# How long will it take to fall?

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• Aug 29th 2006, 04:20 AM
CaptainBlack
Quote:

Originally Posted by ThePerfectHacker
There is a simple formula you can use.
If you are standing at $s_0$ and you throw an upward or downward (this makes it positive or negative) with accelaration of gravity $g$. Then the function that tells you the height after $t$ seconds is:
$S(t)=-\frac{1}{2}gt^2+v_0 t+s_0$
Of course in a vacuum.
---
Topsquark, may I ask you two questions?

It is true that when an object travels further up its effect from gravity diminishes. So what equation describes this, a cubic?

If the initial speed is not too great it is an arc of a sinusoid.

RonL
• Aug 29th 2006, 05:54 AM
topsquark
Quote:

Originally Posted by ThePerfectHacker
There is a simple formula you can use.
If you are standing at $s_0$ and you throw an upward or downward (this makes it positive or negative) with accelaration of gravity $g$. Then the function that tells you the height after $t$ seconds is:
$S(t)=-\frac{1}{2}gt^2+v_0 t+s_0$
Of course in a vacuum.
---
Topsquark, may I ask you two questions?

It is true that when an object travels further up its effect from gravity diminishes. So what equation describes this, a cubic?

Assuming Newtonian gravity (which should suffice) for an object launched from the surface of the Earth we get the equation
$0 = (1/2)mv^2 - (1/2)mv_0^2 - \frac{GmM}{R+y} + \frac{GmM}{R}$
from energy conservation, where v0 is the initial (launch) speed, M is the mass of the Earth, R is the radius of the Earth, G is the Universal Gravitation constant, m is the mass of the object, and y is the height above the Earth. (Note: $g = \frac{GM}{R^2}$) Or we could use Newton's 2nd:
$\frac{GmM}{(R+y)^2} = ma$ (Positive downward.)

Neither equation is trivial to solve for y(t). I can't find my notes on how I did it, but I seem to recall solving the energy equation for y(t) at some point in the past. However, I don't recall getting any great inspiration from it either.

-Dan

The more I stare at the energy equation, the more I doubt I solved it. I may have solved a first or second approximation where y is small.
• Aug 29th 2006, 06:13 AM
topsquark
Quote:

Originally Posted by Quick
Thanks dan, you get a +rep :). I have no idea what how to calculate most of the stuff but it's always nice to see what would be involved...

Thank you, though I can't claim credit...most of it came from one of my textbooks.

I was thinking that if you really wanted to do the project you could still model the equations even if you didn't have the derivations. The quadratic resistance case is a pretty ugly (oxymoron?) equation but easy enough to program on a fancy calculator, or in Excel if you wanted to make a graph.

-Dan
• Aug 29th 2006, 07:24 AM
CaptainBlack
Quote:

Originally Posted by topsquark
Assuming Newtonian gravity (which should suffice) for an object launched from the surface of the Earth we get the equation
$0 = (1/2)mv^2 - (1/2)mv_0^2 - \frac{GmM}{R+y} + \frac{GmM}{R}$
from energy conservation, where v0 is the initial (launch) speed, M is the mass of the Earth, R is the radius of the Earth, G is the Universal Gravitation constant, m is the mass of the object, and y is the height above the Earth. (Note: $g = \frac{GM}{R^2}$) Or we could use Newton's 2nd:
$\frac{GmM}{(R+y)^2} = ma$ (Positive downward.)

Neither equation is trivial to solve for y(t). I can't find my notes on how I did it, but I seem to recall solving the energy equation for y(t) at some point in the past. However, I don't recall getting any great inspiration from it either.

-Dan

The more I stare at the energy equation, the more I doubt I solved it. I may have solved a first or second approximation where y is small.

You shouldn't need to solve it, it is a projection of a conic onto the "up"
axis, because it is a degenerate inverse square law orbit.

RonL
• Aug 29th 2006, 10:21 AM
topsquark
Quote:

Originally Posted by CaptainBlack
You shouldn't need to solve it, it is a projection of a conic onto the "up"
axis, because it is a degenerate inverse square law orbit.

