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Starship Troopers / The Moon Is A Harsh Mistress / Time Enough For Love
1297 pages, Hardcover
First published January 1, 1991
308 people want to read
About the author
Robert A. Heinlein
1,053 books10.5k followersRobert Anson Heinlein was an American science fiction author, aeronautical engineer, and naval officer. Sometimes called the "dean of science fiction writers", he was among the first to emphasize scientific accuracy in his fiction, and was thus a pioneer of the subgenre of hard science fiction. His published works, both fiction and non-fiction, express admiration for competence and emphasize the value of critical thinking. His plots often posed provocative situations which challenged conventional social mores. His work continues to have an influence on the science-fiction genre, and on modern culture more generally.
Heinlein became one of the first American science-fiction writers to break into mainstream magazines such as The Saturday Evening Post in the late 1940s. He was one of the best-selling science-fiction novelists for many decades, and he, Isaac Asimov, and Arthur C. Clarke are often considered the "Big Three" of English-language science fiction authors. Notable Heinlein works include Stranger in a Strange Land, Starship Troopers (which helped mold the space marine and mecha archetypes) and The Moon Is a Harsh Mistress. His work sometimes had controversial aspects, such as plural marriage in The Moon Is a Harsh Mistress, militarism in Starship Troopers and technologically competent women characters who were formidable, yet often stereotypically feminine—such as Friday.
Heinlein used his science fiction as a way to explore provocative social and political ideas and to speculate how progress in science and engineering might shape the future of politics, race, religion, and sex. Within the framework of his science-fiction stories, Heinlein repeatedly addressed certain social themes: the importance of individual liberty and self-reliance, the nature of sexual relationships, the obligation individuals owe to their societies, the influence of organized religion on culture and government, and the tendency of society to repress nonconformist thought. He also speculated on the influence of space travel on human cultural practices.
Heinlein was named the first Science Fiction Writers Grand Master in 1974. Four of his novels won Hugo Awards. In addition, fifty years after publication, seven of his works were awarded "Retro Hugos"—awards given retrospectively for works that were published before the Hugo Awards came into existence. In his fiction, Heinlein coined terms that have become part of the English language, including grok, waldo and speculative fiction, as well as popularizing existing terms like "TANSTAAFL", "pay it forward", and "space marine". He also anticipated mechanical computer-aided design with "Drafting Dan" and described a modern version of a waterbed in his novel Beyond This Horizon.
Also wrote under Pen names: Anson McDonald, Lyle Monroe, Caleb Saunders, John Riverside and Simon York.
Heinlein became one of the first American science-fiction writers to break into mainstream magazines such as The Saturday Evening Post in the late 1940s. He was one of the best-selling science-fiction novelists for many decades, and he, Isaac Asimov, and Arthur C. Clarke are often considered the "Big Three" of English-language science fiction authors. Notable Heinlein works include Stranger in a Strange Land, Starship Troopers (which helped mold the space marine and mecha archetypes) and The Moon Is a Harsh Mistress. His work sometimes had controversial aspects, such as plural marriage in The Moon Is a Harsh Mistress, militarism in Starship Troopers and technologically competent women characters who were formidable, yet often stereotypically feminine—such as Friday.
Heinlein used his science fiction as a way to explore provocative social and political ideas and to speculate how progress in science and engineering might shape the future of politics, race, religion, and sex. Within the framework of his science-fiction stories, Heinlein repeatedly addressed certain social themes: the importance of individual liberty and self-reliance, the nature of sexual relationships, the obligation individuals owe to their societies, the influence of organized religion on culture and government, and the tendency of society to repress nonconformist thought. He also speculated on the influence of space travel on human cultural practices.
Heinlein was named the first Science Fiction Writers Grand Master in 1974. Four of his novels won Hugo Awards. In addition, fifty years after publication, seven of his works were awarded "Retro Hugos"—awards given retrospectively for works that were published before the Hugo Awards came into existence. In his fiction, Heinlein coined terms that have become part of the English language, including grok, waldo and speculative fiction, as well as popularizing existing terms like "TANSTAAFL", "pay it forward", and "space marine". He also anticipated mechanical computer-aided design with "Drafting Dan" and described a modern version of a waterbed in his novel Beyond This Horizon.
Also wrote under Pen names: Anson McDonald, Lyle Monroe, Caleb Saunders, John Riverside and Simon York.
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January 6, 2019
It got better.
