Gravity where acceleration is changing

[okkvlt]

I have a question about gravity:

How do i formulate an equation that incorporates the change in distance between two objects- in other words, where the acceleration due to gravity is changing as the distance changes, instead of simply where the acceleration is held constant. Here's how far I've gotten:

y=-.5ax^2+sx+d
where y is location(distance) and x is time. s is initial velocity and d is initial location and a is acceleration due to gravity.
But just one simple change and things become much more complicated.
Suppose i want to include the change in acceleration.
I use a=gm/d^2 where a is the acceleration, g is the gravitational constant, m is the mass of the planet, and d is distance.
Which means d^2y/dx^2=gm/y^2, so y'' is a function of y and not a function of x. That's my problem. How do i integrate the derivative when the derivative is a function of the dependent variable? I need to know the function of x that equals y in order to integrate, but i need to integrate to find the function of x that equals y. So I am kind of left going in circles trying to solve an impossible problem.

I can't figure out how to integrate this.
I could try integrating to find dx/dy, which would be the inverse of velocity(time/distance instead of distance/time), and manipulate that to get something meaningful, but I am unsure whether that's even possible.
I don't think i can simply replace a with gm/y^2, but I am not sure because this is really confusing me. when i differentiate implicitly to find velocity after replacing a with gm/y^2, i get dy/dx=f[x,y]. Now i have to differentiate velocity=dy/dx to find acceleration=d^2y/dx^2 and verify whether the second derivative equals gm/y^2. But how do i find the second derivative when the first derivative is a function of both variables? this seems to be related to the problem that y'' is a function of y, and y is a function of both x and y''. That's why i don't think i can simply replace a with gm/y^2. The problem is, whenever i would evaluate y at a moment in x after replacing a with gm/y^2, it will be as if the acceleration at that moment in time has been the acceleration at all previous times. But then again, I am not sure.
I think what's happening here is the function itself actually changes into another function as the variables change. This is really confusing me.

The only way i figure i can solve this is by splitting space(splitting time would be simpler because i wouldn't have to solve a million quadratics, but i don't think incrementalizing time would be as accurate.) into increments. I would solve the first problem, then plug the variables and its derivatives at the end of that increment into the next problem, adjust for acceleration using gm/y^2, solve the next problem, and then keep repeating that over and over until i go insane from mindless calculations, knowing that i haven't actually solved the dynamics of the problem. there's no freaking way I am going to attempt that.

 

[Hootenanny]

The equation

y^{\prime\prime}= \frac{gm}{y^2}

is a special case of a second order non-linear ode, which can be solved quite straight forwardly to determine y' (speed). However, one cannot solve this ode to obtain a closed form solution for y (distance) as a function of time. One can however, obtain a closed form solution for x (time) in terms of y (distance).

Let

\frac{dy}{dx} = v\left(y\right)

Then by the chain rule

\frac{d^2 y}{dx^2} = \frac{d}{dx}v = \frac{dv}{dy}\frac{dy}{dx} = \frac{dv}{dy}v

Hence the ode becomes,

\frac{dv}{dy}v = \frac{gm}{y^2}

Which is first order and separable. We can integrate to determine the velocity as a function of distance.

\Rightarrow v\left(y\right) = \frac{dy}{dx} = -\frac{gm}{y} + c

This first order ode is again separable,

\frac{dy}{dx}\frac{1}{\left(gm/y\right)+c} = -1

\frac{dy}{dx}\left(\frac{1}{c} - \frac{gm}{c\left(gm + cy\right)}\right) = -1

Integrating

\Rightarrow \frac{y}{c} - \frac{gm}{c^2}\ln\left|gm+cy\right| = c^\prime-x

Here you should be able to see that one cannot solve this equation to obtain y=y(x). However, we have determined time as a function of distance as promised.

I hope that this was helpful.

 

[ibc]

Hence the ode becomes,

\frac{dv}{dy}v = \frac{gm}{y^2}

Which is first order and separable. We can integrate to determine the velocity as a function of distance.

\Rightarrow v\left(y\right) = \frac{dy}{dx} = -\frac{gm}{y} + c

didn't you mean, \frac{v^2}{2} on the left side?

 

[Hootenanny]

didn't you mean, \frac{v^2}{2} on the left side?

Whoops! Thanks for pointing that out, I'll correct my post above now.

 

[Hootenanny]

Correcting my post #2 above (I thought it looked a little too nice),

\Rightarrow v\left(y\right) = \frac{dy}{dx} = \sqrt{2c -\frac{2gm}{y}}

This first order ode is again separable,

\int\frac{dy}{ \sqrt{2c -\frac{2gm}{y}}} = \int dx

Which integrates (horrendously) to,

x = \frac{\sqrt{cy}\left(cy-gm\right) + gm\ln\left|2\left(\sqrt{cy} + \sqrt{cy-gm}\right)\right|\sqrt{-(gm) + cy}}{c^{3/2}\sqrt{2}\sqrt{cy - gm}}

Which definitely doesn't have a closed form solution for y=y(x).

My apologies again for the mistake.

 

[schroder]

Which integrates (horrendously) to,

x = \frac{\sqrt{cy}\left(cy-gm\right) + gm\ln\left|2\left(\sqrt{cy} + \sqrt{cy-gm}\right)\right|\sqrt{-(gm) + cy}}{c^{3/2}\sqrt{2}\sqrt{cy - gm}}

Which definitely doesn't have a closed form solution for y=y(x).

QUOTE]

That integral is compelling reason to treat g as a constant, at least over short distances, even though we know it is variable! Over long distances, we can get good approximations by taking an average value of g, but it would be much better if we had a neat compact expression to calculate g even as distance varies. It seems to me that a simpler expression can be written, as there are many instances where one variable mutually affects another. A case in point is air resistance affecting the velocity of a falling body, and of course the velocity affects air resistance. Yet we have no trouble writing a simple DE which takes this all into account. Why can’t we do the same for gravitational potential energy and distance, or can we?