# How structurally strong would an object need to be to survive a trip through an ergosphere?

Let us assume we have a Kerr metric (rotating) black hole of 10 Sol-masses, spinning at 300 revs/s (as measured by a rest frame observer). What yield strength (assume that yield strength is limiting here in order to prevent failures in compression) would be required for the structural members of:

1. a 1km radius sphere, or
2. a 1km/side cube

in order for it to survive the gravitational forces of a prograde passage through the ergosphere, along a course in plane with and secant to the equator that passes through a point halfway between the ergosphere's equatorial radius and the event horizon's equatorial radius?

Assume the object is traveling at a constant 1km/s before entering the significant influence of the black hole, if that matters whatsoever.

(If the 1km object is too large to fit into the ergosphere of the given black hole, please let me know.)

This question asks for hard science. All answers to this question should be backed up by equations, empirical evidence, scientific papers, other citations, etc. Answers that do not satisfy this requirement might be removed. See the tag description for more information.

• You’re assuming that the ergosphere is more than a km deep. Also, gravity’s influence extends forever so your last sentence is ill-defined. – JDługosz Nov 6 '16 at 9:59
• @JDługosz -- would a 100m, or even a 1m, object be more reasonable? – Shalvenay Nov 6 '16 at 14:05
• a 1km object will serve totally different role in a story than 1m object, so it depends what these are going to actually be in your world. 1km probes are not reasonable. 1m spaceships aren't, either. And if you are not creating a world for a story, then you asked in wrong place. – Mołot Nov 6 '16 at 14:34
• @Mołot -- 1km is a representative spaceship, 1m is a representative probe – Shalvenay Nov 6 '16 at 14:39
• About your last edit - I think I remember reading that ergosphere starts where event horizon of non-rotating black hole would start, and on the equator can be half as deep, max. "Astronomers have found black holes with event horizons ranging from 6 miles to the size of our solar system." - so 1km can be believeable. And by the way, I do believe that size might matter, if you have any specific black hole size in mind, please share it with us :) – Mołot Nov 6 '16 at 14:45

If you are expecting someone to solve the Kerr metric equations, you probably need to hire a professional mathematician; but if you want an approximation, we can make that happen. Lets start with some simple results and eventually work our way to advanced results.

# Objective

Our objective is to calculate the maximum one dimensional stress tensor acting in the direction along the axis tangential to the assumed spherical surface of the black hole. Since all our forces come from the gravity of the black hole, they will all be acting in this direction, so I'm not going to use any vectors.

# Assumptions

• $\text{M}_{b}$ is the mass of the black hole and it equals $1.99\times10^{31} \text{ kg}$ (10 times the mass of the sun).
• The object in question is a 1 km long, 100m radius cylindrical rod, with the rod aligned in the direction of a radial line from the center of the black hole outwards. The object has constant density $\rho = 1$, its just not that important right now.
• The object is 'suspended' with it's midpoint centered at 3/4 the Schwarzschild radius of the black hole.

Given by $$r_s = \frac{2GM}{c^2}$$ where G is the universal gravitation constant ($6.67\times10^{-11}\frac{\text{N}\cdot\text{m}^2}{\text{kg}^2}$), M is the mass of the black hole, and c is the speed of light ($3.00\times10^{8}\frac{\text{m}}{\text{s}}$). Therefore $$r_s = \frac{2\cdot 6.67\times10^{-11} \cdot 1.99\times10^{31}}{(3.00\times10^{8})^2} = 29500 \text{m}.$$ Therefore, with some rounding, the near-hole point of our rod is at 22 km, the far-hole point is at 23km.

# Force of gravity as a function of distance from near-hole point

Let us define a coordinate system in one dimension with $l = 0$ as the near-hole point, and $l = 1000$ (in meters) as the far-hole point of our rod. We will calculate the force of gravity on each infinitesimally small slice of the rod as a function of it's $l$ coordinate.