March 16, 2023
[Chat's ability to do physics has improved considerably in the Feb 13 release. Here's an example. Trigger warnings: contains 60s-style sexism and algebra.]
mannyrayner: I have another little physics problem. First, are you familiar with Robert Heinlein's novel "The Moon is a Harsh Mistress"?
ChatGPT: Yes, I'm familiar with Robert Heinlein's novel "The Moon is a Harsh Mistress." It's a science fiction novel that explores themes of revolution, politics, and lunar colonization.
mannyrayner: Okay, now there is an incident which occurs early on in the book, and I'm wondering about its physical plausibility. The narrator has just med Wyoming Knott, a woman he later falls in love with and marries, and he's said something that annoys her a great deal, so much that she stamps her foot. They are on the Moon, and in the book this causes her to sail into the air.
Making reasonable assumptions (Ms Knott is described as tall, slim, athletic, and so angry that she stamps as hard as she can; she is in Lunar gravity), would she really leave the ground? If so, approximately how high would she ascend, and how long would it take before she touched ground again?
ChatGPT: Assuming that Wyoming Knott is a tall, slim, athletic woman and that she stamps her foot with enough force, it is plausible that she could leave the ground on the Moon, which has a surface gravity that is approximately 1/6th that of Earth's.
To estimate how high she would ascend, we can use the conservation of energy. When Wyoming Knott stamps her foot on the lunar surface, she imparts a certain amount of kinetic energy to herself, which is converted into potential energy as she rises into the air. The maximum height she reaches would be the point at which all of her initial kinetic energy is converted into potential energy at the top of her trajectory. At this point, her velocity would be zero.
The kinetic energy of Wyoming's foot-stamp can be estimated as:
KE = (1/2) * m * v^2
where m is the mass of Wyoming's foot, and v is the velocity of her foot-stamp. Assuming a mass of 0.5 kg for Wyoming's foot and a velocity of 1 m/s for her foot-stamp (which is a reasonable estimate based on the description in the book), the kinetic energy imparted by the foot-stamp would be approximately 0.25 J.
This kinetic energy is converted into potential energy as Wyoming rises into the air. The potential energy of an object at a certain height above the surface of the Moon can be calculated as:
PE = m * g * h
where m is the mass of the object, g is the acceleration due to gravity on the Moon (1.62 m/s^2), and h is the height above the surface.
Setting the kinetic energy equal to the potential energy and solving for h, we get:
h = KE / (m * g) = 0.25 J / (0.5 kg * 1.62 m/s^2) = 0.15 meters
Therefore, Wyoming would ascend to a maximum height of approximately 0.15 meters (or 15 centimeters) above the lunar surface.
To estimate how long it would take for her to touch the ground again, we can use the kinematic equation:
h = (1/2) * g * t^2
where h is the height, g is the acceleration due to gravity on the Moon, and t is the time.
Solving for t, we get:
t = sqrt(2h/g) = sqrt(2 * 0.15 m / 1.62 m/s^2) = 0.36 seconds
Therefore, it would take Wyoming approximately 0.36 seconds to touch the ground again after she stamps her foot and rises into the air on the Moon.
mannyrayner: The calculation seems correct in principle! But when calculating the energy, should you not use the mass of her whole leg, rather than just her foot?
ChatGPT: You are correct, it would be more accurate to use the mass of her entire leg instead of just her foot when calculating the kinetic energy. Assuming a mass of 5 kg for her entire leg, the kinetic energy would be:
KE = (1/2) * m * v^2 = (1/2) * 5 kg * (1 m/s)^2 = 2.5 J
Using this value instead, the maximum height Wyoming would reach would be:
h = KE / (m * g) = 2.5 J / (5 kg * 1.62 m/s^2) = 0.31 meters
So she would ascend to a maximum height of approximately 0.31 meters (or 31 centimeters) above the lunar surface. The time it would take for her to touch the ground again would still be approximately 0.36 seconds, calculated using the same kinematic equation.
mannyrayner: Could you show me the details for the calculation of the time?
ChatGPT: Sure! To calculate the time it takes for Wyoming to fall back to the surface of the Moon, we can use the kinematic equation:
h = (1/2) * g * t^2
where h is the height, g is the acceleration due to gravity on the Moon (1.62 m/s^2), and t is the time. Rearranging this equation to solve for t gives:
t = sqrt(2h / g)
Plugging in the maximum height Wyoming reaches (0.31 meters) and the acceleration due to gravity on the Moon (1.62 m/s^2) gives:
t = sqrt(2(0.31 m) / (1.62 m/s^2)) = 0.36 seconds
So it would take approximately 0.36 seconds for Wyoming to fall back to the surface of the Moon after she jumps up.
