Re: Cube Planet

On 2005-07-28, mralls@xxxxxxxxxxxxxx <mralls@xxxxxxxxxxxxxx> wrote:
> If one had the engineering capablilities to make a cube planet of
> roughly the size and mass of and chemical and biologiogical make up
> Earth, how long would it stay square before gravitational forces pulled
> it into something sphere-ish?

This is probably better asked in rec.arts.sf.science. I'm crossposting
there and redirecting follow-ups.

Without actually doing any _real_ calculations (i.e. fluid-dynamic
simulations) (because I'm lazy and don't want to look up material
properties and write the code), I would estimate on the order of
hours, as rock is rather plastic, fluid even, at those scales.
Consider the liquidity of magma...

A _minimum_ time can be set by the falling time from the corner of the cube
to the current surface radius.

R = 6378 km
V = 1.09 * 10^21 m^3
L = 1.03 * 10^7 m = 1.03 * 10^4 km = 10300 km
Corner to center = sqrt(3) * L / 2 = 8920 km
dx = falling distance = 2542 km
t = sqrt(2 * dx / a)

If we take a = g = 9.8 m /s^2 we get 720 seconds = 12 minutes.
Changing the acceleration by small amounts affects this little --
Dividing it by a factor of four merely doubles the time.

We can also look at the time for sound waves (and hence any disturbance)
to propagate from one end to the other. Take v = 13 km/s.
10300 km / 13 km/s = 792 s, again on the same order of minutes.

Now these are very definitely minimum bounds though, not taking into account
viscosity, and the need to push apart or flow over rock in the way. But
that shouldn't change the timescale that much -- I would guess that would
roughly double or triple it.

There is going to be a lot of energy changed from potential energy to
kinetic (and heat through friction/viscosity) during this process. The
losses will slow things down considerably from that naive estimate, as
energy goes into heating rather than reducing the radius. But heating
will of course make the rock more fluid, so flowing faster. In fact,
I bet you get enough energy to make the entire surface fluid, if not
vaporize portions of it.

Hmm. Let me do the calculations for that at least.

M = 6 Ã? 10^24 kg
rho = 5.5*10^3 kg / m^3

Now, calculating the GPE (gravitational potential energy) of either
shape is very tricky unless we assume a uniform density. This isn't
terribly realistic, but then again, neither is a cubical Earth. In
fact, we can't calculate a realistic distribution, because we end up
with the spherical one.

So: GPE for spherical, constant density:

Handy fact: viewed from the outside, the attraction due to a spherical
shell (and hence sphere) is the same as that mass concentrated at the
central point. Viewed from the inside, there is no effect. So we can
imagine bringing infinitesimal shells "in from infinity" to build
Earth.

A simple calculation shows that the potential energy of a mass m
at radius r away from one of mass M is - G m M / r

M(r) = rho 4*pi*r^3 / 3
m(r) dr = d[M(r)]/dr * dr = rho * 4 * pi * r^2 dr

GPE = \int dPE = \int PE(r) dr = - \int G M(r) * m(r) * dr / r
= - G rho^2 * 16 * pi^2 \int r^4 dr / 6 = - G 16 pi^2 rho^2 R^5 / 15

rho = M / (4 * pi * R^3 / 3)

GPE = -3 G M^2 / 5 R
= -2.24 * 10^32 Joules

For cube, constant density:

here we take a different tack, because we can't rely on the same effects
for non-spherical shells.

rho = M/V = M/L^3

GPE = -1/2 \int \int dv dv' G rho^2 / (2 |r - r'|)
= -G rho^2/4 \int dx dy dz dx' dy' dz'
((x-x')^2 + (y - y')^2 + (z - z')^2)^(-1/2)

But it's far more complicated. I went through about a third (evaluating
x and x')[1] and gave up before finding
http://arxiv.org/abs/astro-ph/0002496
which gives

- 30.117 G rho^2 a^5, where a = L/2

rho^2 = M^2 / L^6; a^5 = L^5 / 32

GPE = -30.117 G M^2 / 32 L
= -2.18 * 10^32 joules

For a difference of 6 * 10^30 Joules.

Assuming uniform heating, this gives 10^6 Joules/kg.
Heat capacity of rock is about 800 J/kg K.

This is an average temperature gain of 1250 K. Some won't heat that
much, others will heat much more. But (a) most of the interior is
already hot and molten, (b) this average increase is far above the
melting point of rocks. So I bet in the hot spots you will get
vaporisation. And the vaporized portions will be unable to keep nearby
material from moving at nearly the minimum quoted above.

So yeah, I'd estimate on the order of a couple hours.

> Before that happened, what would things be like on the surface of that
> planet? As you approached the corners, what would happen to gravity?

Before that happened, it would seem like a giant earthquake. Then
you'ld get cooked and/or crushed to death.

The corners would seem like giant mountains, poking out of the
atmosphere. The gravity. would be "turned" a bit relative to the cube
faces, pointing towards the center, the turning getting stronger as you
go away from the center of the face.

> Could you have oceans that flowed over the corners?

No.

--
Aaron Denney
-><-

[1]
Evaluating \int dx first, x' constant, and (y-y)'^2 + (z-z')^2 = a^2

\int_0^L dx ((x - x')^2 + a^2)^(-1/2) =
\int_0^x' dx ((x - x')^2 + a^2)^(-1/2) +
\int_x'^L dx ((x - x')^2 + a^2)^(-1/2) =
\int_0^x' dx (x^2 + a^2)^(-1/2) + \int_0^(L-x') dx ((x - x')^2 + a^2)^(-1/2) =
[log (x + sqrt(x^2 + a^2)]_0^x' + [log (x + sqrt(x^2 + a^2)]_0^(L - x') =
log (x' + sqrt(x'^2 + a^2)) + log ((L - x') + sqrt((L - x')^2 + a^2) - 2 log(a)

Now let's do x' (and write x for x'). Symmetry turns this into

2 \int log (x + sqrt(x^2 + a^2)) - log (a) dx
= -2 L log (a) + 2 \int log (x + sqrt(x^2 + a^2)) dx

substitute u = x + sqrt(x^2 + a^2)
(u - x) = sqrt(x^2 + a^2)
u^2 - 2ux + x^2 = x^2 + a^2
2ux = u^2 - a^2
x = (u^2 - a^2) / 2u
dx = du (2u 2u - 2(u^2 - a^2))/ 4u^2 = (1/2) du (u^2 + a^2)/u^2

\int (1 + a^2/u^2) log (u) du =
\int log (u) du + \int a^2 u^(-2) log (u) du =
[u log u - u]_a^(L + sqrt(L^2 + a^2))
+ a^2 * [ - u^(-1) log u - (u)^(-1) ]_a^(L + sqrt(L^2 + a^2)) =
(L + sqrt(L^2 + a^2)) log (L + sqrt(L^2 + a^2)) - (L + sqrt(L^2 + a^2))
- a log a + a
+ a^2 * ( - (L + sqrt(L^2 + a^2))^(-1) log (L + sqrt(L^2 + a^2))
- (L + sqrt(L^2 + a^2))^-1))
+ a log a + a
.

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