Asteroids 175 km in size strike earth at random times, but on average once every

Question

Asteroids 175 km in size strike earth at random times, but on average once every 200,000,000 years. What are the odds that one will hit during the next 50 years?
Secondly, consider these asteroids strike at random locations. If the next asteroid strikes tomorrow and creates a crater 175 km in diameter what are the odds you will be somewhere in the crater zone when the asteroid hits? Asteroids 175 km in size strike earth at random times, but on average once every 200,000,000 years. What are the odds that one will hit during the next 50 years?
Secondly, consider these asteroids strike at random locations. If the next asteroid strikes tomorrow and creates a crater 175 km in diameter what are the odds you will be somewhere in the crater zone when the asteroid hits?
Secondly, consider these asteroids strike at random locations. If the next asteroid strikes tomorrow and creates a crater 175 km in diameter what are the odds you will be somewhere in the crater zone when the asteroid hits?

Explanation / Answer

First define some general terms.
L = angular momentum
I = moment of inertia
w = angular velocity

Angular momentum of the earth is its moment of inertia times its angular rotation rate.
Learth = Ie*we

The asteroid will also have an angular momentum given by the same formula.
Lasteroid = Ia*wa
Ia = m*r^2 ....... moment of inertia for the asteroid about the center of the earth
wa = v/r .......... angular velocity of the asteroid
Lasteroid = (m*r^2)(v/r) = m*v*r

Conservation of momentum says that hte momentum before collision must equal that of after. We will define W as the new angular rotation rate and ignore the mass of the asteroid for the moment of inertia after collision so it will be the same as before collision.
Lbefore = Learth + Lasteroid
Lafter = Ie*W
Ie*we + m*v*r = Ie*W

We can now get the formula for the fractional change in angular speed.
m*v*r = Ie*[W - we]
[W - we] = m*v*r/Ie
Fractional change = [W - we]/we = m*v*r/(le*we)

Now just plug in numbers:
M = mass of earth = 6x10^24 kg
r = radius of the earth = 6x10^6 m
P = period of rotation of the earth = 86400 seconds

Ie = (2/5)M*r^2 = (2/5) * 6x10^24 * (6x10^6)^2
= 8.64 * 10^37
we = 2*pi/P = 7.27x10^(-5)
Ie*we = 6.28x10^33

m = 6.0x10^20 kg
v = 3x10^4 m/s
m*v*r = 1.08x10^32

Fractional change =1.08x10^32 / 6.28x10^33
Fractional change = 0.01719

So, earth's day after collision would be

86400 - (86400 * 0.01719)
= 86400 - 1485.216 seconds
= 84914.784
= 23.58 hours

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