Universe
Universe
11th Edition
ISBN: 9781319039448
Author: Robert Geller, Roger Freedman, William J. Kaufmann
Publisher: W. H. Freeman
bartleby

Videos

Question
Book Icon
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.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
Two resistors of resistances R1 and R2, with R2>R1, are connected to a voltage source with voltage V0. When the resistors are connected in series, the current is Is. When the resistors are connected in parallel, the current Ip from the source is equal to 10Is. Let r be the ratio R1/R2. Find r. I know you have to find the equations for V for both situations and relate them, I'm just struggling to do so. Please explain all steps, thank you.
Bheem and Ram, jump off either side of a bridge while holding opposite ends of a rope and swing back and forth under the bridge to save a child while avoiding a fire. Looking at the swing of just Bheem, we can approximate him as a simple pendulum with a period of motion of 5.59 s. How long is the pendulum ? When Bheem swings, he goes a full distance, from side to side, of 10.2 m.  What is his maximum velocity?  What is his maximum acceleration?
The position of a 0.300 kg object attached to a spring is described by x=0.271 m ⋅ cos(0.512π⋅rad/s ⋅t) (Assume t is in seconds.) Find the amplitude of the motion. Find the spring constant. Find the position of the object at t = 0.324 s. Find the object's velocity at t = 0.324 s.
Knowledge Booster
Background pattern image
Physics
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
Physics for Scientists and Engineers: Foundations...
Physics
ISBN:9781133939146
Author:Katz, Debora M.
Publisher:Cengage Learning
Text book image
The Solar System
Physics
ISBN:9781305804562
Author:Seeds
Publisher:Cengage
Text book image
Foundations of Astronomy (MindTap Course List)
Physics
ISBN:9781337399920
Author:Michael A. Seeds, Dana Backman
Publisher:Cengage Learning
Text book image
Stars and Galaxies (MindTap Course List)
Physics
ISBN:9781337399944
Author:Michael A. Seeds
Publisher:Cengage Learning
Text book image
The Solar System
Physics
ISBN:9781337672252
Author:The Solar System
Publisher:Cengage
Text book image
Stars and Galaxies
Physics
ISBN:9781305120785
Author:Michael A. Seeds, Dana Backman
Publisher:Cengage Learning
Kepler's Three Laws Explained; Author: PhysicsHigh;https://www.youtube.com/watch?v=kyR6EO_RMKE;License: Standard YouTube License, CC-BY