- 09 Feb '15 13:03 / 1 editSo I was imagining a magic acceleration platform that would accelerate at a constant rate of 1.2 G's, 11.76 M/sec^2 or 38.4 M/sec^2, take your pick.

So this device is hooked to a space-worthy vehicle and can accelerate straight up, no curved flight to make it easier. So the thing that obviously pops up when trying to figure out at what altitude does it achieve escape velocity or how much time it takes to get to that point and how fast would it be going when it passes the moon, for instance.

Clearly, when it takes off, you subtract Earth's gravity, 32 F/S^2 or 9.8 M/S^2 so it is only going up at a rate of 0.2 G. But higher up, Earth's gravity is going down, so at 2 R (about 4000 miles up or 6400 Km) Earth's gravity is 1/4th of the surface.

So how do you combine those two to get a formula that can calculate the altitude it achieves escape velocity or the velocity after X amount of seconds, at infinity, the craft will achieve its full accel, 1.2 G's So it starts at 0.2 G's and ends at 1.2.

What would the formula be for velocity at a certain altitude, velocity after a certain time?

I assume this kind of problem was solved by Isaac himself but I would like to learn how to handle this situation. - 09 Feb '15 17:44 / 1 edit

I think I know what you are trying to say, but I believe you are confusing some things which makes it harder to explain. Firstly, you state that the platform accelerates at a*Originally posted by sonhouse***So I was imagining a magic acceleration platform that would accelerate at a constant rate of 1.2 G's, 11.76 M/sec^2 or 38.4 M/sec^2, take your pick.**

So this device is hooked to a space-worthy vehicle and can accelerate straight up, no curved flight to make it easier. So the thing that obviously pops up when trying to figure out at what altitude does it a ...[text shortened]... d of problem was solved by Isaac himself but I would like to learn how to handle this situation.**constant**rate of 1.2 G's, but then you later state that the acceleration of platform is**changing**. There is a contradiction there. If the acceleration of the body is constant it is simple to determine the velocity of the body at any time from (neglecting all physical interactions between the body and it environment)

V = v + a*t

Where V = Velocity at time "t" [distance/time]

v = initial velocity [distance/time]

a = acceleration [distance/time²]

t = time [time]

However, what I think you mean to say is the platform can supply a constant**Force**( call it a thrust force ) and in the absence of gravity that force will cause the body to accelerate at a constant rate of 11.76 m/s²? - 09 Feb '15 19:41

Constant thrust with no expenditure of fuel so it would be like as if it were powered by a laser from the ground. So it is a constant 1.2 G, like you said, 11.76 M/sec^2. The variable here is Earth gravity, so at around 6400 km up, gravity is 1/4th of surface gravity or about 2.45 M/sec^2. So that means that at that altitude, your net acceleration is 9.31 M/second^2, when the net acceleration started out at 1.96 M/Sec^2.*Originally posted by joe shmo***I think I know what you are trying to say, but I believe you are confusing some things which makes it harder to explain. Firstly, you state that the platform accelerates at a [b]constant**rate of 1.2 G's, but then you later state that the acceleration of platform is**changing**. There is a contradiction there. If the acceleration of the body is con ...[text shortened]... bsence of gravity that force will cause the body to accelerate at a constant rate of 11.76 m/s²?[/b]

I would like a formula or set of formulae that gives the distance traveled after X amount of seconds, how high you get in that time and max velocity at any point on that curve. No fancy curves to deal with, just a straight up flight following a radius line of Earth. - 09 Feb '15 22:47

Ah.*Originally posted by sonhouse***Constant thrust with no expenditure of fuel so it would be like as if it were powered by a laser from the ground. So it is a constant 1.2 G, like you said, 11.76 M/sec^2.**

You're not having a constant acceleration, you're having a constant*force*pointing up. And a decreasing force (gravity) going down, so a (asymptotically) varying resulting acceleration going up.

I'm not sure how realistic this can be made, and I'm*certainly*not going to do the math, but the above should help a bit. - 09 Feb '15 23:39

Well, lets use Newtons Second Law to try and determine said formula, which states that a body's acceleration is determined by the net forces exerted on it and its mass.*Originally posted by sonhouse***Constant thrust with no expenditure of fuel so it would be like as if it were powered by a laser from the ground. So it is a constant 1.2 G, like you said, 11.76 M/sec^2. The variable here is Earth gravity, so at around 6400 km up, gravity is 1/4th of surface gravity or about 2.45 M/sec^2. So that means that at that altitude, your net acceleration is 9.31 M ...[text shortened]... curve. No fancy curves to deal with, just a straight up flight following a radius line of Earth.**