[You can evidently nitpick this a good deal. 1 m/s seems like a conservative estimate for how fast Wyoming moves her leg; some of the kinetic energy will be lost; and how can the total time off the ground be the same in the two versions of the calculation? All the same, Chat seems to have a good grasp of the basic issues.
I was surprised it didn't say anything about the sexism. But maybe it's also a fan of this classic SF novel, where the real hero is the AI.]
__________________
[Update, Mar 14 2023]
I was curious to see how well DALL-E would succeed at illustrating this scene. Here's what it came up with in response to the prompt "A cartoon in the style of a Marvel comic. A beautiful woman in a tight-fitting dress with a flared skirt is indoors in a room on the Moon. She angrily stamps her foot, causing her to fly half a meter into the air. To one side, we see a man with Hispanic features laughing at her."
__________________
[Update, Mar 15 2023. I just signed up for the new GPT-4 based release of ChatGPT and tried the problem again. Although Chat still finds the scenario quite challenging, it seems to me that the new version is considerably better.
What impressed me most is that at one point it carries out a calculation, arrives at a physically absurd result (Wyoming bounces over a hundred feet into the air), and of its own accord decides it has made a mistake and goes back to the beginning. I have never seen it do this kind of thing before.
Here's the dialogue from the point where Chat starts responding to the question.]
ChatGPT: To determine if it's physically reasonable for Wyoming to bounce into the air and estimate the height she might reach, let's consider some factors and make a few assumptions.
Assumptions:
Lunar gravity is about 1/6th of Earth's gravity, which is roughly 1.625 m/s².
Wyoming is slim and athletic, so let's assume she weighs 60 kg on Earth.
She stamps her foot with maximum force.
To find the height she might reach, we'll use the following kinematic equation:
h = (v² - u²) / 2a
Where:
h is the height
v is the final velocity (0 m/s at the peak of her jump)
u is the initial velocity (which we need to find)
a is the acceleration due to gravity (1.625 m/s² in lunar gravity)
First, we need to estimate the force she generates while stamping her foot. This is difficult to estimate precisely, as it would depend on the specifics of her muscle strength and how hard she's stamping. However, let's assume she can generate a force equivalent to 3 times her body weight on Earth while stamping her foot down (a reasonable assumption for a strong, athletic person).
Force = mass * acceleration
Force = 60 kg * 9.81 m/s² (Earth's gravity)
Force = 588.6 N (Newtons)
Now, considering she generates 3 times this force:
Force_stamping = 3 * 588.6 N
Force_stamping = 1,765.8 N
Now we need to find the net force acting on her in lunar gravity:
Net_force = Force_stamping - (mass_lunar * lunar_gravity)
mass_lunar = 60 kg / 6 (Since weight is 1/6th on the Moon)
mass_lunar = 10 kg
Net_force = 1,765.8 N - (10 kg * 1.625 m/s²)
Net_force = 1,765.8 N - 16.25 N
Net_force = 1,749.55 N
Now, we can find the initial velocity (u) using Newton's second law:
Net_force = mass_lunar * acceleration_due_to_stamping
1,749.55 N = 10 kg * acceleration_due_to_stamping
acceleration_due_to_stamping = 174.955 m/s²
Now, we can find the time it takes for her to reach maximum height:
a_total = acceleration_due_to_stamping - lunar_gravity
a_total = 174.955 m/s² - 1.625 m/s²
a_total = 173.33 m/s²
t = (v - u) / a_total
t = (0 - u) / 173.33 m/s²
u = -173.33 m/s² * t (We need to find 't')
We can use another equation of motion to find 't':
s = ut + (1/2)at²
s = 0 (Since she starts and ends at the ground)
0 = -173.33t + 0.5(173.33)t
t = 2 (approximately)
Now we can find 'u':
u = -173.33 m/s² * 2
u = -346.66 m/s
Finally, we can find the height she reaches (h):
h = (v² - u²) / 2a
h = (0 - (-346.66 m/s)²) / 2(-1.625 m/s²)
h = (0 - 120,000 m²
mannyrayner:Please continue
ChatGPT: I apologize for the interruption. Continuing the calculation:
h = (0 - 120,000 m²/s²) / 2(-1.625 m/s²)
h = 120,000 m²/s² / 3.25 m/s²
h ≈ 36,923 m
This result is not physically reasonable, as it suggests Wyoming would bounce to an extreme height, which is not what the book implies. There was an error in my previous calculations. Let's reassess the situation.
