the cavity is:

$$ \frac{Gm’}{(R-d)^2} = \eta_1 g_0 = \eta_1 \frac{GM}{R^2} \quad \dots(1) $$

At Point B (Opposite Side):

Distance from center $O$ is $R$. Distance from cavity center $O’$ is $R + d$.

$$ g_B = \frac{GM}{R^2} – \frac{Gm’}{(R+d)^2} $$

We are given that the reduction here is $\eta_2 g_0$. Thus:

$$ \frac{Gm’}{(R+d)^2} = \eta_2 g_0 = \eta_2 \frac{GM}{R^2} \quad \dots(2) $$

2. Solving for Geometry

Dividing equation (1) by equation (2):

$$ \frac{\frac{Gm’}{(R-d)^2}}{\frac{Gm’}{(R+d)^2}} = \frac{\eta_1}{\eta_2} \implies \left( \frac{R+d}{R-d} \right)^2 = \frac{\eta_1}{\eta_2} $$

Taking the square root:

$$ \frac{R+d}{R-d} = \sqrt{\frac{\eta_1}{\eta_2}} $$

Let $k = \sqrt{\frac{\eta_1}{\eta_2}}$. Applying componendo and dividendo:

$$ \frac{(R+d) – (R-d)}{(R+d) + (R-d)} = \frac{k-1}{k+1} \implies \frac{2d}{2R} = \frac{\sqrt{\eta_1} – \sqrt{\eta_2}}{\sqrt{\eta_1} + \sqrt{\eta_2}} $$ $$ d = R \left( \frac{\sqrt{\eta_1} – \sqrt{\eta_2}}{\sqrt{\eta_1} + \sqrt{\eta_2}} \right) $$

3. Finding Depth and Radius

Depth of the cavity center: The question asks for the depth, which corresponds to the distance from the surface, i.e., $R – d$.

$$ \text{Depth} = R – d = R – R \left( \frac{\sqrt{\eta_1} – \sqrt{\eta_2}}{\sqrt{\eta_1} + \sqrt{\eta_2}} \right) $$ $$ \text{Depth} = R \left[ 1 – \frac{\sqrt{\eta_1} – \sqrt{\eta_2}}{\sqrt{\eta_1} + \sqrt{\eta_2}} \right] = R \left[ \frac{\sqrt{\eta_1} + \sqrt{\eta_2} – \sqrt{\eta_1} + \sqrt{\eta_2}}{\sqrt{\eta_1} + \sqrt{\eta_2}} \right] $$

Radius of the cavity ($r$): From equation (1):

$$ \frac{Gm’}{(R-d)^2} = \eta_1 \frac{GM}{R^2} $$

Substitute $m’ = M(r/R)^3$:

$$ \frac{GM (r/R)^3}{(R-d)^2} = \eta_1 \frac{GM}{R^2} \implies \left(\frac{r}{R}\right)^3 = \eta_1 \frac{(R-d)^2}{R^2} = \eta_1 \left( \frac{R-d}{R} \right)^2 $$

From our depth calculation, $\frac{R-d}{R} = \frac{2\sqrt{\eta_2}}{\sqrt{\eta_1} + \sqrt{\eta_2}}$. Substituting this back:

$$ \left(\frac{r}{R}\right)^3 = \eta_1 \left[ \frac{2\sqrt{\eta_2}}{\sqrt{\eta_1} + \sqrt{\eta_2}} \right]^2 = \frac{4\eta_1 \eta_2}{(\sqrt{\eta_1} + \sqrt{\eta_2})^2} $$