Annoying Questions I'd Like Answered...

hyzmarca · Post by **hyzmarca** » Mon Jun 16, 2014 8:55 pm

That's also questionable, especially when you get into ancestor worship.

Catholics, for example, have Saints. They're dead and they're not gods, but you can pray to them and they can intercede on your behalf.

Prak · Post by **Prak** » Mon Jun 16, 2014 9:14 pm

Ok, related question--are the eight immortals of Chinese Myth more ascended gods or heros?

Grek · Post by **Grek** » Tue Jun 17, 2014 9:34 am

They're definitely more like heroes. A western equivalent would be the Argonauts, or the Knights of the Round Table.

Hicks · Post by **Hicks** » Tue Jun 17, 2014 5:54 pm

So I have a maths question that is just beyond me. I know the specific factors that need to be taken into account, but I just do not know how to math them until I get an answer.

The senario is that the moon has disappeared from its orbit in the night sky and is now on a collision course for earth. Relevant information: the moon has an initial velocity of 1,028 m/s toward earth, has an equatorial surface gravity of 1.6249 m/s, has an equatorial radius of 1,738.14 km, and starts 385,000 km from the surface of the earth; the earth has an effective equatorial surface gravity of 9.832 m/s, and an equatorial radius of 6,378.137 km. Solve for the time it takes for the moon/earth system to smash into each other.

Other salient information: the gravity of the earth on the moon and otherwise is 1/3,643.6681031006th the surface gravity on each other during that first second. At this is point I am reduced to writing a slightly less than 385,000 line Excel spreadsheet to count how far both bodies fall into each other; I would much rather grow as a person and learn how to math hammer these kinds of problems, if anyone could help.

rampaging-poet · Post by **rampaging-poet** » Tue Jun 17, 2014 9:47 pm

That's a differential equation. The equatorial surface gravities and radii can be used to get the masses of the Earth and Moon. Once you have those, you can substitute them into the universal gravitation formula to get the force on the Earth from the Moon and vice-versa. Forces become accelerations, and acceleration is the second derivative of position. In coordinates where the center of the Earth is immovable and the Moon is to the right of the Earth, you should end up with an ordinary differential equation of the form:

(d^2x/dt^2) = - (Gm/x^2 + GM/x^2)

where x is the position of the moon, m is the mass of the Moon, M is the mass of the Earth, and G is the universal gravitational constant. Solving the ODE yields an expression for the position of the Moon as a function of time in terms of its initial position and velocity. Substitute those in from the problem, then set x(t) = (Radius of Earth + Radius of Moon) to solve for the collision time.

Your Excel spreadsheet solution is pretty close to what's actually going on. You basically stumbled upon a first-order Euler solver, which is a simple way of approximating solutions to differential equations that works well enough for most purposes. The error in that approximation is proportionate to both the number of time steps you iterate over and the size of those steps.

DSMatticus · Post by **DSMatticus** » Tue Jun 17, 2014 10:03 pm

First: the earth is not actually a stationary target because it has an orbit. It will be in a different position at t=1 than it was at t=0, and that means that the acceleration caused by gravity at t=1 is towards a different point in space than it was at t=0. It is almost certainly not accurate to model gravity as simply accelerating a one-dimensional closing velocity. But I recommend ignoring that because tackling that problem is where I would just model this shit on a computer.

Second: the last paragraph confuses me. The problem does indeed have to deal with variable gravity, but gravity isn't linearly variable with respect to time here - it's nonlinearly variable with respect to distance. And distance in this problem is itself already nonlinearly variable with respect to time (because acceleration). I can't tell if you already knew that and just chose to describe things a little oddly, or were genuinely confused about the problem in some way. Either way, onto the helpful bit: you need the actual formula for gravity, which is F1=F2=(G*m1*m2)/(r^2). The combined acceleration is then (F1/m1)+(F2/m2). Or, written out: [(G*m1*m2)/(m1*r^2)]+[(G*m1*m2)/(m2*r^2)]. (G is the gravitational constant, m1 is the mass of earth, m2 is the mass of the moon, r is the distance between the two centers).

Then you have:
1) An initial distance between their centers (385,000,000m).
2) A final distance for which you want the time of impact (1,738,140m + 6,378,137m).
3) An initial closing velocity (1028 m/s).
4) A function for the acceleration of the closing velocity (f(r) = [(G*m2)/(m1*r^2)]+[(G*m1)/(r^2)]).

That is technically all the information you need to solve the problem and fuck it I've been ninja'd.

That aside, this is one of those problems that can be solved fairly accurately to within an arbitrarily small fraction (okay, that's not entirely true) of a second by a computer program that fits on a piece of notebook paper. My first instinct is to do exactly what you were going to do in excel, except not actually in excel and instead in something more powerful.

