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Astronomers estimate that a 2.0-km-diameter asteroid collides with the Earth onc

ID: 1342654 • Letter: A

Question

Astronomers estimate that a 2.0-km-diameter asteroid collides with the Earth once every million years. The collision could pose a threat to life on Earth. Assume a spherical asteroid has a mass of 3240 kg for each cubic meter of volume and moves toward the Earth at 17.0 km/s .

a.How much destructive energy could be released when it embeds itself in the Earth?

b. For comparison, a nuclear bomb could release about 4.0×1016J. How many such bombs would have to explode simultaneously to release the destructive energy of the asteroid collision with the Earth?

PLEASE explain all steps and where numbers came from!!! I am having so much trouble with this problem

Thanks!!!

Explanation / Answer

kinetic energy to be released in Collison, KE = 1/2mv^2

volume of a regular sphere = 4/3*pi*r^3,
where r = 2000 /2 = 1000 m
density = 3300kg/m^3
mass, m = Density * Volume
m = 3300 *  4/3 * pi * 1000^3 Kg

K.E = 1/2 *m* (17*10^3)^2
K.E = 1/2 *3300 *  4/3 * pi * 1000^3* (17*10^3)^2 J
K.E = 1.997 *10^21 J

Destructive energy could be released, = 1.997 *10^21 J

(B)
No of Bombs needed to explode simulteneously which would release same energy = (1.997 *10^21) / (4.0*10^16)
No of Bombs needed to explode simulteneously which would release same energy = 49925

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