RonL

(Chuckles) You know, because it's a projectile motion problem rather than a classical orbital problem I never thought of that... :o Good call!

-Dan
• Aug 29th 2006, 11:02 AM
CaptainBlack
Quote:

Originally Posted by topsquark
(Chuckles) You know, because it's a projectile motion problem rather than a classical orbital problem I never thought of that... :o Good call!

-Dan

Yes I think its quite neat :cool:

RonL
• Aug 29th 2006, 12:21 PM
topsquark
First approximation to the energy equation
$0 = (1/2)mv^2 - (1/2)mv_0^2 - \frac{GmM}{R+y} + \frac{GmM}{R}$

Note that this is the first integral of the Newton's 2nd equation in the previous post where this appears. (5 posts ago)

Solve this equation for v:
$v = \pm \sqrt{\frac{2GM}{R+y}+\left ( v_0^2-\frac{2GM}{R} \right ) }$

v is positive for an object moving upward and negative for moving downward, so we can split the solution into two pieces for upward and downward motion.

For simplicity, take $a = 2GM$ (NOT the acceleration!) and $b = v_0^2-\frac{2GM}{R}$. This equation is then
$v = \pm \frac{dy}{dt} = \pm \sqrt{\frac{a}{R+y}+b }$

Take the first approximation to this equation where y is considered to be small. (ie. keep y to first order, y << R.) This equation becomes an approximate linear first order equation:
$\frac{dy}{dt} \pm \frac{a}{2R^2 \sqrt{a+\frac{b}{R}}} y = \pm \sqrt{\frac{a}{R}+b}$

Letting $c = \frac{a}{2R^2 \sqrt{a+\frac{b}{R}}}$ and $d = \sqrt{\frac{a}{R}+b}$ we see that
$\frac{dy}{dt} \pm cy = \pm d$
has the solution
(*) $y(t) = Ae^{Bt}+D$

Consider now the upward motion solution. (Use the "+" signs.) Inserting this into equation * we get that $B = -c$ and $D = \frac{d}{c}$, so
$y(t) = Ae^{-ct} + \frac{d}{c}$.

Now $v = \frac{dy}{dt}(0) = v_0$. Thus $A = - \frac{v_0}{c}$.

$y(t) = - \frac{v_0}{c}e^{-ct} + \frac{d}{c}$.

Back-subbing to get to the original variables:
$y(t) = - \frac{2R^2v_0 \sqrt{a+\frac{b}{R}}}{a} exp \left [ -\frac{a}{2R^2 \sqrt{a+\frac{b}{R}}}t \right ]$ + $\frac{2R^2 \sqrt{\frac{a}{R}+b} \sqrt{a+\frac{b}{R}}}{a}$

$y(t) = - \frac{R^2v_0 \sqrt{2GM+\frac{v_0^2-\frac{2GM}{R}}{R}}}{GM}$ $exp \left [ -\frac{GM}{R^2 \sqrt{2GM+\frac{v_0^2-\frac{2GM}{R}}{R}}}t \right ]$ + $\frac{R^2v_0 \sqrt{2GM+\frac{v_0^2-\frac{2GM}{R}}{R}}}{GM}$

$y(t) = \frac{Rv_0 \sqrt{2GMR^2+Rv_0^2-2GM}}{GM}$ $\left ( 1 - exp \left [ -\frac{GM}{R \sqrt{2GMR^2+Rv_0^2-2GM}}t \right ] \right )$

To get the rest of the motion, solve y(T) for maximum height, then use the solution to the equation $\frac{dy}{dt} - \frac{a}{2R^2 \sqrt{a+\frac{b}{R}}} y = - \sqrt{\frac{a}{R}+b}$ with the initial condition $\frac{dy}{dt}(T) = 0$. (Or simply consider that the equation for y(t) for the whole motion will be symmetric about t = T.)

As I mentioned previously, it doesn't give me any great inspirations, though it's a nice exercise in applied differential equations.

-Dan

I was just thinking. I did this analysis a few years ago and it never occured to me until now that the solution needs to be forced to be continuous about the time t = T. That could add some interesting complications in solving for the downward motion equation. I'm not going to bother with it. Consider this as a problem "for the interested student." :)
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