We should look at the problem from an energy perspective. We can assume that the energy she imparts to the ground is equal to the potential energy she gains at the peak of her jump. We can use the following equation for potential energy:
PE = m * g * h
Where:
PE is the potential energy
m is the mass (in lunar gravity, which is 10 kg)
g is the lunar gravity (1.625 m/s²)
h is the height
We still need to estimate the force she generates while stamping her foot. Let's use a more reasonable assumption, such as she generates a force equivalent to 6 times her body weight on the Moon while stamping her foot down.
Force_stamping = 6 * 10 kg * 1.625 m/s²
Force_stamping = 97.5 N
The work done (energy) by Wyoming while stamping her foot can be calculated as:
Work = Force_stamping * distance
Let's assume that the distance her foot moves during the stamping action is about 0.1 meters:
Work = 97.5 N * 0.1 m
Work = 9.75 J (Joules)
Now we can equate this work to the potential energy she gains at the peak of her jump:
PE = 9.75 J
m * g * h = 9.75 J
We can solve for height (h):
h = 9.75 J / (10 kg * 1.625 m/s²)
h ≈ 0.6 m
So, using these assumptions, it is physically reasonable for Wyoming to bounce into the air when she stamps her foot in lunar gravity, and she could potentially rise about 0.6 meters (about 2 feet) above the ground.
mannyrayner: Your second version is as you say much better than the first one, and it arrives at a result similar to the one in the book, where we read that Wyoming bounces "half a meter" into the air. However, I think I see some issues with your calculation:
1. You estimate
Force_stamping = 6 * 10 kg * 1.625 m/s²
based on the assumption that Wyoming stamps down with six times her lunar body weight. But should this not be 6 * 60 kg * 1.625 m/s², i.e. 6 * (her mass) * (lunar gravity)?
2. You say "Let's assume that the distance her foot moves during the stamping action is about 0.1 meters". But in fact her foot will be stationary on the ground as she stamps down. What is moving is surely her body, as her leg straightens up by flexing at the knee, and you need to estimate how much it will rise during the process of straightening her leg?
ChatGPT: You're correct, there are some issues with my previous calculation. Let's address those and revise the calculation.
You're right that I should have used her Earth mass (60 kg) instead of her lunar mass (10 kg) for the force calculation. Let's correct that:
Force_stamping = 6 * 60 kg * 1.625 m/s²
Force_stamping = 585 N
Indeed, I made an incorrect assumption about the distance her foot moves during the stamping action. Instead, let's consider the distance her center of mass moves upward as her leg straightens.
Let's assume her leg length is about 1 meter, and when she stamps her foot, her leg straightens and her center of mass rises by 0.3 meters (a rough estimate considering her leg's straightening action).
Now, we can calculate the work done (energy) by Wyoming while stamping her foot:
Work = Force_stamping * distance
Work = 585 N * 0.3 m
Work = 175.5 J (Joules)
We can equate this work to the potential energy she gains at the peak of her jump:
PE = 175.5 J
m_lunar * g * h = 175.5 J
Where:
m_lunar is her mass in lunar gravity (10 kg)
g is the lunar gravity (1.625 m/s²)
h is the height
We can solve for height (h):
h = 175.5 J / (10 kg * 1.625 m/s²)
h ≈ 10.8 m
This result, while still higher than what's described in the book, is more reasonable than the previous attempt. However, it's important to note that these calculations are based on rough estimates and assumptions, which could impact the final result.
In reality, the energy transfer might not be as efficient, and some energy might be lost to internal deformation or other factors, resulting in a lower height. Additionally, the choice of 6 times her body weight for the stamping force is somewhat arbitrary, and different values could yield different results.
Overall, it's plausible for Wyoming to bounce into the air when she stamps her foot in lunar gravity, but the exact height might differ based on the assumptions made.
mannyrayner: A couple of comments:
1. You estimate that her center of mass will rise by 0.3 m. But this seems like far too much. When she begins stamping down, she will be supported by her other leg, i.e. the one that is not stamping, and this will only be slightly flexed if she stamps in a normal way. So her stamping leg can also only be slightly flexed, and her center of mass can rise a smaller distance. It seems necessary to estimate how much her legs are flexed at the start of the stamp, i.e. the angle subtended at the knee joint.
