EBK LOOSE-LEAF VERSION OF UNIVERSE
EBK LOOSE-LEAF VERSION OF UNIVERSE
11th Edition
ISBN: 9781319227975
Author: KAUFMANN
Publisher: VST
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Chapter 7, Problem 12Q

(a)

To determine

The kinetic energy of an asteroid having radius 2 km with an average density of 2500 kg/m3 striking the Earth with 25 km/s.

(a)

Expert Solution
Check Mark

Answer to Problem 12Q

Solution:

The kinetic energy of the asteroid is found out to be 3.27×1015 J.

Explanation of Solution

Given data:

The velocity of the asteroid hitting the Earth is 25 m/s.

The radius of the asteroid is 2 km.

The density of the asteroid is 2500 kg/m3

Formula used:

The mass of the asteroid can be calculated by the following expression:

m=ρV

Here, m represents the mass of the asteroid, ρ represents the density of the asteroid, and V represents the volume of the asteroid.

The expression for the volume of a sphere is:

V=43πr3

Here, r represents the radius.

Conversion formula from kilometer to meter is:

1 km = 1000 m

The kinetic energy is calculated by the below expression:

K=12mv2

Here, K represents the kinetic energy, m represents the mass of the body, and v represents the velocity of the body.

Explanation:

Recall the expression for calculating the volume.

V=43πr3

Substitute 1 km for r and also use the conversion formula.

V=43π(1 km×(1000 m1 km))3=43π(1000 m)3

Recall the expression of mass.

m=ρV

Substitute 43π(1000 m)3 for V and 2500 kg/m3 for ρ.

m=ρV=ρ(43πr3)=(2500 kg/m3)(43π(1000 m)3)=1.047×1013 kg

The mass of asteroid is 1.047×1013 kg.

Recall the expression for the kinetic energy.

K=12mv2

Substitute 25 km/s for v, 1.047×1013 kg for m, and use the conversion formula:

K=12mv2=12(1.047×1013 kg)(25 km/s×1000 m1 km)2=3.27×1021 J

Conclusion:

Thus, the kinetic energy of the asteroid is found out to be 3.27×1021 J.

(b)

To determine

The comparison between the energy released by the impact of asteroid mentioned in sub-part (a) to the energy released by a 20-kiloton nuclear weapon, which was similar to the nuclear weapon dropped on Hiroshima.

(b)

Expert Solution
Check Mark

Answer to Problem 12Q

Solution:

The energy released during nuclear destruction is 40 times smaller than the energy released during the asteroid strike.

Explanation of Solution

Given data:

The energy released by the nuclear weapon is 20-kilotons.

Formula used:

1 kiloton of TNT emits 4.2×1012 J of energy.

The expression of energy released in joules is:

EJoules=EkN×(4.2×1012J)

Explanation:

From sub-part (a), the value of kinetic energy of the asteroid is 3.27×1015 J.

Recall the expression of energy released in joules.

EJoules=EkN×(4.2×1012J)

Substitute 20 kN for EkN.

EJoules=(20 kN)×(4.2×1012J)=8.4×1013 J

Refer to the value of energy released by the impact of an asteroid that is 3.27×1021 J.

The energy released in nuclear attack is 8.4×1013 J.

Taking the ratio of both the above energies,

3.27×1021 J8.4×1013 J=3.89×1074×107

Conclusion:

Therefore, the kinetic energy of the asteroid striking the Earth’s surface is 4×107 times greater than the energy released during the nuclear attack in Hiroshima.

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