Acceleration (d^2x/dt^2) as a function of x; x as a function of t.

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Hi, I'm Unit and I'm a grade 11 student. This is my first post so I thought I'd introduce myself.

I thought of this problem and I can't figure it out, regarding an acceleration which changes with x position. However, I'm trying to determine an equation (or a system of equations) describing the motion of a massive body with respect to time.

Imagine this: there is a body of big mass M (like the earth or something) and some initial D distance away, is a smaller, very less-massive body m. They are in space, far from any other forces. The moon is attracted towards the earth. Let's assume that the earth is rigidly fixed in place (I know that both the earth and the moon would move.)

[tex]F_g = m \frac{d^2 x}{dt^2} = \frac{GMm}{x^2} [/tex]

[tex]\frac{d^2 x}{dt^2} = GMx^{-2}[/tex]

The problem right away is that x is on both sides. Is there something like the opposite of implicit differentiation?
An idea I had immediately was to use a chain-rule-esque kind of approach:

[tex]\frac{da}{dx} \cdot \frac{dx}{dt}[/tex]

where

[tex]\frac{da}{dx} = -2GMx^{-3}[/tex].

And once I had da/dt, I could integrate three times and get x in terms of t.
But then the problem requires dx/dt, which is velocity, but the only known velocity is v = 0 when x = 0 and t = 0 and that is not helpful.

So my question is, what is the position of the moon as a function of time, where the only expression I can find is one for the acceleration, which is dependent on the x position?

p.s. This is not a homework question! I was looking at the moon one morning and I thought, what would happen if it stopped? (i.e. v = 0 with respect to the earth). I'm really sorry if this is not in the right forum but it's more of a calculus problem than a physics problem, even though it comes from thinking about physics. I hope you don't mind.
 
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Answers and Replies

  • #2
gabbagabbahey
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[tex]\frac{d^2 x}{dt^2} = GMx^{-2}[/tex]
The usual method of solving this DE is to utilize the chain rule by noting that

[tex]\frac{d^2 x}{dt^2}=\frac{dv}{dt}=\frac{dx}{dt}\frac{dv}{dx}=v\frac{dv}{dx}=\frac{GM}{x^2}[/tex]

Which gives you a separable first order differential equation for [itex]v=\frac{dx}{dt}[/itex]. You then solve for [itex]v[/itex] as a function of [itex]x[/itex] and obtain another separable 1st order DE for [itex]x[/itex].
 
  • #3
diazona
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And once I had da/dt, I could integrate three times and get x in terms of t.
But then the problem requires dx/dt, which is velocity, but the only known velocity is v = 0 when x = 0 and t = 0 and that is not helpful.
Integrating twice would give you velocity. "jerk" integrates to acceleration integrates to velocity integrates to position (yes "jerk" is the actual word for the derivative of acceleration ;-)

Although you don't need to take da/dt first, necessarily. A procedure I've seen used in several textbooks is to multiply by v = dx/dt on both sides:
[tex]v \frac{\mathrm{d}v}{\mathrm{d}t} = \frac{GM}{x^2}\frac{\mathrm{d}x}{\mathrm{d}t}[/tex]

[tex]\frac{\mathrm{d}}{\mathrm{d}t}\frac{v^2}{2} = -\frac{\mathrm{d}}{\mathrm{d}t}\frac{GM}{x}[/tex]
which I guess is more or less the same thing gabbagabbahey explained.
 
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Hi, thank you both for your kind replies,

I understand diazona's reply better, but I'm just unsure of how you integrated both sides with respect to different things. I was always under the impression that you could only integrate both sides with respect to the same thing, but what I see you did was:

L.S.:

[tex]\frac{d}{dt} \int v dv [/tex] with respect to v so it became [tex]\frac{d}{dt} \frac{1}{2} v^2 [/tex]

R.S.:

[tex]\frac{d}{dt} GM \int \frac{dx}{x^2}[/tex] with respect to x so it became [tex]-\frac{d}{dt} \frac{GM}{x}[/tex]

So really, am I actually allowed to integrate with respect to different things?

And now, (forgive me) am I allowed to simply remove the two time-derivatives? Or, do I integrate both sides with respect to time?

Thanks! :smile:
 

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