## Fermi Questions C

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### Re: Fermi Questions C

How far does an air molecule (assume at STP) travel on average before it hits another air molecule, in meters?
22 liters in a mole 6E23/22 is 3E22. Assuming a 1 dimentional movement the cube root of 3E22 is about 7E7. Because a liter is 1/10 of a meter, 7E8 air molecules in a straight line 1 meter. 1/7E8 is around E-9 so -9
I saw both mentiones of -9 and -7 as a answer (3 nm and 68). I think it might be 68 but I can't understand the math behind it. Can you confirm what the answer is?
I believe that it would be $\frac{1}{n \sigma}$, where $n$ is the number density and $\sigma$ is the cross sectional area of an air molecule (going to make the rough assumption that diatomic oxygen and nitrogen have relatively similar sizes).

$n$ is about $2.5 \times 10^{25} \mathrm{m^{-3}}$ (you can calculate this using the Ideal Gas Law if you don't have it memorized). The cross sectional area of diatomic oxygen or nitrogen is about $4 \times 10^{-19} \mathrm{m^2}$.

Ok thanks! Also oops forgot to set up a question

Lets say you gained the ability to harvest salt from the ocean on a mass scale. If you take the salt and compacted it into a 1 meter radius sphere, what is it's escape velocity
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### Re: Fermi Questions C

22 liters in a mole 6E23/22 is 3E22. Assuming a 1 dimentional movement the cube root of 3E22 is about 7E7. Because a liter is 1/10 of a meter, 7E8 air molecules in a straight line 1 meter. 1/7E8 is around E-9 so -9
I saw both mentiones of -9 and -7 as a answer (3 nm and 68). I think it might be 68 but I can't understand the math behind it. Can you confirm what the answer is?
I believe that it would be $\frac{1}{n \sigma}$, where $n$ is the number density and $\sigma$ is the cross sectional area of an air molecule (going to make the rough assumption that diatomic oxygen and nitrogen have relatively similar sizes).

$n$ is about $2.5 \times 10^{25} \mathrm{m^{-3}}$ (you can calculate this using the Ideal Gas Law if you don't have it memorized). The cross sectional area of diatomic oxygen or nitrogen is about $4 \times 10^{-19} \mathrm{m^2}$.

Ok thanks! Also oops forgot to set up a question

Lets say you gained the ability to harvest salt from the ocean on a mass scale. If you take the salt and compacted it into a 1 meter radius sphere, what is it's escape velocity
Volume of the oceans is about 18, so the mass is about 21. Guess that 1% of the ocean's mass is salt, so the salt mass is 19. Escape velocity is sqrt(2GM/r). G is -10, M is 19, r is 1. Escape velocity would be 4.
According to [url=https://www.reddit.com/r/askscience/comments/1cyizs/if_all_the_salt_in_the_oceans_was_removed_and/]this reddit post[/url], there's about 4.9 x 10^22 grams of salt = 4.9 x 10^19 kg of salt. [url=http://www.wolframalpha.com/input/?i=sqrt(2*6.67*10%5E(-11)*4.9*10%5E(19)%2F1)]Wolfram Alpha[/url] gives an answer of 5 overall.
How many complete rotations has the fastest pulsar made in the time between the extinction of the dinosaurs and the birth of Newton?
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### Re: Fermi Questions C

How many complete rotations has the fastest pulsar made in the time between the extinction of the dinosaurs and the birth of Newton?
The number of years between Newton's death and the present is negligible compared to the K-T extinction event, 65 mya (I think: this could just be one of those numbers the public loves to misquote). 7E7 years = what, 1 to 2E15 seconds?

I have no idea how many rotations a fast pulsar makes per second, but I think 40 rps was a number I encountered somewhere for pulsars? So my final answer is [b]17[/b].
PSR J1748-2446ad spins at [b]716 hz[/b]. My K-T extinction estimate was correct, so my answer is off by a magnitude from the real answer, [b]18.[/b]
RuBisCO is the most abundant enzyme in the entire biosphere. RuBisCO fixes carbon dioxide into ribulose bisphosphate, but can also fix diatomic oxygen into RuBP in a wasteful process known as photorespiration. How many kilograms of O2 are metabolized by RuBisCO within the entire biosphere in one second?
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### Re: Fermi Questions C

RuBisCO is the most abundant enzyme in the entire biosphere. RuBisCO fixes carbon dioxide into ribulose bisphosphate, but can also fix diatomic oxygen into RuBP in a wasteful process known as photorespiration. How many kilograms of O2 are metabolized by RuBisCO within the entire biosphere in one second?
RuBisCO I think is in the leaves. Trees make up the majority of leaves. There are about E12 trees. I have no idea how many RuBisCO are in a leaf. I'm gonna assume a tree uses 100 calories a day (thier large but don't spend energy moving and whatnot) so E14 calories of energy are used per day. There are 5 Cal of energy in a gram of glucose so 2E13 grams of glucose have to be produced. About 200 grams per mole 6E34 glucose molecules. I think it takes 6 cycles for 1 molecule (I suck at bio) so 6 o2 will be fixed or 3E35. Or 5E11 moles of 02 or about 2E10 grams of o2 or 2E7 kg per day or 2 per second.
Can't find anything on the answer, and my answer seems off. Can you confirm what the answer is?

