Help with a math problem?

Well, I hope I'm not boring you with all this, but I found two things:

(1) An analytic way to solve this equation for velocity as a function of the radius (An analytic solution for position as a function of time is a whole other ball game).

(2) From the first result above, I found the way I approximated the position does not produce a good model. So throw it out.

Starting with a form of Equation (6) again:

d²r/dt² = a - G*M/r² ...6'

Making the following substitutions:

dv/dt = d²r/dt² ...7

v= dr/dt ...8 → dt = dr/v ...8'

Sub ...7 & ...8' → ...6'

d²r/dt² = a - G*M/r² → dv/dt = a - G*M/r² → v*dv = a*dr - G*M/r²*dr ...9

Then integrate (9) with the initial conditions v= v_o, and r_o = R

⌠v*dv = a⌠dr - G*M⌠1/r²*dr

½*v²- ½*v_o² = a(r - R) - G*M(1/r - 1/R) →

v = √{2*[a(r - R) - G*M(1/r - 1/R) + ½*v_o²]} ...10

v: velocity
a: acceleration (in your example 11.76 m/s², but it can be anything now as long as the units are consistent)
v_o: initial velocity ( if you assume it starts from rest at the surface that whole last term "½*v_o²" under the square root drops out)
R: radius of the Earth
M: mass of the Earth
G: gravitational constant

You can use equation (10) to predict the velocity as a function of distance of the body from the center of the earth.

Have fun!

Originally posted by joe shmo
Well, I hope I'm not boring you with all this, but I found two things:

(1) An analytic way to solve this equation for velocity as a function of the radius (An analytic solution for position as a function of time is a whole other ball game).

(2) From the first result above, I found the way I approximated the position does not produce a good model. So t ...[text shortened]... velocity as a function of distance of the body from the [b]center of the earth.

Have fun![/b]

Thanks for the help. Will play with that tomorrow! What about the other ball game? Result as function of time?

Originally posted by sonhouse
Thanks for the help. Will play with that tomorrow! What about the other ball game? Result as function of time?

Well, the form of the equation lends itself to a simple method of solution by integration (the differential equation is separable, one of the easiest types of differential equations to solve in general) However,... the difficulty in integrating a function is dependent on the complexity of the integrand (another name for the function that is to be integrated) itself.

v = √{2*[a(r - R) - G*M(1/r - 1/R) + ½*v_o²]} →

dr/dt = √{2*[a(r - R) - G*M(1/r - 1/R) + ½*v_o²]} →

1/[√{2*[a(r - R) - G*M(1/r - 1/R) + ½*v_o²]}]*dr = dt

⌠1/[√{2*[a(r - R) - G*M(1/r - 1/R) + ½*v_o²]}]*dr = ⌠dt

The right side of the equation is a simple integration, The left half...on the other hand... because it is very complicated may very well be impossible to perform by algebraic/analytic means (I wouldn't even know where to begin)

The resulting form of the position function (if it could even be found) would undoubtedly be an monstrosity. On top of that, it would still need to be algebraically rearranged to an explicit form: r = f(t), which may be also be an impossibility.

I may try to cook up another numerical approach later, but i once again will not be able to validate it.

Originally posted by sonhouse
Thanks for the help. Will play with that tomorrow! What about the other ball game? Result as function of time?

Post Edited Away: I don't know which way is down or up anymore!! There is a sign error for one of the terms: edited equation is below.

v = √{2*[a(r - R) + G*M(1/r - 1/R) + ½*v_o²]}

Originally posted by joe shmo
Post Edited Away: I don't know which way is down or up anymore!! There is a sign error for one of the terms: edited equation is below.

v = √{2*[a(r - R) + G*M(1/r - 1/R) + ½*v_o²]}

Eureka! I believe I have found a reasonable approximation for R(t). 2 days in the making, its a good thing I'm currently unemployed... 😞

Using the equation above I I iterated about 80,000 times with the following procedure:

First step was to add 100 meter increments 80,000 times successively to the radius of the earth in meters

v = √{2*[a((r+δr) - R) + G*M(1/(r+δr) - 1/R)]}

Calculate the velocity at each increment using the above equation taking v_o = 0

Next, I assumed:

δr = v_o*δt + 1/2*(δv/δt)*δt² for relatively ( on planetary scales 100 meters isn't much) small δr

which I then solve for δt

δt = δr/(v_o + ½δv)

and thus the time at the current radius is just:

t = t_o + δt

Lather, rinse, repeat about 80,000 times. Write a program to pull time and position data at regularly spaced intervals ( about 80 points in total)

Plot and Trend

R(t) = 0.0008*t^3 + 0.12*t^2 + 292.43*t + 6373718 [meters]

I compared the second derivative w/respect to time of this equation at the associated radius, with the acceleration as a function of the radius as follows

R(2000 s) = 0.0008*2000^3 + 0.12*2000^2 + 292.43*2000 + 6373718

= 13838578 [meters]

R"(t) = 0.0048*t + .24 [meters/second²]

R"(2000) = 0.0048*2000 + .24

= 9.84 [m/s²]

From our original Working Equation (6)

d²r/dt² = a - G*M/r²

Using: a = 11.76 m/s²
G = 6.673e-11 [N*m²/kg²]
M = 5.97219e 24 [kg]
R = 13838578 [m] ( radius calculated above )

d²r/dt² = 11.76 - 6.673e-11*5.97219e 24/13838578^2

= 9.69 [m/s²]

about a 1.5% error....not bad all thing considered!