Hicks · Post by **Hicks** » Tue Jun 17, 2014 11:49 pm

I am so glad I resisted my first inclination to do this all by hand. So this is what I came up with and after 5 hours of debugging an Excel spreadsheet:


Constants	Values
Moon gravity at equator	1.6249 m/s
Moon radius at equator	1738140 m
Earth gravity at equator	9.832 m/s
Earth radius at equator	6378137 m


Operation	Formula
Seconds Elapsed	Old Seconds Elapsed + 1
Moon Accelleration (m/s)	Old Moon Acceleration + Earth gravity on Moon
Earth Accelleration (m/s)	Old Earth Acceleration + Moon gravity on Earth
Moon gravity divisor	((Current distance between Earth and Moon + Moon Radius + Earth Radius) / Moon Radius)^2
Earth gravity divisor	((Current distance between Earth and Moon + Moon Radius + Earth Radius) / Earth Radius)^2
Moon gravity on Earth (m/s)	Earth Gravity / Current Moon Gravity Divisor
Earth gravity on Moon (m/s)	Earth Gravity / Current Earth Gravity Divisor
Distance between Earth and Moon Surfaces (m)	Old Distance between Earth and Moon Surfaces + Current Earth Acceleration + Current Moon Acceleration
Moon Roche Limit	IF earth gravity on Moon >= Moon Gravity, TRUE, ELSE FALSE


Seconds Elapsed	Moon Acceleration (m/s)	Earth Acceleration (m/s)	Moon gravity divisor	Earth gravity divisor	Moon gravity on Earth (m/s)	Earth gravity on Moon (m/s)	Distance between Earth and Moon Surfaces (m)	Moon Roche Limit
0	1028	0	51153.16901	3798.869418	3.17654E-05	0.002588138	385000000	FALSE
1	1028.002588	3.17654E-05	51152.90148	3798.84955	3.17655E-05	0.002588152	384998972	FALSE
218206	7039.091247	73.77681638	81.49238635	6.051999129	0.019939286	1.624587147	7574457.429	FALSE
218207	7040.715834	73.79675567	81.41850235	6.046512163	0.01995738	1.626061395	7567342.917	TRUE
219127	9825.023115	107.9698065	21.82265628	1.620650747	0.074459313	6.066698837	3396.389835	TRUE
219128	9831.089814	108.0442659	21.7692636	1.616685561	0.074641937	6.081578408	-6542.744244	TRUE

Yes Virginia, that is 219,128+1 lines of code; one line for each second of simulation. I still think It could be wrong, as I never figured out how to account for the mass of both bodies, but that is immaterial as I now have not one but two formulas to massage out sweet, sweet data. This was all for a 10th level adventure where the PCs must stop the moon from falling, and I wanted to be able to honestly and impartially tell them when the moon hits the planet. I'm also probably wrong about the Roche Limit, but I figured if earth's gravity ever superseded lunar surface gravity, the moon would break up into innumerable micro and macroscopic chunks that would be nearly impossible to interdict, a "better" option at that point of no return may be for the PCs to take their last 15 minutes trying to figure out how to move the Earth away from the Moon.

DSMatticus · Post by **DSMatticus** » Wed Jun 18, 2014 4:18 am

Five hours? Ouch. This is ten minute java noobcode:

Code: Select all

public class Tutorial &#123;
	public static void main&#40;String&#91;&#93; args&#41; &#123;
		double gravConstant = 6.67384E-11;
		double massEarth = 5.97219E24;
		double massMoon = 7.34767309E22;
		double radiusEarth = 6378137.00;
		double radiusMoon = 1738140.00;
		double distance = 385000000.00;
		double velocity = 1028.00;
		double timeInterval = 1.00;
		double timeElapsed = 0.00;

		while &#40;distance > radiusEarth + radiusMoon&#41; &#123;
			distance = &#40;distance - velocity * timeInterval&#41;;
			
			double accelEarth = &#40;gravConstant * massMoon&#41; / &#40;distance * distance&#41;;
			double accelMoon = &#40;gravConstant * massEarth&#41; / &#40;distance * distance&#41;;
			velocity = &#40;velocity + &#40;accelEarth * timeInterval&#41; + &#40;accelMoon * timeInterval&#41;&#41;;

			timeElapsed = &#40;timeElapsed + timeInterval&#41;;
		&#125;

		System.out.println&#40;"Result&#58; " + &#40;timeElapsed - timeInterval&#41;
				+ " seconds."&#41;;
	&#125;
&#125;

That is the solution as I imagine it would be written by an intro to code monkeying 101 student. It's not the least bit professional, and I feel dirty having written it, but it's functional and avoids using any complicated concepts like "functions" and "objects" and "+=". That seems like the sort of thing you could learn to do in an afternoon - it's just variables, arithmetic, and a loop. But I may be insane and wildly out of touch, hell if I know. It also doesn't give a spreadsheet full of data, but on the other hand can be easily modified to do all sorts of things pretty easily.

The answer it gives is 213,699 seconds. That's close enough to your's that I'm gonna call it a success.

rampaging-poet · Post by **rampaging-poet** » Wed Jun 18, 2014 4:52 am

I just ran a similar program in FreePascal. It yields a time for physical collision of 213,719 seconds, which is probably slightly closer because I used 10-byte Extended floating point numbers instead of 8-byte Double floating points. I also calculated the time to reach the rigid-satelite Roche limit (per Wikipedia). That was only 154,181 seconds.
EDIT: Oops, missed a cube root in my Roche radius. One moment while I recalculate that. ... There we go. The real time to reach the Roche limit was 213,575 seconds.

