Grex Science Conference

View Responses

Grex Science Item 20: Hey, where'd you get that number?
Entered by russ on Fri Mar 14 03:10:28 UTC 1997:

When you're having a discussion, it's not usually appropriate to fill your
page with equations, or ask people exactly how they arrived at a figure.
So this is the item for showing your work, or asking other people to.

6 responses total.

#1 of 6 by russ on Fri Mar 14 05:52:55 1997:

How I got the 4000 mile figure in item 19:
1 day = 24 hours
6 months = 182 days.
182 * 24 = 4368.
So, at 1 mile per hour, an object travels over 4000 miles in 6 months.

#2 of 6 by russ on Fri Mar 14 05:52:56 1997:

How I got the 400 lb of TNT fgure in item 19 (which is an under-estimate):
20 pounds is roughly 9 kilograms.  Escape velocity at Earth's surface
is about 11 kilometers per second, while orbital velocity is about 8
kilometers per second.  (I did the first number on the back of a virtual
envelope, whose bits I have misplaced, and I wrongly used 7 km/sec.)
Assuming impact at right angles, the sum of the squares of the velocities
is 185 km^2/sec^2 (original figure: 170 km^2/sec^2).  Dividing by 2 and
multiplying by the mass gives 832 (765) megajoules.  TNT has about 4.2 MJ
per kilogram of explosive energy, so the impact energy is roughly equal
to 198 kg/435 lb (182 kg/400 lb) of TNT.

#3 of 6 by russ on Sat Mar 15 03:44:08 1997:

How I got the 4000 mile figure in item 19:
1 day = 24 hours
6 months = 182 days.
182 * 24 = 4368.
So, at 1 mile per hour, an object travels over 4000 miles in 6 months.

#4 of 6 by russ on Sat Mar 15 03:44:09 1997:

How I got the 400 lb of TNT fgure in item 19 (which is an under-estimate):
20 pounds is roughly 9 kilograms.  Escape velocity at Earth's surface
is about 11 kilometers per second, while orbital velocity is about 8
kilometers per second.  (I did the first number on the back of a virtual
envelope, whose bits I have misplaced, and I wrongly used 7 km/sec.)
Assuming impact at right angles, the sum of the squares of the velocities
is 185 km^2/sec^2 (original figure: 170 km^2/sec^2).  Dividing by 2 and
multiplying by the mass gives 832 (765) megajoules.  TNT has about 4.2 MJ
per kilogram of explosive energy, so the impact energy is roughly equal
to 198 kg/435 lb (182 kg/400 lb) of TNT.

#5 of 6 by russ on Sat Mar 15 20:18:53 1997:

Here's the back of the envelope stuff I promised in item 19.
How much can you move an asteroid with a megaton?
 
First, what's a megaton?  It's the explosive energy of a million tons
of TNT.  The way the figure appears to be derived is to use metric
tons, assume that TNT yields 1000 calories/gram when it detonates,
and round the conversion factor of 4.184 joules/calorie to 4.2.
So, 10^6 tons * 10^6 g/ton * 1000 cal/g * 4.2 J/cal = 4.2*10^15
Joules per megaton.  In CGS units, it's 4.2*10^22 ergs.
 
From what I hear, the energy from a nuclear blast radiates largely
as X-rays.  X-rays are absorbed fairly well by most heavier elements.
The rest of the energy comes out as gamma rays, neutrons, and hot
plasma.  Almost all of these are going to be soaked up by the first
couple meters or so of the surface; the gammas and neutrons will go
deep, the X-rays will stop in the first inch or three, and the
plasma (which arrives late) will mix with the material evaporated
by the X-rays.
 
It's time to make some assumptions.  Let us assume:
 
-       A 1-km spherical object with a density of 2.5 grams/cc.
        The total mass is 1.31 billion tons.
-       A 1-megaton explosion 500 meters from the surface.
-       The absorbed energy evaporates a layer 15 cm thick and
        the expanding material converts 30% of the thermal
        energy to kinetic energy.
 
From 500 meters away from its surface, a 500-meter radius sphere 
appears to be 60 degrees (pi/3 radians) across.  If I haven't messed
up my integration, it covers 6.7% of the total sphere visible from the
point of the explosion.  So, of the 4.2*10^15 joules of energy from
the bomb, 2.8*10^14 joules hit the target.
 
The area of the sphere from which the bomb is visible extends to 60
degrees on either side of center, or 1/4 of its total area.  The total
area is 7.85*10^5 square meters.  The evaporated material goes 15 cm
(0.15 meters) deep and has a mass of 375 kilograms/m^2, for a total
evaporated mass of 2.9*10^8 kilograms (290 thousand tons).
 
30% of the 2.8*10^14 joules of absorbed energy, or 1.7*10^13 joules,
goes into the expansion of the evaporated material.  E = 1/2 m v^2, so:
 
v = sqrt( 2 * E / M ) = sqrt( 2 * 1.7*10^13 kg-m^2/sec^2 / 2.9*10^8 kg )
  = sqrt( 1.17*10^5 m^2/sec^2 )
  = 340 m/sec
 
Momentum = M * v = 2.9*10^8 * 3.4*10^2 = 9.9*10^10 kg-m/sec^2.
 
Not all the material goes off in the same direction, so the net momentum
transferred back to the rock is some fraction of that, greater than 1/2
and less than 1.  Call it 3/4.  Transferring that much momentum to an
object of 1.31*10^12 kilograms changes its velocity by a whopping...
 
        ... 0.056 meters per second (about 2.2 inches/second).
 
One MPH is about 1.5 feet per second.  You'd need about 9 1-megaton
bombs, according to this model, to change the baseline 1 kilometer wide
asteroid's speed by that much.  You could put the bombs closer, but that
would apply larger forces to less of the surface, risking a fracture
into smaller pieces which might be more hazardous.  A single, larger bomb
might do the job.
 
If anyone has corrections or other models, I'd appreciate seeing them.

#6 of 6 by russ on Tue Mar 18 03:46:31 1997:

In #5 I made a couple errors and omissions.  I omitted the breakdown of
some units in the calculation of v:  the units "kg-m^2/sec^2" are equal
to joules.  In the calculation of momentum, I mis-stated the units.  The
units are not kg-m/sec^2, they are kg-m/sec (kilogram-meters per second).
(Momentum is the product of mass and velocity and is expressed in those units.)

Response not possible - You must register and login before posting.