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grav
2007-Oct-07, 01:01 AM
What is the formula for the time of descent of an object falling from some original distance from a body to a closer distance, beginning at rest relative to the body, classically, using the inverse square law? What about with some initial relative speed?

Kaptain K
2007-Oct-07, 01:41 AM
s0-s1=v0+at^2/2

Sorry, I'm not too good with scientific notation on a computer. :(

publius
2007-Oct-07, 01:59 AM
What I'm sure grav is asking for is the radial free fall taking into account the inverse square variation of g. That is, we're not assuming constant acceleration. That involves solving this differential equation:

d^2x/dt^2 = -k/x^2.

See here, for a solution (first one that popped up with Google!):

http://www.mathpages.com/rr/s4-03/4-03.htm

While not all that complex in the grand scheme, the solution is a bit more complex that most simple physics results.

Grav, it turns out that this Newtonian result holds in Schwarzschild (an amazing "coincidence" some call it, but it's really more than that) in a fashion. The time will the *proper time* of the free faller, not the coordinate time of the Schwarzschild observer. But the "r" will be the Schwarzschild observer's notion of distance, and not the free faller's. :)

That does not hold in the general case of arbitrary mass distributions, mind you, it's just a rather neat property of the high spherical symmetry.

-Richard

Kaptain K
2007-Oct-07, 02:11 AM
Oops! :doh:

grav
2007-Oct-07, 02:13 AM
Thanks a lot, Richard. :) I'll probably have to read through that link a few times, though, before I can make anything out of it.

grav
2007-Oct-07, 02:15 AM
Thanks to you also for trying, Kaptain K. :)

Tobin Dax
2007-Oct-07, 02:19 AM
For a constant acceleration, Kaptain K was correct: y = v0*t + (1/2 at^2). Here, y is the distance the object falls, v0 is the initial velocity, a is the acceleration, and t is the time. If accleration is a function of time or height, this equation will still work, but you need to use calculus instead of algebra to solve it.

grant hutchison
2007-Oct-07, 02:21 AM
It's a calculation that featured in Arthur C Clarke's short story, Jupiter V.
Clarke pointed out that the time to fall from a given distance all the way to the centre, under the inverse-square law of a point mass, is 1/(4*sqrt(2)) times the orbital period at that distance.
More importantly for his story, he also worked out the time to fall from Amalthea's distance to Jupiter's cloud-tops.

Grant Hutchison

KaiYeves
2007-Oct-07, 02:53 AM
The MythBusters once mentioned the fastest a human can go while falling, but I forgot the exact speed.

publius
2007-Oct-07, 03:02 AM
The MythBusters once mentioned the fastest a human can go while falling, but I forgot the exact speed.

We're not taking air resistance into account here. :) Just pure Newtonian gravity of a point mass. If we did take that into account, along with inverse square, the equations of motion get intractable rather quickly.

In a mechanics course, I did indeed have to solve some equations of motion with air resistance, but assuming constant 'g'. There, the air resistance modifies the basic parabolic result that a firing a projectile at a 45 degree angle gives you the maximum horizontal range. The trajectory (under whatever conditions we were assuming) became a logarithmic curve, rather than a parabola. Air resistance can kill horizontal velocity and the end of the trajectory approaches a vertical drop.

Another thing was to take into account the variation of air resistance with height (but still not gravity itself, the 1/x^2 term just messes things up quickly). And add Coriolis terms there as well, for good measure. I forget if the latter was completely tractable.

-Richard

frankuitaalst
2007-Oct-07, 12:24 PM
See here, for a solution (first one that popped up with Google!):

http://www.mathpages.com/rr/s4-03/4-03.htm

Nice formula !
In case of an exact given initial velocity the formula can be reduced to the form : s = a * (t0-t)^2/3 where , given an initial distance the initial velocity does have to be equal to some precise value . a is a parameter depending upon G , m and so.

grav
2007-Oct-07, 02:04 PM
Um, thanks for the replies, but I'm afraid I'm still not getting this. I have run it out on a computer simulation and found that

vf^2 - vo^2 = 2GM * [1/d' - 1/d]

Pretty simple, but I still cannot find it for the time yet. I have not found it on Google either yet, surprisely, since I would think it to be a pretty common formula for a precise time of descent, and I can't understand that link Richard gave, which seems to also deal with trajectories and so forth, nor the one in that last post. I'm just working on something simple, or so I thought, and I just want the formula for straight freefall. Someone please help me. :) Can anyone just please, please write out the classical formula in a straightforward manner like the one above? ;)

grant hutchison
2007-Oct-07, 02:54 PM
Starting from zero velocity, at a distance D from the centre of mass, and falling to a distance d, you first need to integrate out the velocity change with distance, using

v = sqrt(2as)

and plugging in the inverse square law representation of a.

