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Chapter 7: Problem 4

An explosion at ground level leaves a crater with a diameter that is proportional to the energy of the explosion raised to the \(\frac{1}{3}\) power; an explosion of 1 megaton of TNT leaves a crater with a \(1 \mathrm{~km}\) diameter. Below Lake Huron in Michigan there appears to be an ancient impact crater with a \(50 \mathrm{~km}\) diameter. What was the kinetic energy associated with that impact, in terms of (a) megatons of TNT (1 megaton yields \(4.2 \times 10^{15} \mathrm{~J}\) ) and (b) Hiroshima bomb equivalents (13 kilotons of TNT each)? (Ancient meteorite or comet impacts may have significantly altered the climate, killing off the dinosaurs and other life-forms.)

Short Answer

Expert verified
Kinetic energy was 125000 megatons of TNT or 9615385 Hiroshima bombs.

Step by step solution

01

Understand the Relationship

We are given that the diameter of the crater is proportional to the energy raised to the \(\frac{1}{3}\) power. This relationship can be expressed as \( D = k \cdot E^{1/3} \), where \( D \) is the diameter of the crater, \( k \) is a proportionality constant, and \( E \) is the energy of the explosion.
02

Determine the Proportionality Constant

Given an explosion of 1 megaton of TNT leaves a crater with a \(1 \mathrm{~km}\) diameter, we plug these values into the equation: \( 1 = k \cdot (1)^{1/3} \). Thus, \( k = 1 \). So the relationship simplifies to \( D = E^{1/3} \).
03

Calculate Energy for Lake Huron Crater

The ancient impact crater under Lake Huron has a \(50 \mathrm{~km}\) diameter. Using the equation \( 50 = E^{1/3} \), we solve for \( E \) by cubing both sides: \( E = 50^3 = 125000 \) megatons of TNT.
04

Convert Megatons to Joules

Since 1 megaton of TNT yields \(4.2 \times 10^{15} \mathrm{~J}\), we have \( 125000 \times 4.2 \times 10^{15} = 5.25 \times 10^{20} \mathrm{~J}\). This is the kinetic energy in joules.
05

Convert Megatons to Hiroshima Bomb Equivalents

1 Hiroshima bomb is equivalent to 13 kilotons of TNT, or \(0.013\) megatons. Therefore, the energy in Hiroshima bomb equivalents is \( \frac{125000}{0.013} \approx 9615385 \) Hiroshima bombs.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Proportionality Constant
Understanding the idea of a proportionality constant is crucial when dealing with equations that describe physical phenomena. In this context, the proportionality constant connects the size of an impact crater to the energy of the explosion that created it.
  • The formula connecting these two quantities is \( D = k \cdot E^{1/3} \).
  • Here, \( D \) represents the diameter of the crater, \( E \) denotes the energy, and \( k \) is the proportionality constant.
By knowing the diameter produced by a specific energy, we can determine \( k \). For the problem at hand, since a 1 megaton explosion results in a 1 km wide crater, we find \( k = 1 \). This simplifies the relationship to \( D = E^{1/3} \), giving a straightforward method to compute the energy from crater diameters.
Conversion of Energy Units
Converting energy units is a fundamental skill when handling different scientific and engineering problems. The energy of explosions is often measured in terms of TNT equivalents or joules, depending on the context.
  • A megaton of TNT is commonly used as a standard unit in impact sciences, equivalent to \( 4.2 \times 10^{15} \) joules.
  • This conversion allows for comparing and relating different energies using a recognizable and standardized metric.
In calculations, transitioning between TNT equivalents and joules makes it easier to express and communicate energy magnitudes clearly.
Impact Crater Analysis
Impact crater analysis is essential in understanding historical events through physical geological features. By analyzing crater sizes, we can infer the energy and scale of ancient impacts.
  • Using the equation \( D = E^{1/3} \), a crater with a 50 km diameter signals a massive energy release during its formation.
  • Solving for \( E \), the energy needed to produce such a large crater equates to \( 50^3 = 125000 \) megatons of TNT.
Events like these offer insight into past impacts that have significantly altered Earth's environment and life.
TNT Equivalents
The concept of TNT equivalents simplifies comparing different energy releases by providing a common reference point. A useful approach is measuring energy using TNT because it's a familiar benchmark.
  • One megaton of TNT (an explosive energy) provides \( 4.2 \times 10^{15} \) joules.
  • Using this conversion, we can calculate enormous energies in manageable terms, helping scientists and engineers effectively communicate ideas.
For instance, the impact crater in Lake Huron corresponds to an energy of \( 5.25 \times 10^{20} \) joules when assessed in TNT equivalents.
Meteorite Impact
Meteorite impacts have historically played a significant role in shaping Earth鈥檚 surface and environment. Studying them provides valuable knowledge about past climate changes and biological extinctions.
  • Events such as meteor impacts have been hypothesized to cause significant changes, such as the mass extinction that possibly wiped out the dinosaurs.
  • By calculating the energy of these impacts, we understand the potential devastations and their aftermath impacts.
The quality of analyses, like determining the ancient Lake Huron crater's formation energy, underscores the colossal effects meteorites once had on our planet.