2. When estimating the potential energy at the peak of her jump, m_lunar * g * h, note that m_lunar is her mass rather than her weight, so it is still 60 kg rather than 10 kg.
ChatGPT: You're right, I made some errors in my previous response. Let's address those and revise the calculations once again.
Estimating the distance her center of mass rises during the stamping action: You're correct that 0.3 meters might be too much. Let's assume that at the start of the stamp, her leg is flexed by about 15 degrees. In this case, the distance her center of mass rises can be estimated using the following formula:
distance = leg_length * (1 - cos(angle))
Assuming a leg length of 1 meter and an angle of 15 degrees:
distance = 1 * (1 - cos(15°))
distance ≈ 0.034 m
You're correct that I should use her actual mass (60 kg) rather than her lunar "weight" (10 kg) when calculating the potential energy. Let's correct that as well.
Now, we can calculate the work done (energy) by Wyoming while stamping her foot:
Work = Force_stamping * distance
Work = 585 N * 0.034 m
Work ≈ 19.89 J (Joules)
We can equate this work to the potential energy she gains at the peak of her jump:
PE = 19.89 J
m * g * h = 19.89 J
Where:
m is her mass (60 kg)
g is the lunar gravity (1.625 m/s²)
h is the height
We can solve for height (h):
h = 19.89 J / (60 kg * 1.625 m/s²)
h ≈ 0.205 m
This result, a height of approximately 0.205 meters (about 8 inches), is much closer to the "half a meter" described in the book. Please keep in mind that these calculations are based on rough estimates and assumptions, and the actual height might differ based on various factors.
mannyrayner: This calculation looks much better to me. As you say, some of the quantities are difficult to estimate.
If you don't mind, I will share your calculations with my physics teacher friend Thomas. He is better than I am at finding errors in this kind of thing!
ChatGPT: I'm glad you find the latest calculation more reasonable. Please feel free to share the calculations with your physics teacher friend, Thomas. I appreciate any feedback or additional insights that may improve the accuracy of the estimation.
mannyrayner: I have another little physics problem. First, are you familiar with Robert Heinlein's novel "The Moon is a Harsh Mistress"?
ChatGPT: Yes, I'm familiar with Robert Heinlein's novel "The Moon is a Harsh Mistress." It's a science fiction novel that explores themes of revolution, politics, and lunar colonization.
mannyrayner: Okay, now there is an incident which occurs early on in the book, and I'm wondering about its physical plausibility. The narrator has just med Wyoming Knott, a woman he later falls in love with and marries, and he's said something that annoys her a great deal, so much that she stamps her foot. They are on the Moon, and in the book this causes her to sail into the air.
Making reasonable assumptions (Ms Knott is described as tall, slim, athletic, and so angry that she stamps as hard as she can; she is in Lunar gravity), would she really leave the ground? If so, approximately how high would she ascend, and how long would it take before she touched ground again?
ChatGPT: Assuming that Wyoming Knott is a tall, slim, athletic woman and that she stamps her foot with enough force, it is plausible that she could leave the ground on the Moon, which has a surface gravity that is approximately 1/6th that of Earth's.
To estimate how high she would ascend, we can use the conservation of energy. When Wyoming Knott stamps her foot on the lunar surface, she imparts a certain amount of kinetic energy to herself, which is converted into potential energy as she rises into the air. The maximum height she reaches would be the point at which all of her initial kinetic energy is converted into potential energy at the top of her trajectory. At this point, her velocity would be zero.
The kinetic energy of Wyoming's foot-stamp can be estimated as:
KE = (1/2) * m * v^2
where m is the mass of Wyoming's foot, and v is the velocity of her foot-stamp. Assuming a mass of 0.5 kg for Wyoming's foot and a velocity of 1 m/s for her foot-stamp (which is a reasonable estimate based on the description in the book), the kinetic energy imparted by the foot-stamp would be approximately 0.25 J.
This kinetic energy is converted into potential energy as Wyoming rises into the air. The potential energy of an object at a certain height above the surface of the Moon can be calculated as:
PE = m * g * h
where m is the mass of the object, g is the acceleration due to gravity on the Moon (1.62 m/s^2), and h is the height above the surface.
