QUESTION: if a planet explodes, which parts spread away faster
JJ
I wanted to ask a question about the physics related to the explosion
of a planet.
Lets say that there was a planet with 2 "layers":
A. solid core (with density something similar to the average density
of our moon), having diameter of around 10000 kilometers
B. around the core, a layer of water vapor, having a diameter of
around 1 kilometer
Although I must say that I am not sure if that core could hold that
water vapor around the planet...in that case sorry because my planet-
calculating-math-skills are not that great.
The question I wanted to ask is about how different components would
act if something blew the planet core into smithereens and the
explosion resulted in pieces of following 3 sizes:
1. many particles less than 5mm in diameter.
2. fewer of those between 1m-100m in diameter.
3. a handfull of those between 100m-1km in diameter.
Also in this scenario the explosion would start from the inside and
all the different size particles would spread away from the center.
The scenario also assumes that the explosive force acting on each of
the particles would be of the same magnitude and that there would be
nothing in the space around the planet affecting the spread of any of
the particles.
With these in mind, would the smallest particles move away from the
center faster than the larger ones?
Or would the larger ones move faster...or would all particle types
move at the same speed?
By the way if you know of a website or a good book that discusses
these types of calculations, I would appreciate hearing about it.
Thanks in advance!
[[Mod. note -- The enery required to disassemble a planet is (much)
more than is necessary to vaporize it, so a scenario in which most
of the remnant is still solid (as you seem to posit) seems implausible.
-- jt]]
Kent Paul Dolan
> [[Mod. note -- The energy required to disassemble
> a planet is (much) more than is necessary to
> vaporize it, so a scenario in which most of the
> remnant is still solid (as you seem to posit)
> seems implausible.
> -- jt]]
???
Since some asteroids are described as "barely bound
collections of rubble", wouldn't "the energy
required to disassemble a planet" be on some sort of
a continuum based on its mass, density, rotation
rate, composition, and so on, rather than always
being "sufficient to vaporize the planet"?
xanthian.
Jonathan Thornburg [remove -animal to reply]
> Since some asteroids are described as "barely bound
> collections of rubble", wouldn't "the energy
> required to disassemble a planet" be on some sort of
> a continuum based on its mass, density, rotation
> rate, composition, and so on
Yes. I'll give a rough calculation below.
> rather than always
> being "sufficient to vaporize the planet"?
Rather than speculate, let's do some calculations to find out:
Even if the planet is a "barely bound collection of rubble, it still
has gravitational binding energy. Let's work out approximately how
much this energy is, and how much that same amount of energy could
heat the planet.
[I'll use LaTeX notation for the math, but it should
still be easy for non-LaTeX people to follow along.]
For simplicity, let's model the planet as a sphere of radius $r$ and
uniform density $\rho$. Then the planet has mass $(4/3) \pi r^3 \rho$.
To within a factor of order unity, the gravitational potential energy
of that much mass at a typical mass-element-to-mass-element distance
of $r$ is $-G m^2/r = -(16/9)\pi G r^5 \rho^2$. A more accurate
calculation (a straightforward elementary-calculus exercise) replaces
that 16/9 with 16/15, i.e., the gravitational potential energy of the
planet is actually $-(16/15)\pi G r^5 \rho^2$.
Notice that this energy is negative, i.e., energy is released when
we assemble the planet (equivalently, we have to put energy in to
disassemble the planet). The magnitude of that energy is
$E = (16/15)\pi G r^5 \rho^2$.
If we put that much energy ($E$) into heating the planet, how hot
would it get? Approximating the specific heat of the rock $C_p$ as
constant, the energy E is related to the temperature change $\Delta T$
by $E = m C_p \Delta T$, so we have $\Delta T = E / (m C_p)$.
Substituting in our expressions for $E$ and $m$, we find
$\Delta T = (4/5) G r^2 \rho/C_p$.
Now let's plug in some numbers for the Earth:
mean density 5.5 grams/cm^3 = 5500 kg/m^3
radius 6400 km = 6.4e6 m
G = 6.67e-11 N m^2/kg^2
C_p for "rock" varies for one mineral to another, but is somewhere
around 900 J/(kg K)
(http://www.engineeringtoolbox.com/specific-heat-solids-d_154.html)
This gives a temperature rise of 13,000 Kelvin. Typical rocks melt
at 900 to 1500 Kelvin, and vaporize at a few thousand Kelvin, so roughly
speaking, the Earth's gravitational binding energy is about 10 times
larger than the energy required to melt the Earth, or about 4 times
larger than the energy required to vaporize the Earth.
In contrast, for the Earth's moon, we have
mean density 3.36 g/cm^3 = 3360 kg/m^3 and radius 1738 km = 1.738e6 m,
so the temperature rise comes out to be only 600 K, i.e., the energy
required to disassemble the moon is only about 1/2 that needed to
melt the moon.
And for Ceres (the largest asteroid), we have radius around
470 km = 4.7e5 m and mean density = 2.1 g/cm^3 = 2100 kg/m^3, so the
temperature rise comes out to be only about 30 Kelvin, which is tiny.
Thus, it's reasonably plausible that blowing up an asteroid or something
the size of the moon would yield rubble, but blowing up the Earth would
be more likely to vaporize it.
--
-- "Jonathan Thornburg [remove -animal to reply]" <jth...@astro.indiana-zebra.edu>
Dept of Astronomy, Indiana University, Bloomington, Indiana, USA
"Washing one's hands of the conflict between the powerful and the
powerless means to side with the powerful, not to be neutral."
-- quote by Freire / poster by Oxfam