Ceres: A Hole from Pole to Pole?

The Kola Super Borehole attempted to drill 15 km into continental basalt. It was stopped at 12.26 km depth by unexpectedly high temperatures ( 180°C instead of the estimated 100°C ) which made the rock soft and plastic, filling the 9 inch 23 cm hole when drill bits were replaced. We can estimate the pressure at the bottom of the hole from the weight of the 2600 kg/m³ granite overburden in 1 gee 9.8m/s² gravity as 2600 × 9.8 × 12260 = 310 MPa (MegaPascal). For comparison, the pressure at the floor of the Challenger Deep is 111 MPa, and sea level air is 101 KPa.

What if we drilled a hole down from a rotation pole on Ceres?

While we are getting some great data from the Dawn orbiter (whose mission has been extended through 2018, with an periapse shift down to 200 km altitude by April 2018!) we can only estimate the stratigraphy inside this large asteroid. Let's be lazy and assume a uniform round ball of equal average density.

Ceres radius is r_0 = 473 km, has a surface gravity of g_0 = 0.28 m/s², and an average density of \rho = 2160 kg/m³. What is the pressure at the bottom of a hole drilled into that? Simply, the total gravitational weight of the mass column above the bottom of the hole. If the density is uniform, the gravity decreases linearly with depth; mass is proportional to radius cubed, gravity is proportional to mass over radius squared, hence gravity is proportional to radius. So, the integrated pressure at depth is given by:

P = \int^{r_0}_r \rho ~ g(r) ~ d r ~ = ~ \int^{r_0}_r \rho ~ g_0 \left( r \over r_0 \right ) ~ d r ~ = ~ \left( { \rho ~ g_0 } \over r_0 \right ) \int^{r_0}_r r ~ d r = \left( { \rho ~ g_0 } \over { 2 r_0 } \right ) ( r{_0}^2 - r^2 )

The pressure at the center of Ceres is ~ ~ ~ \rho ~ g_0 ~ r_0 / 2 = 2160 * 0.28 * 473000 / 2 = 143 MPa, less than half of the pressure at the bottom of the Kola Super Borehole, and only 30% higher than the pressure on the floor of the Challenger Deep.

So, the most excellent Stupid Guy Trick of the third millenium will be drilling a large-bore hole from pole to pole through Ceres, lining it with strong material, pumping out all the air, jumping in one end, and popping out the other end about 70 minutes later. Don't touch the walls; peak velocity at the center is 364 m/s, or about 800 miles per hour for the hopelessly archaic (the perversely archaic can compute that in root BTUs per carat). Miles will not be permitted on Ceres.

t = \pi \sqrt{ r_0 / g_0 } ~ end to end, v = \sqrt{ r_0 ~ g_0 } ~ at the center.

For comparison, Kola Super Borehole pressure occurs less than 40 km deep on Mars, not even going all the way through the crust. Why would anyone want to drill a deep recreation hole there?

Semi-practically speaking

The mass of a sphere is M = 4 \pi \rho r^3 / 3 so ~ g = G M / r^2 = 4 \pi G \rho ~ r / 3 . Thus ~ t = \pi \sqrt{ r_0 / g_0 } ~=~ \sqrt{ 3 \pi / 4 G \rho ~ } , so ~ G = 3 \pi / ( 4 \rho ~ t^2 ) . If the shape and the density and the oscillation period is precisely known, with computer modelled tweaks for feedback-controlled centering forces, material shear and compression, and other perturbations, we can estimate G to high accuracy. A huge quartz sphere in the asteroid belt, manufactured out of asteroidal material with an axial hole, might someday help us measure G to better than the current 5 decimal places. Since the system is axially symmetric, electric and magnetic fields can be nulled out.