Setting the kinetic energy equal to the potential energy and solving for h, we get:
h = KE / (m * g) = 0.25 J / (0.5 kg * 1.62 m/s^2) = 0.15 meters
Therefore, Wyoming would ascend to a maximum height of approximately 0.15 meters (or 15 centimeters) above the lunar surface.
To estimate how long it would take for her to touch the ground again, we can use the kinematic equation:
h = (1/2) * g * t^2
where h is the height, g is the acceleration due to gravity on the Moon, and t is the time.
Solving for t, we get:
t = sqrt(2h/g) = sqrt(2 * 0.15 m / 1.62 m/s^2) = 0.36 seconds
Therefore, it would take Wyoming approximately 0.36 seconds to touch the ground again after she stamps her foot and rises into the air on the Moon.
mannyrayner: The calculation seems correct in principle! But when calculating the energy, should you not use the mass of her whole leg, rather than just her foot?
ChatGPT: You are correct, it would be more accurate to use the mass of her entire leg instead of just her foot when calculating the kinetic energy. Assuming a mass of 5 kg for her entire leg, the kinetic energy would be:
KE = (1/2) * m * v^2 = (1/2) * 5 kg * (1 m/s)^2 = 2.5 J
Using this value instead, the maximum height Wyoming would reach would be:
h = KE / (m * g) = 2.5 J / (5 kg * 1.62 m/s^2) = 0.31 meters
So she would ascend to a maximum height of approximately 0.31 meters (or 31 centimeters) above the lunar surface. The time it would take for her to touch the ground again would still be approximately 0.36 seconds, calculated using the same kinematic equation.
mannyrayner: Could you show me the details for the calculation of the time?
ChatGPT: Sure! To calculate the time it takes for Wyoming to fall back to the surface of the Moon, we can use the kinematic equation:
h = (1/2) * g * t^2
where h is the height, g is the acceleration due to gravity on the Moon (1.62 m/s^2), and t is the time. Rearranging this equation to solve for t gives:
t = sqrt(2h / g)
Plugging in the maximum height Wyoming reaches (0.31 meters) and the acceleration due to gravity on the Moon (1.62 m/s^2) gives:
t = sqrt(2(0.31 m) / (1.62 m/s^2)) = 0.36 seconds
So it would take approximately 0.36 seconds for Wyoming to fall back to the surface of the Moon after she jumps up.
[You can evidently nitpick this a good deal. 1 m/s seems like a conservative estimate for how fast Wyoming moves her leg; some of the kinetic energy will be lost; and how can the total time off the ground be the same in the two versions of the calculation? All the same, Chat seems to have a good grasp of the basic issues.
I was surprised it didn't say anything about the sexism. But maybe it's also a fan of this classic SF novel, where the real hero is the AI.]
__________________
[Update, Mar 14 2023]
I was curious to see how well DALL-E would succeed at illustrating this scene. Here's what it came up with in response to the prompt "A cartoon in the style of a Marvel comic. A beautiful woman in a tight-fitting dress with a flared skirt is indoors in a room on the Moon. She angrily stamps her foot, causing her to fly half a meter into the air. To one side, we see a man with Hispanic features laughing at her."
__________________
[Update, Mar 15 2023. I just signed up for the new GPT-4 based release of ChatGPT and tried the problem again. Although Chat still finds the scenario quite challenging, it seems to me that the new version is considerably better.
What impressed me most is that at one point it carries out a calculation, arrives at a physically absurd result (Wyoming bounces over a hundred feet into the air), and of its own accord decides it has made a mistake and goes back to the beginning. I have never seen it do this kind of thing before.
Here's the dialogue from the point where Chat starts responding to the question.]
ChatGPT: To determine if it's physically reasonable for Wyoming to bounce into the air and estimate the height she might reach, let's consider some factors and make a few assumptions.
Assumptions:
Lunar gravity is about 1/6th of Earth's gravity, which is roughly 1.625 m/s².
Wyoming is slim and athletic, so let's assume she weighs 60 kg on Earth.
She stamps her foot with maximum force.
To find the height she might reach, we'll use the following kinematic equation:
h = (v² - u²) / 2a
Where:
h is the height
v is the final velocity (0 m/s at the peak of her jump)
u is the initial velocity (which we need to find)
a is the acceleration due to gravity (1.625 m/s² in lunar gravity)
First, we need to estimate the force she generates while stamping her foot. This is difficult to estimate precisely, as it would depend on the specifics of her muscle strength and how hard she's stamping. However, let's assume she can generate a force equivalent to 3 times her body weight on Earth while stamping her foot down (a reasonable assumption for a strong, athletic person).