That gives you

v = sqrt(2GM[D-d]/[Dd])

Then integrate inverse velocity with respect to distance to get the time of fall:

t = sqrt(1/[2GM]) ∫ sqrt([Dr]/[D-r]) dr

There's a bit of fun with solving the integral, and then it shakes down to:

t = sqrt(D³/[2GM]){π/2 - asin(sqrt[d/D]) + sqrt([d/D][1-d/D])}

If d = 0 (that is, a fall to the centre of gravity), the above simplifies to

t = π/(2sqrt).sqrt(D³/[GM])

Hence Clarke's use of the simple relationship between this fall time and orbital period in his story.

Grant Hutchison

grav
2007-Oct-07, 05:38 PM
Ouch! Thank you very much, Grant. :) Wow. That's not nearly quite as simple as I was hoping for. Oh, well. I wouldn't expect you to, seeing how complex that can get, so it might be and probably is too much too ask, but you wouldn't also happen to know how that would work out with an initial velocity toward the center of mass by any chance, would you?

grant hutchison
2007-Oct-07, 05:47 PM
... but you wouldn't happen to know how that would work out with an initial velocity toward the center of mass by any chance, would you?OK, enough is enough.
Who are you, and what have you done with the real grav? The one who always works everything out for himself?
:D

Grant Hutchison

grav
2007-Oct-07, 05:58 PM
I wouldn't expect you to, seeing how complex that can get, so it might be and probably is too much too ask, but you wouldn't also happen to know how that would work out with an initial velocity toward the center of mass by any chance, would you?Oh, wait. I think I might have a way. As long as the original speed is less than the escape velocity at the original distance, one can find the distance an object would have to be dropped from to gain the speed at the distance we are starting with by substituting vo=0 into the earlier formula for the velocities, vf^2-vo^2 = 2GM*[1/d-1/D] to find D, where d here would be the original starting distance in this case and vf is the original velocity. Then one can just find the time to fall from D to the original d, and then also from D to the new d, and subtract them. Does this sound plausible?

grav
2007-Oct-07, 06:02 PM
OK, enough is enough.
Who are you, and what have you done with the real grav? The one who always works everything out for himself?
:D

Grant HutchisonSometimes my brain hurts too much. :doh:

hhEb09'1
2007-Oct-07, 06:04 PM
Oh, wait. I think I might have a way. As long as the original speed is less than the escape velocity at the original distance, one can find the distance an object would have to be dropped from to gain the speed at the distance we are starting with by substituting vo=0 into the earlier formula for the velocities, vf^2-vo^2 = 2GM*[1/d-1/D] to find D, where d here would be the original starting distance in this case and vf is the original velocity. Then one can just find the time to fall from D to the original d, and then also from D to the new d, and subtract them. Does this sound plausible?Absolutely :)

But maybe you can start to see why people study calculus :)
Sometimes my brain hurts too much. :doh:
that too

grav
2007-Oct-07, 06:10 PM
Absolutely :)

But maybe you can start to see why people study calculus :)

that too:)

astromark
2007-Oct-07, 06:14 PM
Ouch! Thank you very much, Grant. :) Wow. That's not nearly quite as simple as I was hoping for. Oh, well. I wouldn't expect you to, seeing how complex that can get, so it might be and probably is too much too ask, but you wouldn't also happen to know how that would work out with an initial velocity toward the center of mass by any chance, would you?

Ouch is right... To much of this mathematics for my feeble brain to get to grip with but,... If you have initial velocity then yes your velocity will increase only as much as terminal velocity would dictate. I have the number 267 km/hr for planet Earth... Do not Quote me. I know not from where that number did cometh. You do the math... One more point. If you have initial velocity other than toward the center of gravity then you might miss it and end up going into orbit.

