MadSci Network: Astronomy
Query:

Re: Minimum size of asteroid or comet needed to break up the Earth?

Date: Wed Jul 7 23:27:14 1999
Posted By: Robert Macke, Grad student, Physics, Washington University
Area of science: Astronomy
ID: 930759053.As
Message:

Dan,

The real issue here is not of the size of the object, but of the energy it 
releases in the collision.  Even a very small asteroid could destroy the 
earth if it traveled fast enough (though that's not very likely at all), 
just as a cannon ball causes much more damage when fired from a cannon than 
it causes if just thrown at its target by hand, because it releases much 
more energy in the target if it travels faster.

So, how much energy do you need to destroy the earth?  Certainly, it is a 
lot more than you need just to wipe out life.  Life exists in a very 
fragile balance that can easily be disrupted worldwide by an object that 
would otherwise just leave a crater the size of Ohio but leave the rest of 
the planet untouched.  You need a lot more energy to split the Earth in 
two.

Since we're talking hypothetical situations here, let's go to extremes.  
(This simplifies the calculation a bit)  Let's say we wish to pull the 
planet apart entirely, separating it dust grain from dust grain. The energy 
it would take to do this is simply the opposite of the energy stored in the 
gravitational collapse of the matter that formed the Earth.  I won't go 
into the mathematics of it, since I am not sure of your calculus 
background, but the concept is to calculate the energy difference for each 
individual particle added to the Earth's surface as it collapses, and I 
assume a planet of constant density (I presume that's a first-order 
difference, and since we're just estimating anyway, that doesn't matter.)
Anyway, the energy is (3/5)*G*M*M/R, where G is the gravitational constant, 
M is the mass of the Earth, and R is the radius of the Earth.  This gives 
us 2.24*10^32 Joules, which is a huge amount.

Now, for the sake of argument, let's say that we have an object impacting 
the Earth at escape velocity.  (That is, the speed it would have if it was 
dropped from rest a very far distance away and permitted simply to fall 
onto the Earth.)  That's something like 11,000 meters per second, which is 
pretty fast.  How much mass would it need to have to produce the kind of 
energy we need in the collision to destroy the planet?
The calculation isn't too difficult.  The energy of the rock is G*M*m/R, 
where little m is the mass of the object, and the other variables are the 
same as before.  This needs to be equal to (3/5)*G*M*M/R in order to 
produce enough energy to destroy the Earth.  Solving the equation for m, 
we see: m=(3/5)*M.
So, the mass of the asteroid is just 3/5 times the mass of the Earth.  In 
other words, you need a planet or large moon to cause that kind of damage!

As for the Moon, the same calculation applies, so you would need something 
about 3/5 times the mass of the Moon to destroy it.  However, since the 
Moon itself is just over 1/85 the mass of the Earth (0.012 times the mass 
of the Earth), it could not cause the destruction of the Earth if it fell 
out of orbit. (It would certainly cause a lot of damage, though)

I should note that, in the early days of the Earth before it cooled, 
scientists believe that it was hit by a large object roughly the size of 
Mars (approx. 1/10 Earth's mass).  This removed a large part of the planet 
and put it into permanent orbit, where it eventually coalesced into the 
Moon.  Nevertheless, the rest of the planet fell back together to become 
Earth as we know it, rather than flying apart completely.  Needless to say, 
if there had been life at that time it would have been completely wiped 
out. :)

I hope this answers your question,

---Bob Macke
S.B. Physics, MIT, 1996
M.A. Physics, Washington University in St. Louis, 1999
 


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