The force of gravity on a mass is $$F = G\frac{m_1m_2}{r^2}.$$ The mass of a slice of the rod (equivalent to the distance derivative of the mass of the rod) is equal to the mass of a circle $\frac{dm}{dl} = \rho \pi (100)^2$. Therefore the distance derivative of the force of gravity on a slice is $$\frac{dF_{slice}}{dl} = 6.67\times10^{-11} \frac{1.99\times10^{31}\cdot \rho \pi (100)^2}{(23000 + l)^2} = \frac{4.17\times10^{25}}{(23000 + l)^2}$$

# Integrate the distance derivative of the force of gravity

To find the net force between points $l = a$ and $l = b$, we integrate the distance derivative of the force of gravity with respect to distance from the near-hole point. $$\int_a^b \frac{4.17\times10^{25}}{(23000 + l)^2} dl = \left.\frac{-4.17\times10^{25}}{23000 + l}\right|^b_a = -4.17\times10^{25}\left(\frac{1}{23000+b}-\frac{1}{23000+a}\right)$$

Solving this for the net force on the entire rod, we get $$-4.17\times10^{25}\left(\frac{1}{23000+1000}-\frac{1}{23000+0}\right) = 7.55\times10^{19} \text{N}.$$

Now that force has to be counteracted by a 'lift' force keeping the rod out of the black hole. For simplification let us assume that the counteracting force acts equally on each slice of the rod, so each a slice of the rod from a to b is pulled out of the black hole with force $$F_{lift} = -7.55\times10^{19}\cdot\frac{b-a}{1000}.$$ Note the force is negative because it is acting in the direction out of the hole.

# Solve for stress at any point in the rod

In this simplification, the highest gravity force will be at the lowest point closest to $l = 0$. Therefore, the stress causing force at any distance $x$ in this rod is going to be the net force of gravity and lift for all slices below it minus the net force of gravity and lift for all points above it. \begin{align}F_{net} =&\left.\frac{-4.17\times10^{25}}{23000 + l}\right|^x_0 - 7.55\times10^{19}\cdot\frac{x-0}{1000}- \left.\frac{-4.17\times10^{25}}{23000 + l}\right|^{1000}_x + 7.55\times10^{19}\cdot\frac{1000-x}{1000} \\ =& 3.55\times10^{21}-\frac{8.34\times10^{25}}{23000+x} + 7.55\times10^{16}\cdot (1000-2x) \end{align}

The net force graph looks like this:

Maximum force is $1.61\times10^{18} \text{ N}$ at $l=500$.

Stress defined as $\sigma = \frac{\text{F}}{\text{A}}$. The cross sectional area is $\pi(100)^2 = 31415 \text{m}^2$, so maximum stress is $$\sigma = \frac{1.61\times10^{18} \text{ N}}{31415 \text{m}^2} = 5.12\times10^{13} \text{Pa}.$$

# Conclusion

The calculation works and produces logical results. Stress should be zero at the ends of the rod (there is nothing to pull away from) and should be maximum in the center. The stress produced is very high, as would be expected 23km from the center of a black hole.

Required yield strength is about 51 TPa. The required material strength is probably not achievable with any known or theoretical material. I can't find anything with a yield strength over 1 TPa, much less 51.

• If the dimensions of the object were smaller, would it be possible for present-day materials to suffice? – Demi Jan 28 '17 at 21:02
• @Demi That would be a new follow-on question asking what the maximum size the object could be using present day materials :) – Tim B Feb 23 '17 at 9:37
• I'm always baffled how you guys know such stuff... – Alexander von Wernherr Feb 23 '17 at 13:45
• @AlexandervonWernherr This one is just Calc II and Statics, plus looking up some stuff about black hole on Wikipedia. So sophomore level for an engineering major :) – kingledion Feb 23 '17 at 13:47
• So sophomore level for an engineering major Great. Now I'm depressed... ;) – Alexander von Wernherr Feb 23 '17 at 13:49