Kaptain K
2007-Oct-07, 07:10 PM
astromark,
As has been said before, terminal velocity only applies when falling through the atmosphere! It is obvious, both from the OP and the responses, that it is not the case here.

frankuitaalst
2007-Oct-07, 07:40 PM
Heres a picture representing the formula .
Right = time ; above = distance to initial distance .
Graph was made for G=m=so=1

grav
2007-Oct-07, 09:08 PM
Um... guys?... knock, knock... still there? I hate to bother you again, but it turns out I actually need a formula for the distance an object will fall to if dropped from some distance above a body, knowing the time. I don't see how to retract that from the formula given for time. Maybe it'll be simpler than that one or something (yeah, right)? Sorry. :shifty:

grav
2007-Oct-07, 09:19 PM
Or maybe if someone knows the formula for the velocity that will be achieved, classically, after a certain duration of time for an object dropped from an initial distance. I can find the final distance from that.

frankuitaalst
2007-Oct-07, 09:22 PM
This formula you want doesn't exist . The formula for given distance then calculate time is shown above and works . It doesn t work the other way .

grav
2007-Oct-07, 09:34 PM
This formula you want doesn't exist . The formula for given distance then calculate time is shown above and works . It doesn t work the other way .That's too bad. :( But I guess I can still run the computer simulation at small intervals of time to find the final distance with the givens, though. And I'm sure the formula for the time will come in handy as well. Thanks everyone for your time and effort. I really appreciate it. :)

grav
2007-Oct-07, 10:31 PM
Well, although I still have to run a computer simulation to find the final distance, that formula for time has already helped me out tremendously. Before, I had to divide the total time up into a million intervals and run them one by one to find the final distance after falling for some time. This took about five to ten seconds. But to get any real accuracy, I would have to divide it up into something like a billion intervals, especially if the falling object approached the center of the gravitating body. This, however, would take a couple of hours, and I'm not patient enough to wait that long each time, especially since it bogs down the computer so I can't do anything else.

But now, with that formula for time, I can find the total time it would take the falling object to reach the center of the body. Then, as long as the duration of time I want to find for is smaller than that, I know the final distance will lie somewhere between zero and the original distance. So I have the computer run it this way. First it finds the time to fall to half the distance. Then it finds the time for that plus or minus a quarter of the distance, depending upon whether the time from the first iteration was larger or smaller than the time I want. Then it adds or subtracts half of that, or an eight of the total distance, and does the same thing again, and so on. After just twenty iterations, I have an accuracy of about one part in a million of the original total distance. With thirty iterations, it's one part in a billion. So now, thanks to you guys, I can have my answer in an instant, with just tens of iterations, instead of millions or billions, with as great an accuracy as I could possibly need. Thanks again. :)

grant hutchison
2007-Oct-07, 10:37 PM
This formula you want doesn't exist . The formula for given distance then calculate time is shown above and works . It doesn t work the other way .Yes, it's the equivalent of reversing Kepler's formula: you can get from distance to time analytically, but you need to iterate to get from time to distance.

Grant Hutchison

publius
2007-Oct-08, 02:42 AM
I was go to second hh/kilopi/whatever-incarnation's hint and make a formal plea for grav to advance himself in his mathematical and physical studies. That is learn calculus. :) Then I was going to suggest H&R (and not even give a plug for a freshman physics text some of my own professors wrote, which hasn't made a dent in H&R's market share :lol: ). Then I was going to suggest an intermediate (undergraduate) mechanics text which will detail Newtonian gravity calculations.

My text was Symon's, 3rd edition. I went to Amazon, and my mind was blown:

http://www.amazon.com/Mechanics-3rd-Keith-R-Symon/dp/0201073927/ref=sr_1_1/002-9791717-8559246?ie=UTF8&s=books&qid=1191810722&sr=1-1

\$157!!!! I knew academic titles were pretty much a racket, but Geez. It's been 20 years, but what are the current prices in the university bookstores now?

The (advanced leve) Bible of EM is Jackson, of course, and heck that's only \$87. I ordered a copy (to have a nice new one) a couple years ago from Amazon. That along with a Schwinger title of the same name are two of the best to have:

That is actually a compilation of his lectures, along with some other authors.

Anyway, \$157 for Symon, I couldn't believe it.

-Richard

Tobin Dax
2007-Oct-08, 04:13 AM
My text was Symon's, 3rd edition. I went to Amazon, and my mind was blown:

http://www.amazon.com/Mechanics-3rd-Keith-R-Symon/dp/0201073927/ref=sr_1_1/002-9791717-8559246?ie=UTF8&s=books&qid=1191810722&sr=1-1

\$157!!!! I knew academic titles were pretty much a racket, but Geez. It's been 20 years, but what are the current prices in the university bookstores now?