Force = mass * acceleration
Force = 60 kg * 9.81 m/s² (Earth's gravity)
Force = 588.6 N (Newtons)
Now, considering she generates 3 times this force:
Force_stamping = 3 * 588.6 N
Force_stamping = 1,765.8 N
Now we need to find the net force acting on her in lunar gravity:
Net_force = Force_stamping - (mass_lunar * lunar_gravity)
mass_lunar = 60 kg / 6 (Since weight is 1/6th on the Moon)
mass_lunar = 10 kg
Net_force = 1,765.8 N - (10 kg * 1.625 m/s²)
Net_force = 1,765.8 N - 16.25 N
Net_force = 1,749.55 N
Now, we can find the initial velocity (u) using Newton's second law:
Net_force = mass_lunar * acceleration_due_to_stamping
1,749.55 N = 10 kg * acceleration_due_to_stamping
acceleration_due_to_stamping = 174.955 m/s²
Now, we can find the time it takes for her to reach maximum height:
a_total = acceleration_due_to_stamping - lunar_gravity
a_total = 174.955 m/s² - 1.625 m/s²
a_total = 173.33 m/s²
t = (v - u) / a_total
t = (0 - u) / 173.33 m/s²
u = -173.33 m/s² * t (We need to find 't')
We can use another equation of motion to find 't':
s = ut + (1/2)at²
s = 0 (Since she starts and ends at the ground)
0 = -173.33t + 0.5(173.33)t
t = 2 (approximately)
Now we can find 'u':
u = -173.33 m/s² * 2
u = -346.66 m/s
Finally, we can find the height she reaches (h):
h = (v² - u²) / 2a
h = (0 - (-346.66 m/s)²) / 2(-1.625 m/s²)
h = (0 - 120,000 m²
mannyrayner:Please continue
ChatGPT: I apologize for the interruption. Continuing the calculation:
h = (0 - 120,000 m²/s²) / 2(-1.625 m/s²)
h = 120,000 m²/s² / 3.25 m/s²
h ≈ 36,923 m
This result is not physically reasonable, as it suggests Wyoming would bounce to an extreme height, which is not what the book implies. There was an error in my previous calculations. Let's reassess the situation.
We should look at the problem from an energy perspective. We can assume that the energy she imparts to the ground is equal to the potential energy she gains at the peak of her jump. We can use the following equation for potential energy:
PE = m * g * h
Where:
PE is the potential energy
m is the mass (in lunar gravity, which is 10 kg)
g is the lunar gravity (1.625 m/s²)
h is the height
We still need to estimate the force she generates while stamping her foot. Let's use a more reasonable assumption, such as she generates a force equivalent to 6 times her body weight on the Moon while stamping her foot down.
Force_stamping = 6 * 10 kg * 1.625 m/s²
Force_stamping = 97.5 N
The work done (energy) by Wyoming while stamping her foot can be calculated as:
Work = Force_stamping * distance
Let's assume that the distance her foot moves during the stamping action is about 0.1 meters:
Work = 97.5 N * 0.1 m
Work = 9.75 J (Joules)
Now we can equate this work to the potential energy she gains at the peak of her jump:
PE = 9.75 J
m * g * h = 9.75 J
We can solve for height (h):
h = 9.75 J / (10 kg * 1.625 m/s²)
h ≈ 0.6 m
So, using these assumptions, it is physically reasonable for Wyoming to bounce into the air when she stamps her foot in lunar gravity, and she could potentially rise about 0.6 meters (about 2 feet) above the ground.
mannyrayner: Your second version is as you say much better than the first one, and it arrives at a result similar to the one in the book, where we read that Wyoming bounces "half a meter" into the air. However, I think I see some issues with your calculation:
1. You estimate
Force_stamping = 6 * 10 kg * 1.625 m/s²
based on the assumption that Wyoming stamps down with six times her lunar body weight. But should this not be 6 * 60 kg * 1.625 m/s², i.e. 6 * (her mass) * (lunar gravity)?
2. You say "Let's assume that the distance her foot moves during the stamping action is about 0.1 meters". But in fact her foot will be stationary on the ground as she stamps down. What is moving is surely her body, as her leg straightens up by flexing at the knee, and you need to estimate how much it will rise during the process of straightening her leg?
