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Weekend Geek: Pennies to Heaven
In case you need to get your mind off the Supreme Court, here’s a fun math problem for the weekend. Imagine a stack of pennies from the surface of the earth to the surface of the moon (equator-to-equator; neglect rotation of both bodies and all other factors that would make this impossible). Use the properties of modern (post-1982) pennies. The distance from the earth to the moon varies a lot, so use the mean (or “semi-major axis”) distance. You may have seen this problem before, but I’m taking it a little further:
1. How many pennies would be required to reach from the earth to the moon?
2. Do that many pennies exist?
3. What’s the mass of all those pennies? What do they weigh on the surface of the earth (not stacked)?
4. What do they weigh on the surface of the moon (not stacked)?
5. It’s often said that if you want to create the impression of a large quantity of something, talk about how long it would be if you stacked those things end-to-end. But if you want to create the opposite impression, talk about how much volume they would occupy stuffed into a box. If all those pennies from the earth to the moon could be packed into a perfectly cubical box, what would be the size of the smallest possible box that could hold them all?
6. Neglecting rotation, if all the pennies were in a single stack, and placed on a scale on the earth (not going to the moon, just out into space), what would the scale read?
7. Similarly, if all the pennies were in a single stack, and placed on a scale on the moon (again into space), what would the scale read?
8. Again neglecting rotation, if the pennies were stacked from the earth to the moon, what would the earth scale read? What would the moon scale read?
9. Still neglecting rotation, is there a limit to what a stack of pennies might weigh on earth, assuming the stack goes on forever? What about on the Moon?
10. Assuming the earth rotates, is there a maximum height of a stack of pennies? (Neglect wind resistance, column slenderness, compressive stresses, and all the other factors that make the stack impossible from a practical standpoint.) What’s the mass of the stack, and what would it weigh on the scale?
11. How strong is the gravitational attraction to the moon of objects on the earth’s surface? How does such a weak attraction cause the tides?
The answers may surprise you. Feel free to guess if you don’t feel like doing the math.
(Answers: 1. As many pennies as the Supreme Court finds are necessary. 2. The pennies exist regardless of the original intent of the Mint. Okay, I guess I need to get my mind off the Supreme Court too. The real answers will be posted on Sunday, unless Ricochet nails them all before then.)
ANSWERS:
1. Approximately 247,554,450,000 (See comment #41).
2. Over 302 billion pennies have been minted between 1983 and 2014. Over 80% of them would be required to reach the Moon, if that many are still in existence (not flattened and embossed as souvenirs, for example). See comment #42.
3. Mass 618,886,000 kg (1,364,387,000 lb). Weight 6,069,416,000 N (1,364,387,000 lb). See comment #43.
4. Weight on the moon 1,003,833,000 N (225,658,850 lb). See comment #44.
5. Cube is 49.08m (161 ft) on a side (see comment #45).
6. Approximately 101,070,700 N (22,720,400 lb), just 1.67% of what they would weigh flat on Earth’s surface (see comment #46).
7. Approximately 4,619,000 N (1,038,300 lb), just 0.46% of what they would weigh flat on the Moon’s surface.
8. Earth scale 100,700,450 N (22,637,200 lb), just 1.66% of what they would weigh flat on Earth’s surface. Moon scale 4,248,800 N (955,100 lb), just 0.42% of what they would weigh flat on the Moon’s surface.
9. The Earth limit is about 102,783,900 N (23,105,500 lb). The moon limit is 4,640,300 N (1,043,100 lb). Note that even though the Moon has 1/6 Earth gravity, the limit on the Moon is about 1/22 of the Earth limit. Due to the Moon’s smaller radius, its gravity field diminishes more quickly than that of the Earth.
10. If the Earth rotates, all pennies have an angular velocity equal to one revolution per day. The centripetal acceleration increases with increasing distance from the Earth while the gravitational acceleration decreases. They are equal at a radius of 42,240,500 m, which is about 22,300 miles above the surface of the Earth (the same altitude of geosynchronous satellites). This requires about 23,593,693,000 pennies with a total mass of 58,984,200 kg and weight of 82,653,600 N on the earth scale (14.29% of what the pennies would weigh flat on Earth).
11. Objects on Earth’s surface are subject to a gravitational attraction of approximately 0.0000329 m/s^2 (about 0.00000335 g), with the moon at its semi-major axis distance. Although this gravity field is quite weak, it is able to produce tides acting on a very large mass of water such as an ocean. No noticeable tides are produced in smaller bodies such as lakes. The largest tide observed in the Great Lakes produces a difference in water level of approximately two inches.
I have not posted the equations for questions 7 -11 in the comments. I can do it tomorrow if anyone is interested.
Published in General
That’s about right.
Edited when I realized I had used Kg, a measure of mass, which does not change. The mass is the same on Earth and the Moon, it is only the weight, the force exerted by gravity on the mass, that changes.
5. Approximately 170.15 feet per side of the cube.
And, I think I’m done playing for tonight.
The fierce electrodynamics of “gravity” would blast the moon into Jupiter, creating a second Venus.