The (advanced leve) Bible of EM is Jackson, of course, and heck that's only \$87. I ordered a copy (to have a nice new one) a couple years ago from Amazon. That along with a Schwinger title of the same name are two of the best to have:

That is actually a compilation of his lectures, along with some other authors.

Anyway, \$157 for Symon, I couldn't believe it.

-Richard

Marion & Thorton's Classical Mechanics text cost me \$120-\$150. It's about 1/3 the volume of HRW/H&R, roughly the size of a mass market paperback, but in hardcover, and 200 pages long, give or take. Definitely the highest price density of any book I've bought for school. Seven years later, I'm still upset about it.

frankuitaalst
2007-Oct-08, 04:47 PM
you can get a good approximation with the following formula , shown in the picture : y = ....t^6+.....
In order to do it well you have to divide t by sqrt(s0^3/4g/m).
The result (r) should be multiplied with s0 (initial distance).

grav
2007-Oct-08, 05:13 PM
t = sqrt(D³/[2GM]){π/2 - asin(sqrt[d/D]) + sqrt([d/D][1-d/D])}

I can't help but notice that the last bit is all in ratios of the new distance to the original distance, so that if one were to fall to half the original distance, this can be found with simply

t = sqrt[D^3/(2GM)] * 1.285398163397

Falling to one third of the original distance would give

t = sqrt[D^3/(2GM)] * 1.426721138915

I wonder if some sort of exact solution for the new distance can be found using this with the ratios in some way, knowing the time. Or maybe with the ratio of the times themselves. From the above, we now know that it takes exactly 1.109944902 times longer to fall to one third of the original distance than to fall to half the distance, regardless of what the original or new distances are or what the strength of the gravitational field may be.

grant hutchison
2007-Oct-08, 08:43 PM
I can't help but notice that the last bit is all in ratios of the new distance to the original distance.Yeah. That'll be because I deliberately laid it out that way, for ease of computation. :)
I could have made it into a simpler-looking formula that would have been more of a faff to calculate.

Grant Hutchison

grav
2007-Oct-08, 09:08 PM
Yeah. That'll be because I deliberately laid it out that way, for ease of computation. :)
I could have made it into a simpler-looking formula that would have been more of a faff to calculate.

Grant HutchisonAh, yes. Ratios are always the easiest to handle, at least for me, and that formula was very comprehensible the way you had it. Thank you for that. It also went straight into the computer very nicely the way it was laid out.

grant hutchison
2007-Oct-08, 09:38 PM
Another way of looking at is that a fall to some given proportion of the starting distance is always achieved in some given proportion of the orbital period at that starting distance.
If you think about this in comparison to the behaviour of an elliptical orbit, you can see the correspondence between the two situations.

Grant Hutchison

joema
2007-Oct-09, 04:03 AM
....I actually need a formula for the distance an object will fall to if dropped from some distance above a body, knowing the time. I don't see how to retract that from the formula given for time..
The formulas and on-line calculators for constant acceleration are here:

http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html

grav
2007-Oct-10, 12:38 AM
I've been thinking about this some more and figured that the equation for the time of descent says nothing about what lies at the center of gravity or any stopping point. Therefore, it should be possible for the same equation to describe an object that falls right through the center point and keep going. Since the gravitational field here is symmetrical, it would take the same time to travel from the center point to -D as it did to travel from D to the center point. Substituting that into the equation, then, we get

pi/2 - asin(sqrt(d/D)) + sqrt((d/D)(1-d/D)) = 2(pi/2)

where d=-D, so

sqrt((-D/D)(1-(-D/D))) - asin(sqrt(-D/D)) = pi/2

sqrt(-2) - asin(sqrt(-1)) = pi/2

asin(i) = i*sqrt(2) - pi/2

Now, UBasic deals with complex numbers, but it only gives me "illegal parameter" when I try to find asin(i). Can anyone tell me if this is an equality? I'm sure it probably is, otherwise I wouldn't have found it, but I'm also wondering about an equation this complicated describing gravity to begin with, even a close approximation using Newtonian applications. It seems strange that a finite mass should produce infinite gravity, even at a singularity, that maybe gravity actually levels off with something other than the inverse square, like approaching light speed at the center point or something, but then, there would be no actual photon sphere surrounding black holes, since it would then lie at the center, which would never actually become reachable, except through a mass, which levels off the gravity anyway. But I don't want to get too much into that here. Any thoughts about the equation above? I'm not very familiar with the workings of complex numbers.

