Fermi Questions C

[Unome]

According to the shell theorem (essentially Gauss's law), the only mass that matters should be within a spherical shell at the radius being considered.

Vector calculus being useful XD

[Name]

Wait a screwed up the question. It is E0 but I should've said something like 3000 where the new mass is around E22 and radius is 1000 where the gravitational force is E1 stronger while on the surface

I feel like this is being overthought. According to the shell theorem (essentially Gauss's law), the only mass that matters should be within a spherical shell at the radius being considered. If you do the math, this gives a linearly decreasing force for a linearly decreasing radius. Thus, just taking the ratio of 5/6 (5000 km compared to around 6000 km) gives you fermi 0.

Wait hmm I checked the math and this seems correct. I probably screwed up something while doing the math earlier. The mass is porportional to volume which is radius cubed and is divided by radius squared so the force is linearly decreasing.

[Tom_MS]

What is the 273rd Fibonacci number?
To clarify, 0 will be considered the first Fibonacci number.

Attempt (Click to Show Hidden Text)

Using an approximation which I will not disclose, it's RIGHT on the edge between 56 and 57, but I'll say 57.

Answer (Click to Show Hidden Text)

Using an online calculator, the 273rd Fibonacci number is 5.06*10^56, so yes. Barely 57. Bwahahaha.

A proton and electron are initially separated by 1 meter. After being released, they end up colliding due to their coulombic attraction. If the final kinetic energy (right before collision) of the two particles is converted almost entirely into rest mass energy of the particles released, how much mass is formed (kg)?

[PM2017]

Hmmm... I sort of made that question so the others would be able to learn the method of approximating fibbonacci numbers... (I'm pretty sure I know what you're doing, but I won't disclose it if you don't want it disclosed.)

[Name]

Hmmm... I sort of made that question so the others would be able to learn the method of approximating fibbonacci numbers... (I'm pretty sure I know what you're doing, but I won't disclose it if you don't want it disclosed.)

Uh it's pretty easy to just google

[whythelongface]

Hmmm... I sort of made that question so the others would be able to learn the method of approximating fibbonacci numbers... (I'm pretty sure I know what you're doing, but I won't disclose it if you don't want it disclosed.)

Uh it's pretty easy to just google

It's just Stirling's Approximation of Factorials. It's not exactly a trade secret.

[PM2017]

Hmmm... I sort of made that question so the others would be able to learn the method of approximating fibbonacci numbers... (I'm pretty sure I know what you're doing, but I won't disclose it if you don't want it disclosed.)

Uh it's pretty easy to just google

It's just Stirling's Approximation of Factorials. It's not exactly a trade secret.

Wait I was thinking of something else... The limit of the ratio between one Fibonacci number and the next is the golden ratio, as you get larger and larger numbers. So you just do (Phi)^(273-1).

[Unome]

Uh it's pretty easy to just google

It's just Stirling's Approximation of Factorials. It's not exactly a trade secret.

Wait I was thinking of something else... The limit of the ratio between one Fibonacci number and the next is the golden ratio, as you get larger and larger numbers. So you just do (Phi)^(273-1).

This is how I would do it.

[whythelongface]

It's just Stirling's Approximation of Factorials. It's not exactly a trade secret.

Wait I was thinking of something else... The limit of the ratio between one Fibonacci number and the next is the golden ratio, as you get larger and larger numbers. So you just do (Phi)^(273-1).

This is how I would do it.

Oops I appear to have read "fibonacci" as "factorial"

Yes, this seems to be the ideal way of doing it.

[Raleway]

Time to restart this thread for now... b/c it'll be relevant next year!

With the hot air coming out of Donald Trump's mouth when he says cofveve, how many times would he need to repeat it in order to boil water? Whoever actually solves this (not you whythelongface) correctly gets a cookie!