In variable form:

Σ**F_net**= m**a**

For the left half of the equation there are two forces acting on the body:

Force of Thrust = T

Force of Gravity = W (which depends on r; the radial distance of the body from the Earth)

Newtons Second Law reads:

T - W = m*a (1)

From newtons law of gravitation:

W = G*m*M/r² (2)

Substitute (2)->(1)

T - G*m*M/r² = m*a (3)

"a" in this case can be written as the radial acceleration of the body in the following way

a = d²r/dt² (4)

Substituting (4)->(3) yeilds

T - G*m*M/r² = m*d²r/dt² (5)

Unfortunately here is where our fun ( as far as I can take it ) comes to an end. In order to find r as a function of time ( the formula that predicts distance for a given time) we have to solve (5) with the conditions the that the initial velocity is zero, and as r→∞;a→11.76 m/s². The equation above is what is known as a Second Order Non-Linear Differential Equation, and may not even be solvable by analytic methods. So from here we need help from the physicists who may know tricks around this road block. - 10 Feb '15 14:55 / 1 edit

Here is a question for the Physicists:*Originally posted by joe shmo***Well, lets use Newtons Second Law to try and determine said formula, which states that a body's acceleration is determined by the net forces exerted on it and its mass.**= m

In variable form:

Σ[b]F_net**a**

For the left half of the equation there are two forces acting on the body:

Force of Thrust = T

Force of Gravity = W (which depends on ...[text shortened]... hods. So from here we need help from the physicists who may know tricks around this road block.[/b]

Can the following technique be used to numerically determine r(t)

From(dividing reference equation 5 by mass):

11.76 - G*M/r² = d²r/dt² (6)

Assume "a" is approximately constant for very small intervals of "t"

Initial Velocity of the body is zero.

body leaves from Earths Surface at R

Calculate "a" at r=R

Calculate dr/dt using a very small increment of time from the following:

dr/dt = V_o + a*δt

Calculate "r" from:

r = r_o + v_o*δt + 1/2*a_o*δt²

Then iterate this process? - 11 Feb '15 16:41

Using the technique above and a spreadsheet I came up with a polynomial approximation for r(t). I'm not sure how much error is associated with it, but its a "formula" that seems to have believable characteristics.*Originally posted by joe shmo***Here is a question for the Physicists:**

Can the following technique be used to numerically determine r(t)

From(dividing reference equation 5 by mass):

11.76 - G*M/r² = d²r/dt² (6)

Assume "a" is approximately constant for very small intervals of "t"

Initial Velocity of the body is zero.

body leaves from Earths Surface at R

Calculate "a" at r= ...[text shortened]... _o + a*δt

Calculate "r" from:

r = r_o + v_o*δt + 1/2*a_o*δt²

Then iterate this process?

nR(t) = 1e-10*t^3 + 1e-07*t^2 - 1e-05*t +1

nR(t) = the number of Earth radii at time "t"

t = time in seconds

**!WARNING! This "formula" is only an approximation of the behavior for the Range of 1-7 Earth radii, and time<3600 seconds (1 hr)** - 11 Feb '15 16:46 / 1 edit

Wow, didn't realize it was so complicated. My engineer buddy here thought it was a simple differential. The last one, 1e-5 * t+1 is 1e-5 *(t+1) I assume?*Originally posted by joe shmo***Using the technique above and a spreadsheet I came up with a polynomial approximation for r(t). I'm not sure how much error is associated with it, but its a "formula" that seems to have believable characteristics.**[/b]

nR(t) = 1e-10*t^3 + 1e-07*t^2 - 1e-05*t +1

nR(t) = the number of Earth radii at time "t"

t = time in seconds

[b]!WARNING! This "for ...[text shortened]... approximation of the behavior for the Range of 1-7 Earth radii, and time<3600 seconds (1 hr)

Setting T = 1000 seconds, seems to work out to 0.21001 Earth radius.

So .21*6400 puts you up 1344 km after 16 2/3 minutes. Can we get that result another way to prove the concept? - 11 Feb '15 17:12

No, its (1e-5*t) + 1*Originally posted by sonhouse***Wow, didn't realize it was so complicated. My engineer buddy here thought it was a simple differential. The last one, 1e-5 * t+1 is 1e-5 *(t+1) I assume?**

Setting T = 1000 seconds, seems to work out to 0.21001 Earth radius.