publius
2007-Oct-10, 01:55 AM
Grav,

Let's have a bit of fun:

This is one of Euler's relations:

e^(ix) = cos x + i sin x

e^(-ix) = cos x - i sin x

Now, use that to solve for sin x in terms of e^(ix) and e^(-ix)

Then, substitute that for this statement:

sin x = y

Now, solve for x in terms of y, and you have something for the arcsin. Be careful. There are some surprises and pitfalls there.

-Richard

grav
2007-Oct-10, 03:18 AM
Well, I have apparently fallen into one of those pitfalls. I know complex numbers can be tricky. I got

e^(ix) = cos x + i sin x
e^(-ix) = cos x - i sin x

sin x = [e^(ix) - cos x]/i
sin x = [cos x - e^(-ix)]/i

sin x = y
x = asin y

y = [e^(i asin y) - cos (asin y)]/i
y = [cos (asin y) - e^(-i asin y)]/i

iy = e^(i asin y) - sqrt(1-y^2)
iy = sqrt(1-y^2) - e^(-i asin y)

y = i

i^2 = e^(i asin i) - sqrt(1-i^2)
i^2 = sqrt(1-i^2) - e^(-i asin i)

-1 = e^(i asin i) - sqrt(2)
-1 = sqrt(2) - 1/e^(i asin i)

e^(i asin i) = sqrt(2) - 1
e^(i asin i) = 1 / (sqrt(2) + 1)

and these are equalities as far as sqrt(2) - 1 and 1 / (sqrt(2) + 1) are concerned, but then, taking e^(i sin i) = sqrt(2) - 1, I get

i asin i = In[sqrt(2) - 1]
asin i = In[sqrt(2) - 1]/i

and substituting that back into the original equation,

asin i = i sqrt(2) - pi/2
In[sqrt(2) - 1]/i = i sqrt(2) - pi/2
In[sqrt(2) - 1] = i^2 sqrt(2) - i pi/2
In[sqrt(2) - 1] = -sqrt(2) - i pi/2
-i pi/2 = In[sqrt(2) - 1] + sqrt(2)

and this is not an equality. The left side is complex while the right side is real. Where did I go wrong?

publius
2007-Oct-10, 03:39 AM
Grav, grav, grav,

You went astray at the very first step: Solve for sin x in terms of the e's:

e^(ix) = cos x + i sin x
e^(-ix) = cos x - i sin x

Subtract the latter from the former:

e^(ix) - e(-ix) = 2i sin x

sin x = [e^(ix) - e^(-ix)]/2i

Now, set sin x = y, recognize that e^(-a) = 1/e^a, and go. This is were you need to be careful and think about things.

-Richard

grav
2007-Oct-10, 04:18 AM
Oh. I saw that cos x would cancel out but it sounded to me like you wanted me to perform them separately or something. Well, starting with what you gave, this time I found

sin x = [e^(ix) - e^(-ix)]/2i

y = sin x

y = [e^(i asin y) - e^(-i asin y)]/2i
2iy = e^(i asin y) - 1/e^(i asin y)

y = i

2 i^2 = [e^(2i asin i) - 1] / e^(i asin i)

z = e^(i asin i)

-2 = (z^2 - 1)/z
-2z = z^2 -1
z^2 + 2z - 1 = 0

So this time I used the quadratic formula to solve for z with

z = [-2 +/- sqrt(4 - 4(-1))] / 2
z = -1 +/- sqrt(8)/2
z = -1 +/- sqrt(2)

e^(i asin i) = -1 +/- sqrt(2)

and if I take the positive version, I just get the same result as I did before, with

e^(i asin i) = sqrt(2) - 1

I'm still not sure what I'm missing here, but I'll keep working on it. Are there multiple solutions or something?

publius
2007-Oct-10, 04:34 AM
e^(i asin i) = sqrt(2) - 1

I'm still not sure what I'm missing here, but I'll keep working on it. Are there multiple solutions or something?

Nope, you done good there. That's the correct (principle) answer, or

asin i = -i*(ln(sqrt(2) - 1) ~ (0.881)i

Now, the ln function, when we go to the complex domain, is multi-valued (big time)........ z = r*e^(iO). O can go around and around, 2pi at the time and get back to the same angle, right?