Hicks · Post by **Hicks** » Wed Jun 18, 2014 6:42 am

From what I can find, 385,000km is the starting distances between surfaces, not distance between the "point sources" for earth and moon gravity, which would be 393,116.277km, which is why I think the spreadsheet predicts a longer time to impact.

rampaging-poet · Post by **rampaging-poet** » Wed Jun 18, 2014 3:08 pm

Ok. Substituting (385,000 km + R_Earth + R_Moon) into my simulation yields 219,221 seconds to the Roche limit and 219,365 seconds until surface-to-surface contact.

radthemad4 · Post by **radthemad4** » Fri Jun 20, 2014 7:09 pm

What're some good books, websites, etc. for grokking English grammar?

Stahlseele · Post by **Stahlseele** » Fri Jun 20, 2014 8:20 pm

err, i'd suspect school grade engrish books that are supposed to teach this stuff? O.o
on the other hand, just try and read some english novels so even if you don't really know what's what, you at least know what's supposed to go where?

radthemad4 · Post by **radthemad4** » Fri Jun 20, 2014 8:51 pm

School grade books could be a good start. I winged it for most of my life as my English has usually been good enough for day to day purposes so I didn't look too closely at the underlying structure.

While I can usually use grammar correctly, correcting other people's grammar (occasional part time job) just gives me headaches and seems to take a lot longer than it should as I keep needing to look up stuff online.

I'd also like to learn about the weird edge cases I may never have considered. I guess I'd like to be proficient enough to say, write some code that can detect wrong grammar (not that I see myself actually needing to do this any time soon).

Additionally, this would probably help with GRE.

...You Lost Me · Post by **...You Lost Me** » Fri Jun 20, 2014 9:08 pm

A junior high grammar book should be good. Once you've got that down, I also highly recommend Strunk & White's "Elements of Style".

Maj · Post by **Maj** » Fri Jun 20, 2014 9:09 pm

What's the goal of this, RTM4?

radthemad4 · Post by **radthemad4** » Fri Jun 20, 2014 9:18 pm

...You Lost Me: Thanks, I'll look up that book.

Maj: To be able to grade English worksheets designed by other people without needing to check online every few minutes. Additionally, there are times when a student writes something that 'feels' wrong to me but turns out to be grammatically correct. Also, it'd probably help with GRE.

Stahlseele · Post by **Stahlseele** » Fri Jun 20, 2014 10:51 pm

if you are not a native speaker that "feels wrong but is actually correct" may never ever go away.
hell, i like to think my engrish is pretty good, and i still can't help but shake my head at some stuff i read.
Ancient History for example, when he writes his unpublishable stuff.

DSMatticus · Post by **DSMatticus** » Fri Jun 20, 2014 11:06 pm

English is a weird language of patchwork bullshit. There is an impressive amount of inconsistency, and as such I'm with stahlseele that the "feels wrong but isn't" thing is probably just going to stick with you forever if English wasn't your first language. It's the Queen's curse for speaking a heathen's tongue.

Prak · Post by **Prak** » Fri Jun 20, 2014 11:32 pm

Dude, "feels wrong but isn't" is a problem native speakers have. Hell, last week I had a fight with my mother over whether or not "Trying to not" was grammatically correct (the fact that Strunk and White backed me up on it in fact being acceptable made me more accepting of "Old white men's guide to english")

Stahlseele · Post by **Stahlseele** » Fri Jun 20, 2014 11:34 pm

how and why would you ever argue that "trying to not" fuck up was incorrect? @.@ *boggles*

Prak · Post by **Prak** » Fri Jun 20, 2014 11:40 pm

Because "not trying to" and "trying to not" are two entirely different statements with different meanings. "not trying to" means that you are making no attempt, "trying to not" means you are making an attempt to not do so something. "Not trying to kill someone" means you are making no attempt to kill someone, "trying to not kill someone" means you are making an attempt to not cause someone's death through your actions. And S&W backs me up on the latter being acceptable since the 14th century.

Pixels · Post by **Pixels** » Fri Jun 20, 2014 11:46 pm

Stahlseele wrote:how and why would you ever argue that "trying to not" fuck up was incorrect? @.@ *boggles*

I had assumed it was an objection about splitting infinitives, which still annoys a non-zero portion of the population.

Stahlseele · Post by **Stahlseele** » Fri Jun 20, 2014 11:46 pm

yes, my question was why and how anybody would argue against that being correct to use in certain cases o.O
my engrish is good enough for me to be able to get the difference there, so i hope that natives would be able to do so too.

Prak · Post by **Prak** » Fri Jun 20, 2014 11:51 pm

Oh, gotcha. Yeah, it's a splitting infinitives thing and, like ending sentences with prepositions, gets old folks' undergarments in a bunch for no fucking reason other than a nun told them it was bad.

(the preposition was recently deemed by word nerds to be acceptable because the prohibition of it was based on latin, where a sentence ended with a preposition would make literally no sense. In English it's perfectly clear what such sentences mean, so it's not a big deal.)