Originally posted by joe shmo
Eureka! I believe I have found a reasonable approximation for R(t). 2 days in the making, its a good thing I'm currently unemployed... 😞

Using the equation above I I iterated about 80,000 times with the following procedure:

First step was to add 100 meter increments 80,000 times successively to the radius of the earth in meters

v = √{2*[a((r+δr) - ...[text shortened]... 24/13838578^2

= 9.69 [m/s²]

about a 1.5% error....not bad all thing considered!

Wow, that is frigging complicated! You would think something like that would not be so complex! Anyone else want to chime in?

Originally posted by sonhouse
Wow, that is frigging complicated! You would think something like that would not be so complex! Anyone else want to chime in?

Btw, the equation:

R(t) = 0.0008*t^3 + 0.12*t^2 + 292.43*t + 6373718 [meters]

is only a "good" approximation for R(t) from 0-2000 seconds, after that it will start to give progressively worse estimates of "R".

Originally posted by joe shmo
Btw, the equation:

R(t) = 0.0008*t^3 + 0.12*t^2 + 292.43*t + 6373718 [meters]

is only a "good" approximation for R(t) from 0-2000 seconds, after that it will start to give progressively worse estimates of "R".

What is the significance of the 6 million meters?

Originally posted by sonhouse
What is the significance of the 6 million meters?

Well, the earths radius vary's (albeit slightly in comparison with the average radius), and its mass is distributed unevenly throughout the body.

The gravitational Force (From Newtons Law of Gravitation) assumes Earth is a body with uniform density, as such it can be represented as a point mass, and also is perfectly spherical.

That number is the solution for the radius of the Earth using 9.81 m/s² as the acceleration due to gravity at the surface, and all the assumptions (simplifications) above.

Originally posted by joe shmo
Well, the earths radius vary's (albeit slightly in comparison with the average radius), and its mass is distributed unevenly throughout the body.

The gravitational Force (From Newtons Law of Gravitation) assumes Earth is a body with uniform density, as such it can be represented as a point mass, and also is perfectly spherical.

That number is the solu ...[text shortened]... the acceleration due to gravity at the surface, and all the assumptions (simplifications) above.

wow, with 7 digit accuracy too🙂

Originally posted by sonhouse
wow, with 7 digit accuracy too🙂

Well, I wouldn't say that it is accurate...but it is precise (perhaps overly so).

Originally posted by joe shmo
Well, I wouldn't say that it is accurate...but it is precise (perhaps overly so).

I should have figured the 6 mil was Earth's radius in meters.
You mentioned being currently unemployed. What do you do when you ARE employed? Where do you live?

Originally posted by sonhouse
I should have figured the 6 mil was Earth's radius in meters.
You mentioned being currently unemployed. What do you do when you ARE employed? Where do you live?

Mechanical Engineering, nothing fancy. I 'm from central PA (Johnstown, home of the 1889 flood)

Your from eastern PA, correct?

Originally posted by joe shmo
Mechanical Engineering, nothing fancy. I 'm from central PA (Johnstown, home of the 1889 flood)

Your from eastern PA, correct?

yes, north of Allentown, a cultural black hole called Slatington.

Johnstown is pretty far east into Pa, I thought. what, 60 miles from Pittsburgh?

I had a temp job in Toledo Ohio, had to drive fairly close to you, hwy 80, goes all the way. I was working at Midwest Microdevices, a semiconductor foundry, working on semiconductor furnaces. Now work in a cleanroom in South Plainfield, NJ. A company called Gulton.

Great job, bad part is the 160 mile round trip commute. I refurbished two MRC sputtering machines, the MRC 603, from the 1980's. One thing I found out, if those machines had been using modern electronics, I would have killed it. There is a water cooling loop that cools the targets and such and one of the hoses popped loose and I was stuck like the kid with his thumb in the dam but got the main computer totally wet with about a half inch of water. So when the water chiller got turned off, we sucked up the water with a vacuum cleaner then with an air gun sprayed the water out of the card cage and such (Intel 8080 based computer!) and then used a fan positioned on top of the card cage to complete the dry out and the next day, fired it up and it came right back. So that old technology does have some good points.

If that had been modern controllers with new CPU's and such, one drop of water could kill it. You know what happens to a cell phone when it gets wet. That was still a traumatic day for sure, had no idea if the computer would come back online. This is the day of single board computers, no hard drive, no floppy, none of that nonsense🙂 Just ram, and not much of that!

Originally posted by sonhouse
yes, north of Allentown, a cultural black hole called Slatington.

Johnstown is pretty far east into Pa, I thought. what, 60 miles from Pittsburgh?

I had a temp job in Toledo Ohio, had to drive fairly close to you, hwy 80, goes all the way. I was working at Midwest Microdevices, a semiconductor foundry, working on semiconductor furnaces. Now work in a ...[text shortened]... ard computers, no hard drive, no floppy, none of that nonsense🙂 Just ram, and not much of that!

Yeah, J-town is about 60-70 miles southeast ish of Pittsburgh.

160 miles is one heck of a commute...I'm currently searching in the area, cant imagine 3 hrs of driving back and forth to work on a good travelling day, let alone on a typical Pennsylvania winter day!

I'm not much of a computer tech head. I just know how to use them (more or less). I didn't really get a chance to dive deeply into control systems and PLC's at my previous place of employment. They always hoped that I would, but had me very busy on mechanical systems (which I prefer to the the jumbled ball of wiring in a digital brain) , and besides there were enough electrical engineers in the plant that could do the that type of work efficiently.

Sounds like you have been in the tech field for quite sometime, and have seen some older electronics. The mill where I worked was 110 years old ( I only spent 2 years there), there was still in operation some real "Dr. Frankenstein laboratory" type electrical equipment running the machines to this day!