-Richard

publius
2007-Oct-10, 04:47 AM
Grav,

I'm probably going to go to bed soon, and I know you get to chompin' at the bit sometimes, so....

I just wanted to show you how to play around with complex numbers and find out what the asin of i was. I didn't say anything about the correctness of your first formula from that freefall time expression. :) You wondered if that gave you the correct expression for asin(i). I wanted to show you how to find asin(i) so you compare.

So, go forth and compare. Think about that time expression. Maybe even look at Grant's expression a little closer.

-Richard

grav
2007-Oct-10, 05:01 AM
Thanks, Richard. From your post before last, I don't see how the In function can vary in itself, but I can see how the asin function can, with 2pi. So we might have something like

e^(i asin i) = sqrt(2)-1

or e^[i (asin i + n 2pi)] = sqrt(2) - 1

where n is any integer, so e^(i asin i) * e^(n 2i pi) = sqrt(2) - 1

and then e^(i asin i) = [sqrt(2) - 1] / e^(n 2i pi)

or, since n can be negative or positive just as easily, e^(i asin i) = [sqrt(2)-1] [e^(n 2i pi)]

Does that sound about right?

grav
2007-Oct-10, 07:11 AM
So, go forth and compare. Think about that time expression. Maybe even look at Grant's expression a little closer.

-RichardAha. I ran back through that formula in a different way. It goes like this.

t = sqrt(D^3/2GM)[pi/2 - asin(sqrt(d/D)) + sqrt((d/D)(1-d/D))]

T (falling to center) = sqrt(D^3/2GM)(pi/2)

t (given) = 2T

sqrt(2GM/D^3)t - pi/2 (knowns)= -asin(sqrt(d/D)) + sqrt((d/D)(1-d/D)) (unknown)

y = sqrt(2GM/D^3)t - pi/2
z = sqrt(d/D)

y = -asin(z) + z*sqrt(1-z^2)

z = sin(x)

y = -asin(sin(x)) + sin(x)*cos(x)
y = -x + sin(2x)/2
2y = -2x + sin(2x)

x' = 2x , y' = 2y

y' = sin(x') - x'

t = 2T = 2*sqrt(D^3/2GM)(pi/2)
y = sqrt(2GM/D^3)t - pi/2
= pi - pi/2
= pi/2
y' = 2y = pi

y' = sin(x') - x'
pi = sin(x') - x'
x' = -pi

x = x'/2 = -pi/2
z = sin(x) = -1

z = sqrt(d/D)
-1 = sqrt(d/D)
1 = d/D

So apparently, using d=-D was incorrect. It will still be a positive D even though it lies on the other side of the origin for the center. As far as the formula for time goes for this, then, this gives us

pi/2 - asin(sqrt(d/D)) + sqrt((d/D)(1-d/D)) = 2(pi/2)
pi/2 - asin(sqrt(1)) + sqrt(1*0) = pi
-asin(1) = pi/2
-pi/2 + n*pi= pi/2

n=1
pi/2 = pi/2

No complex results, and it also looks like the asin function, then, can add integer multiples of n*pi instead of n*2pi.

grav
2007-Oct-10, 01:35 PM
pi/2 - asin(sqrt(d/D)) + sqrt((d/D)(1-d/D)) = 2(pi/2)
pi/2 - asin(sqrt(1)) + sqrt(1*0) = pi
-asin(1) = pi/2
-pi/2 + n*pi= pi/2

n=1
pi/2 = pi/2

No complex results, and it also looks like the asin function, then, can add integer multiples of n*pi instead of n*2pi.Whew. I'm glad that didn't jibe with my concept of complex numbers representing a physical impossibility. It would be difficult to think that the universe is capable of working through two sets of impossibilities to make something possible, although I know it's still mathematically possible, but I'm thinking the universe is much more direct with the simplicity of nature. But that last result still concerns me. One doesn't add integer multiples of pi for the asin function, which would make a positive value become negative, but 2*pi, which keeps it the same. It only works for t=2T because sine for pi at 180 degrees is the same as sine for zero degrees, because -0 = 0. But how would that work out for something in between, where the time to fall to d is t, so the time to fall to -d is 2T-t? I'll attempt to work that out later.

mugaliens
2007-Oct-10, 03:47 PM
At least two of you did mention terminal velocity, but what wasn't mentioned was the fact that terminal velocity decreases as the object falls because the air is more dense.

