Answer to Question #97995 in Classical Mechanics for NILAKSHA GHOSH

Question

The energy delivered by an asteroid impact.

(a) A small asteroid weighs about 1012 kg. If its density is twice that of water, roughly what is the size (diameter) of the asteroid, in meters?

(b) The asteroid travels at a speed of about 10 km s−1 relative to the earth. How much kinetic energy (in Joules) would be released if the asteroid hit the earth?

(c) Convert the previous result to Calories.

(d) Suppose the asteroid lands in a roughly circular lake, about 10 km across and 10 m deep. If all the asteroid’s kinetic energy is converted into heat energy of the water in the lake, by how many degrees Centigrade will the lake warm-up?

(e) If the asteroid had been going twice as fast, how would this affect the amount of energy released on impact?

Answer 1

a)

$$m=2\rho V=2\rho \frac{4}{3}\pi r^3= \frac{1}{3}\pi \rho d^3$$

$$3(10^{12})=\pi (1000) d^3$$

$$d=985\ m$$

b)

$$K=0.5mv^2$$

$$K=0.5(10^{12})(10000)^2=5\cdot10^{19}J$$

c)

$$K=(0.238846)5\cdot10^{19}=1.2\cdot10^{19}cal$$

d)

$$K=Q=mc\Delta T=0.25\pi d^2hc\Delta T$$

$$5\cdot10^{19}=0.25\pi (10000)^2(10)(4200)\Delta T$$

$$\Delta T=1.5\cdot10^{7}\degree C$$

e) The amount of energy released on impact increases by 4 times:

$$K'=4K=20\cdot10^{19}J=2\cdot10^{20}J$$