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Most popular questions from this chapter

A single force acts on a \(3.0 \mathrm{~kg}\) particle-like object whose position is given by \(x=3.0 t-4.0 t^{2}+1.0 t^{3}\), with \(x\) in meters and \(t\) in seconds. Find the work done by the force from \(t=0\) to \(t=4.0 \mathrm{~s}\).

To push a \(25.0 \mathrm{~kg}\) crate up a frictionless incline, angled at \(25.0^{\circ}\) to the horizontal, a worker exerts a force of \(209 \mathrm{~N}\) parallel to the incline. As the crate slides \(1.50 \mathrm{~m}\), how much work is done on the crate by (a) the worker's applied force, (b) the gravitational force on the crate, and (c) the normal force exerted by the incline on the crate? (d) What is the total work done on the crate?

How much work is done by a force \(\vec{F}=(2 x \mathrm{~N}) \hat{\mathrm{i}}+(3 \mathrm{~N}) \hat{\mathrm{j}}\), with \(x\) in meters, that moves a particle from a position \(\vec{r}_{i}=\) \((2 \mathrm{~m}) \hat{\mathrm{i}}+(3 \mathrm{~m}) \hat{\mathrm{j}}\) to a position \(\vec{r}_{f}=-(4 \mathrm{~m}) \hat{\mathrm{i}}-(3 \mathrm{~m}) \hat{\mathrm{j}}\) ?

A force \(\vec{F}=(2.00 \hat{1}+9.00 \hat{\mathrm{j}}+5.30 \hat{\mathrm{k}}) \mathrm{N}\) acts on a \(2.90 \mathrm{~kg}\) object that moves in time interval \(2.10 \mathrm{~s}\) from an initial position \(\vec{r}_{1}=(2.70 \hat{i}-2.90 \hat{j}+5.50 \hat{k}) \mathrm{m}\) to a final position \(\overrightarrow{\vec{r}}_{2}=\) \((-4.10 \hat{\mathrm{i}}+3.30 \hat{\mathrm{j}}+5.40 \mathrm{k}) \mathrm{m}\). Find (a) the work done on the object by the force in that time interval, (b) the average power due to the force during that time interval, and (c) the angle between vectors \(\vec{r}_{1}\) and \(\vec{r}_{2}\).

A force \(\vec{F}=(3.00 \mathrm{~N}) \hat{\mathrm{i}}+(7.00 \mathrm{~N}) \hat{\mathrm{j}}+(7.00 \mathrm{~N}) \hat{\mathrm{k}}\) acts on a \(2.00 \mathrm{~kg}\) mobile object that moves from an initial position of \(\vec{d}_{i}=(3.00 \mathrm{~m}) \hat{\mathrm{i}}-(2.00 \mathrm{~m}) \hat{\mathrm{j}}+(5.00 \mathrm{~m}) \hat{\mathrm{k}}\) to a final position of \(\vec{d}_{f}=-(5.00 \mathrm{~m}) \hat{\mathrm{i}}+(4.00 \mathrm{~m}) \hat{\mathrm{j}}+(7.00 \mathrm{~m}) \hat{\mathrm{k}}\) in \(4.00 \mathrm{~s}\). Find (a) the work done on the object by the force in the \(4.00 \mathrm{~s}\) interval, (b) the average power due to the force during that interval, and (c) the angle between vectors \(\vec{d}_{i}\) and \(\vec{d}_{f}\).
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