So, it's really not at all a straightforward calculation.

What they used to do with bomb ballistics is drop a bunch of them in known aerodynamic conditions then reduce the calculations down the number of parameters in the equations then publish them in the tables, interpolating for missing data.

The initial parameters aren't difficult, and include density altitude at the point of release, calibrated velocity at the point of release, the vertical distance of fall, and the winds between what they're observed on the ground and at altitude.

grav
2007-Oct-10, 10:55 PM
Okay, I've finally figured it out. It seems pretty simple in the end, as usual. If the time to fall to d is t, and to the center is T, then the time to fall to -d is 2T - t. So this becomes

2T - t = 2 * sqrt(D^3/2GM) (pi/2) - sqrt(D^3/2GM) [pi/2 - asin(sqrt(d/D)) + sqrt((d/D)(1-d/D))]

2T - t = sqrt(D^3/2GM) [2(pi/2) -pi/2 + asin(sqrt(d/D)) - sqrt((d/D)(1-d/D))]

2T - t = sqrt(D^3/2GM) [pi/2 + asin(sqrt(d/D)) - sqrt((d/D)(1-d/D))]

So the time to fall to d is

t(d) = sqrt(D^3/2GM) [pi/2 - asin(sqrt(d/D)) + sqrt((d/D)(1-d/D))]

while the time to fall to -d is

t(-d) = sqrt(D^3/2GM) [pi/2 + asin(sqrt(d/D)) - sqrt((d/D)(1-d/D))]

Notice the equations are identical except for the last two terms which now carry opposite signs. So for this to work for -d by using the same original formula for the time of descent, in the case of asin, then, we would still use the positive value for d, but take the negative value for the square root. That gives a negative value for asin as well. We would also do the same thing with the last term, using d but taking the negative value for the square root there as well. It's still kind of strange how that works out, though, where we don't use the negative value for d but we do take the negative value for the square roots. I'm not really sure why it plays out that way, except for one thing perhaps, that there's really no such thing as a negative distance, only with direction, so I can understand that part of it. But it's almost like, instead of using just -d for the negative distance, we are using (-1)^2*d, which still gives us a positive d for the distance, but in the formula, it plays out like

t(-d) = sqrt(D^3/2GM) [pi/2 - asin(sqrt([(-1)^2*d]/D)) + sqrt(([(-1)^2*d]/D)(1-[(-1)^2*d/D))]

t(-d) = sqrt(D^3/2GM) [pi/2 - asin((-1)*sqrt(d/D)) + (-1)*sqrt((d/D)(1-1*d/D))]

t(-d) = sqrt(D^3/2GM) [pi/2 + asin((sqrt(d/D)) - sqrt((d/D)(1-d/D))]

publius
2007-Oct-10, 11:35 PM
Grav,

:clap:

Congratulations, and I hope you see my purpose. You like to see things for yourself and I was hoping to get you to see it for yourself. Multi-valued functions like square roots and asin (turn a sine vertically -- there are an infinite number of values for -1 < x < 1) play tricks like this all the time. In many problems, you end up solving for "pieces" of the problem. This expression is valid from a to b, but from b to c, we must use a different expression. And that's the trouble with "blindly following formulae". The insight into all this only comes with practice of *deriving* those formula and appreciating what inclusions and exlcusions you make during each step of the process.

Some times both values of such functions might work and so you've got more than one answer. And sometimes they don't, and you have to pick which one is the real one.

Consider this x = 2. Square it, x^2 = 4. Now take the square root, and we have x = +2 or -2. We gained a solution that wasn't there when we started.

When you solve equations, and that includes differential equations, intermediate steps that get you to the answer can add solutions like that. The final answer includes solutions that weren't part of the original statement. You've got to be aware of that. You remember that mess of sinh's and cosh's I went through for the Rockets in Rindler? Remember how I had 4 solutions and had to figure out which one was the proper one?

And in other cases, you can loose solutions. For example, you might divide through by something. That operation can *loose* solutions because you can't divide by zero. If that expression happened to be identically zero for some solution, you are in essence throwing it away, because the following steps depend on it not being zero.

-Richard

grav
2007-Oct-11, 12:19 AM
Thanks, Richard. That was a fun (and educational) exercise. :)