ChatGPT: You're correct, there are some issues with my previous calculation. Let's address those and revise the calculation.
You're right that I should have used her Earth mass (60 kg) instead of her lunar mass (10 kg) for the force calculation. Let's correct that:
Force_stamping = 6 * 60 kg * 1.625 m/s²
Force_stamping = 585 N
Indeed, I made an incorrect assumption about the distance her foot moves during the stamping action. Instead, let's consider the distance her center of mass moves upward as her leg straightens.
Let's assume her leg length is about 1 meter, and when she stamps her foot, her leg straightens and her center of mass rises by 0.3 meters (a rough estimate considering her leg's straightening action).
Now, we can calculate the work done (energy) by Wyoming while stamping her foot:
Work = Force_stamping * distance
Work = 585 N * 0.3 m
Work = 175.5 J (Joules)
We can equate this work to the potential energy she gains at the peak of her jump:
PE = 175.5 J
m_lunar * g * h = 175.5 J
Where:
m_lunar is her mass in lunar gravity (10 kg)
g is the lunar gravity (1.625 m/s²)
h is the height
We can solve for height (h):
h = 175.5 J / (10 kg * 1.625 m/s²)
h ≈ 10.8 m
This result, while still higher than what's described in the book, is more reasonable than the previous attempt. However, it's important to note that these calculations are based on rough estimates and assumptions, which could impact the final result.
In reality, the energy transfer might not be as efficient, and some energy might be lost to internal deformation or other factors, resulting in a lower height. Additionally, the choice of 6 times her body weight for the stamping force is somewhat arbitrary, and different values could yield different results.
Overall, it's plausible for Wyoming to bounce into the air when she stamps her foot in lunar gravity, but the exact height might differ based on the assumptions made.
mannyrayner: A couple of comments:
1. You estimate that her center of mass will rise by 0.3 m. But this seems like far too much. When she begins stamping down, she will be supported by her other leg, i.e. the one that is not stamping, and this will only be slightly flexed if she stamps in a normal way. So her stamping leg can also only be slightly flexed, and her center of mass can rise a smaller distance. It seems necessary to estimate how much her legs are flexed at the start of the stamp, i.e. the angle subtended at the knee joint.
2. When estimating the potential energy at the peak of her jump, m_lunar * g * h, note that m_lunar is her mass rather than her weight, so it is still 60 kg rather than 10 kg.
ChatGPT: You're right, I made some errors in my previous response. Let's address those and revise the calculations once again.
Estimating the distance her center of mass rises during the stamping action: You're correct that 0.3 meters might be too much. Let's assume that at the start of the stamp, her leg is flexed by about 15 degrees. In this case, the distance her center of mass rises can be estimated using the following formula:
distance = leg_length * (1 - cos(angle))
Assuming a leg length of 1 meter and an angle of 15 degrees:
distance = 1 * (1 - cos(15°))
distance ≈ 0.034 m
You're correct that I should use her actual mass (60 kg) rather than her lunar "weight" (10 kg) when calculating the potential energy. Let's correct that as well.
Now, we can calculate the work done (energy) by Wyoming while stamping her foot:
Work = Force_stamping * distance
Work = 585 N * 0.034 m
Work ≈ 19.89 J (Joules)
We can equate this work to the potential energy she gains at the peak of her jump:
PE = 19.89 J
m * g * h = 19.89 J
Where:
m is her mass (60 kg)
g is the lunar gravity (1.625 m/s²)
h is the height
We can solve for height (h):
h = 19.89 J / (60 kg * 1.625 m/s²)
h ≈ 0.205 m
This result, a height of approximately 0.205 meters (about 8 inches), is much closer to the "half a meter" described in the book. Please keep in mind that these calculations are based on rough estimates and assumptions, and the actual height might differ based on various factors.
mannyrayner: This calculation looks much better to me. As you say, some of the quantities are difficult to estimate.
If you don't mind, I will share your calculations with my physics teacher friend Thomas. He is better than I am at finding errors in this kind of thing!
ChatGPT: I'm glad you find the latest calculation more reasonable. Please feel free to share the calculations with your physics teacher friend, Thomas. I appreciate any feedback or additional insights that may improve the accuracy of the estimation.
December 19, 2021
Read all many times over, love the concept of sentient computers.
September 7, 2014
All great stories! Btw the movie starship troopers is nothing like this awesome book. I don't think they even read the book.
November 16, 2015
Holy crap! What trio. Read it
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