Question

On August 10, 1972, a large meteorite skipped across the atmosphere above the western United States and western Canada, much like a stone skipped across water. The accompanying fireball was so bright that it could be seen in the daytime sky and was brighter than the usual meteorite trail. The meteorite's mass was about 4.4 × 106 kg; it's speed was about 14 km/s. Had it entered the atmosphere vertically, it would have hit Earth's surface with about the same speed. (a) Calculate the meteorite's loss of energy (as a positive number, in joules) that would have been associated with the vertical impact. (b) Express the energy as a multiple of the explosive energy of 1 megaton of TNT, which is 4.2 × 1015 J. (c) The energy associated with the atomic bomb explosion over Hiroshima was equivalent to 13 kilotons of TNT. To how many Hiroshima bombs would the meteorite impact have been equivalent?

Answer 1

(a)
`4.31\cdot 10^(14) J`

The energy lost by the meteorite is equal to its initial kinetic energy, since the meteorite comes to a stop, therefore it loses all its energy:

`E_(lost)=K=(1)/(2)mv^2`

where

`m=4.4\cdot 10^6 kg` is the mass of the meteorite

`v=14 km/s = 14000 m/s` is the meteorite's speed

Substituting into the formula, we find

`K=(1)/(2)(4.4\cdot 10^6 kg)(14000 m/s)^2=4.31\cdot 10^(14) J`

(b) 0.1 Megaton

1 Megaton (MT) corresponds to:

`1 MT = 4.2\cdot 10^(15)J`

So, we can find how many MT the energy of the meteorite corresponds to by using the following proportion:

`1 MT: 4.2 \cdot 10^(15) J = x : 4.31\cdot 10^(14) J`

Solving for x we find:

`x=(1 MT \cdot 4.31\cdot 10^(14)J)/(4.2\cdot 10^(15) J)=0.1 MT`

(c) 7.9 Hiroshima bombs

1 kiloton (kT) corresponds to 1/1000 of megaton, so:

`1 kT=(4.2\cdot 10^(15)J)/(1000)=4.2\cdot 10^(12) J`

As before, we can find how many kiloton the energy of the meteorite corresponds to by using the following proportion:

`1 kT: 4.2 \cdot 10^(12) J = x : 4.31\cdot 10^(14) J`

Solving for x we find:

`x=(1 kT \cdot 4.31\cdot 10^(14)J)/(4.2\cdot 10^(12) J)=103 kT`

The energy of the Hiroshima bomb was 13 kiloton, so the impact of the meteorite corresponds to:

`n=(103 kT)/(13 kT)=7.9`

so, to 7.9 Hiroshima bombs.