So .21*6400 puts you up 1344 km after 16 2/3 minutes. Can we get that result another way to prove the concept? - 11 Feb '15 18:07 / 2 edits

Ok, let's see how much that changes things. Yep, certainly does. That would be 1.2 Earth Radii, or about 7700 Km up. Or maybe not.*Originally posted by joe shmo***No, its (1e-5*t) + 1**

That answer 1.2R, is that from the surface of Earth or from the center? If so, we are back to 1300 km high. - 11 Feb '15 18:26 / 3 edits

You are at 1.2 radii above the earths surface*Originally posted by sonhouse***Ok, let's see how much that changes things. Yep, certainly does. That would be 1.2 Earth Radii, or about 7700 Km up. Or maybe not.**

That answer 1.2R, is that from the surface of Earth or from the center? If so, we are back to 1300 km high.

**EDIT: Nevermind, you are 1300 km above the surface at 1000 seconds, 0r 1.2 radii above the earths center** - 11 Feb '15 19:40

Actually its best if you hold off on the whole thing, something is off in the model. I'll look for the error ( it may be the method itself for the solution, I cooked it up myself...not peer reviewed as a legitimate approach).*Originally posted by joe shmo***You are at 1.2 radii above the earths surface**[/b]

[b]EDIT: Nevermind, you are 1300 km above the surface at 1000 seconds, 0r 1.2 radii above the earths center - 11 Feb '15 19:59 / 2 edits

I tried an algebraic averaging thing, like starting at the Earth's surface, accel is 0.2 and then at 1 R, 6400 Km up where it would be accel at 1.2G minus 0.25g or .95 g averaged, and come out with a time to 6400 Km of 1515 seconds.*Originally posted by joe shmo***You are at 1.2 radii above the earths surface**[/b]

[b]EDIT: Nevermind, you are 1300 km above the surface at 1000 seconds, 0r 1.2 radii above the earths center

Using 1515 seconds with your method, I come out at 10,000 odd Km up.

What am I doing wrong? - 12 Feb '15 00:27 / 1 edit

That is 10,000 km from the*Originally posted by sonhouse***I tried an algebraic averaging thing, like starting at the Earth's surface, accel is 0.2 and then at 1 R, 6400 Km up where it would be accel at 1.2G minus 0.25g or .95 g averaged, and come out with a time to 6400 Km of 1515 seconds.**

Using 1515 seconds with your method, I come out at 10,000 odd Km up.

What am I doing wrong?**center**of the earth. To get how far up from the surface at 1515 seconds you have to subtract the radius of the Earth. 10,000 [km]-6400[km] = 3600[km]

And besides that, I think something in my formula is wrong...so act as if I hadn't shown it to you for now.

As for the averaging that you are doing: If the distance function were linear you could do that, but its not...so don't bother

Its like saying average the following 7 numbers

1, 2, 3 ,4 ,5, 6, 7

Avg = (1+2+3+4+5+6+7)/7 = 4

4 is (for lack of a better word) "geographically" in the middle of the set ( 3 numbers to its left, 3 numbers to its right)

Now take the squares of those same 7 numbers, the new set is

1,4,9,16,25,36,49

Avg = (1+4+9+16+25+36+49)/7 = 20

Clearly the mean is not "geographically" in the middle of the set anymore ( between 16 and 20, 4 numbers to its left, 3 to its right)

The distance function ignoring all this changing gravity stuff and having a constant acceleration is like the squared set, not the linear set. - 12 Feb '15 16:29

Also... you are on the right track with the "averaging" concept. My method for attacking the problem is basically the same idea, with the exception that I do something similar to that "averaging" a couple thousand times in 6400 km, where you did it once.*Originally posted by sonhouse***I tried an algebraic averaging thing, like starting at the Earth's surface, accel is 0.2 and then at 1 R, 6400 Km up where it would be accel at 1.2G minus 0.25g or .95 g averaged, and come out with a time to 6400 Km of 1515 seconds.**

Using 1515 seconds with your method, I come out at 10,000 odd Km up.

What am I doing wrong?

I thought my model was wrong because as time gets large, the body's acceleration should asymptotically approach 11.76 m/s^2. which it does in the calculation part (the excel spreadsheet). However, thinking about it...as long as the radius is growing in time that is necessarily going to happen.

Another problem is that the polynomial curve fit equation I found (I am now supposing) has no physical meaning, or the result is just flat out wrong. I tried to take its second derivative w/respect to time (dependent variable) and I get an equation that is a linear function of time,which means it diverges to ∞ as t grows, but it should approach a constant (11.76) as t grows . I could have just as easily fit the curve with a 6th order polynomial and its second derivative would have been a 4th order equation, again divergent to infinity as t grows. The whole thing has me quite confused...

Can anyone out there help clarify these issues?