At the current rate this thread is moving at (starting from the first post) how many posts will there be in the time it takes for sound to travel from here to Sagittarius A*
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whythelongface
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### Re: Fermi Questions C

RuBisCO is the most abundant enzyme in the entire biosphere. RuBisCO fixes carbon dioxide into ribulose bisphosphate, but can also fix diatomic oxygen into RuBP in a wasteful process known as photorespiration. How many kilograms of O2 are metabolized by RuBisCO within the entire biosphere in one second?
RuBisCO I think is in the leaves. Trees make up the majority of leaves. There are about E12 trees. I have no idea how many RuBisCO are in a leaf. I'm gonna assume a tree uses 100 calories a day (thier large but don't spend energy moving and whatnot) so E14 calories of energy are used per day. There are 5 Cal of energy in a gram of glucose so 2E13 grams of glucose have to be produced. About 200 grams per mole 6E34 glucose molecules. I think it takes 6 cycles for 1 molecule (I suck at bio) so 6 o2 will be fixed or 3E35. Or 5E11 moles of 02 or about 2E10 grams of o2 or 2E7 kg per day or 2 per second.
Can't find anything on the answer, and my answer seems off. Can you confirm what the answer is?
I cannot confirm because I have not calculated it myself and am far too lazy to do so, but I'll give you some numbers:
"For every person on Earth, there are around five kilograms of RuBisCO in the biosphere."
"At ambient levels of carbon dioxide and oxygen, the ratio of the reactions is about 4 to 1, which results in a net carbon dioxide fixation of only 3.5."
I can't find the data for rate of carbon fixation, but if you really wanted to, you probably could.
WEST WINDSOR-PLAINSBORO HIGH SCHOOL SOUTH '18
EMORY UNIVERSITY '22
SONT 2017 5th Place Medalist [Microbe Mission]

"One little Sciolyer left all alone,
He went out and hanged himself and then there were none."

Congratulations to WW-P South for winning 14th place at Nationals!

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### Re: Fermi Questions C

RuBisCO is the most abundant enzyme in the entire biosphere. RuBisCO fixes carbon dioxide into ribulose bisphosphate, but can also fix diatomic oxygen into RuBP in a wasteful process known as photorespiration. How many kilograms of O2 are metabolized by RuBisCO within the entire biosphere in one second?
RuBisCO I think is in the leaves. Trees make up the majority of leaves. There are about E12 trees. I have no idea how many RuBisCO are in a leaf. I'm gonna assume a tree uses 100 calories a day (thier large but don't spend energy moving and whatnot) so E14 calories of energy are used per day. There are 5 Cal of energy in a gram of glucose so 2E13 grams of glucose have to be produced. About 200 grams per mole 6E34 glucose molecules. I think it takes 6 cycles for 1 molecule (I suck at bio) so 6 o2 will be fixed or 3E35. Or 5E11 moles of 02 or about 2E10 grams of o2 or 2E7 kg per day or 2 per second.
Can't find anything on the answer, and my answer seems off. Can you confirm what the answer is?
I cannot confirm because I have not calculated it myself and am far too lazy to do so, but I'll give you some numbers:
"For every person on Earth, there are around five kilograms of RuBisCO in the biosphere."
"At ambient levels of carbon dioxide and oxygen, the ratio of the reactions is about 4 to 1, which results in a net carbon dioxide fixation of only 3.5."
I can't find the data for rate of carbon fixation, but if you really wanted to, you probably could.
I can't find the rate and really don't want to search more but the number is probably significantly higher then 2
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### Re: Fermi Questions C

Well, might as well prep for next year.

How many leaves are in Manhattan?

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### Re: Fermi Questions C

Well, might as well prep for next year.

How many leaves are in Manhattan?
2E5 leaves per tree and maybe E5 trees in Manhattan so E10
It says 5 million trees in NYC with 8 percent in Manhattan so 4E5 trees in Manhattan so actually E11
At the current rate this thread is moving at (starting from the first post) how many posts will there be in the time it takes for sound to travel from here to Sagittarius A*
Distance to Sagittarius A is about E18 km or 21 m. Speed of sound about 300m/s so 3E18 seconds or E15 hour or 5E13 days. Post rate works out to about 1/day so 14 posts
By what magnitude is the electrostatic force of a eletron and proton greater then the gravitational force exerted from each other
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### Re: Fermi Questions C

But sound can't travel in space
Electrostatic force is approximately 10^10 * 10^-20 * 10^-20 and gravitational is approximately 10^-6 * 10^-26 * 10^-30, dividing gives 10^32
Actually around 10^39 (whoops)
How many full-scale models of the Titanic would you have to lay to build a bridge across the Atlantic Ocean (from around Maine to Spain)

edit: Or Unome can post his answer before I finish editing whoops
Last edited by UTF-8 U+6211 U+662F on May 1st, 2018, 12:36 pm, edited 2 times in total.

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### Re: Fermi Questions C

By what magnitude is the electrostatic force of a eletron and proton greater then the gravitational force exerted from each other
Assumed distance of 1 m for simplicity. Gravitational force = (6.7E-11)(1E-27)(2E-24) = ~1.2E-61. Electrical force = (9E9)(1.6E-19)(1.6E-19) = 2.7E-28. E/G = 2E33 so Fermi Answer: 33.
Confirmed the measurements via the internet, however I accidentally used grams instead of kg for the masses :oops:
How many electrons flow through an iPhone 8 over the course of one battery charge?
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