As best I can recall, gravity on the moon is roughly 1/6 that of the earth. When you ask what the stack would weigh on the different bodies, do you mean what would be within the moon’s gravity well as opposed to what’s within the earth’s gravity well? In other words, you have a stack extending from the earth to the moon. What percentage of that stack would be attracted to the moon, and what to the earth?
That’s sort of what he is asking in 8. In 6 and 7, you’re dealing with the square of the distance rule, and because you have so many pennies stacked up, you’re basically getting into calculus to get the right equation that will allow you to quickly calculate the weights from the first penny to the last.
Surface gravity on the Moon is 0.1654 g, so just under 1/6th of Earth gravity.
I got a D in calculus. And my professor was being generous.
But this is your chance to shine. You just have to dust off your brains and go looking for the information you need.
Like a new penny?
I just checked my pocket. I have four one cent coins. Two are from this year, one from 2010, and one from 1978. So, there are definitely still at least some of the older cent coins out there.
Arahant—I strongly suspected you of doing what it was my first instinct to do: make up a lot of numbers. Then I looked more carefully and thought: Holy Moley, he actually KNOWS?!?! Why are you guys so freakin’ smart? Is this a gender thing? (I can say that, right?) And why is it I have so much trouble disregarding rotation of both bodies and all other factors that would make this impossible?
No, it’s a STEM mind thing. They do come in female flavors. I’d do it myself if I wasn’t feeling so frustrated and depressed at mother nature and needed to get on with studying for a bar exam.
Oh thank God(-dess) Amy. Big relief for an old feminista…(What is a STEM mind?) (Other than—“the mind I don’t have.”)
It’s a bit like the Potter test for porn, but basically, a mind that looks at the world and sees numbers and dimensions and force diagrams. The physical world is a problem to be solved or engineered around, and math is the beautiful language to express one’s thoughts in. It’s the stereotypical geek — and I’m proud to be one, married to another one, and most of my friends are the same way.
Okay, then yes—“the mind I don’t have” came pretty close. My step-son thinks this way. Over supper one night, he began explaining a problem in theoretical math and he explained it so beautifully that for just a brief moment, I sort of got it. Very, very cool. And really useful—people like you can make life a whole lot better (mysteriously) for people like me!
Why is kg a measure of mass, while lb is a measure of weight? I thought either unit of measure could be used to describe either mass or weight, so long as you specify the gravitational force causing the weight.
Here’s how big a geek I am: when A Beautiful Mind first showed Dr. Nash visualizing the equations that he wrote, I nearly burst into tears. I wish, so desperately, that I could somehow let people see what I see when I do what I do. Because it is breathtaking.
If you have a very particular kind of mind.
The proper American system for mass is the slug which is equal to 32.2 lbs at mean sea level. (my inner Geek)
First, kilograms and pounds avoirdupois do tend to be used imprecisely as measures of both mass and weight (force calculated as mass x gravitational acceleration) here on Earth. But, kilograms are a measure of mass. As Vectorman pointed out, in the Imperial system of weights and measures, the slug is the unit of mass. The pound (avoirdupois) is a measure of force. Force units are measured in a unit which shows mass unit x distance unit/time unit^2. So, a pound is actually the force used to accelerate one slug to a speed of one foot per second in one second.
There is a corresponding force unit in the metric system. it is called the Newton. It is translated into mass x distance / time^2 as kg x m / s^2.
So, I should have converted from kilograms to Newtons, but that would have required multiplying by Earth gravity of 9.807 m/s^2, and instead I just used 2.2 lbs/kg.
I should also present four other caveats on my figures.
Last thoughts on the matter:
In other words, I only even tried the really, really easy questions.
Pounds are properly a measure of weight (force), but they are often used for mass as well. On Earth we weigh objects to determine their mass. When other accelerations are involved, we often have to multiply or divide by g(subscript C)* in order to convert from pounds-mass (lbm) to pounds-force (lbf) and vice versa. Occasionally slugs are used as a mass unit (1 slug = 32.2 lbm), but most people have never heard of them, or think you can kill them with salt.
In SI metric, we have kilograms for mass and Newtons for force, and there is no g(c) to worry about, so engineering calculations become simpler. 1 N = 1kg x 1 m/s^2, so 1 kg weighs about 9.8 N on Earth.
Thanks also to Arahant and Vectorman for their explanations.
* g(c)= (32.2 lbm-ft)/(lbf-s^2)
These are very close to what I got. Everything depends on the Earth to Moon distance that you use. I used the distance from Wikipedia. From what I can tell it appears to be the distance from the center of the Earth to the center of the Moon, so I subtracted the equatorial radii of the two bodies, and came up with an answer about 9 billion lower.
Blue States are you saying if asked by certain parties to add 2+2 SCOUS may not necessarily come up with 4?
Yep. Of course, to get that many cent coins, we would have to use ones that have mostly been circulated for years, and the wear may be such that that nine billion and extra would be required. Luckily, you specified doing this as a simple problem.
Pikers. You forgot to take into account that the earth doesn’t revolve in the ecliptic, and that both the earth and the moon are ovoid.
Put up your numbers, son.
I did mention my D in calculus, right?
As I said earlier, the first five questions only require multiplication. division, and one application of the third root of a number.