Integrating the inverse square law

  • Thread starter haaj86
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  • #1
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Hi, I am trying to solve the following problem (but note you can skip the details and look at the last equation and help me with the integral): a small planet (mass m) is found at rest relative to the Sun (mass M) with a separation distance [tex]x_{0}[/tex]. Consider now the small planet move in response to the gravitational force of the Sun (the sun is assumed so massive such that it doesnt move) what is the position function of the small planet as a function of time t? So,

[tex]m\frac{dv}{dt}=-\frac{GMm}{x^{2}}[/tex]

[tex]\int^{v}_{0}{mvdv}=-\int^{x}_{x_{0}}{\frac{GMm}{x^{2}}}[/tex]

[tex]\frac{1}{2}mv^{2}=GMm(\frac{1}{x}-\frac{1}{x_{0}})[/tex]

Therefore,
[tex]\frac{dx}{dt}= \pm[2GM(\frac{1}{x}-\frac{1}{x_{0}})]^{\frac{1}{2}}[/tex]

We take the minus sign because we know that the velocity is in the negative direction (the planet moves toward the sun), hence:
[tex]t=-\frac{1}{\sqrt{2GM}}\int^{x}_{x_{0}}{(\frac{1}{x'}-\frac{1}{x_{0}})^{-\frac{1}{2}}}dx'[/tex]
So far I haven't found this integral solved in any text book, please if the calculation is long and you can't type it then tell me where to find it.
 

Answers and Replies

  • #2
I'm not sure I followed your very first step. You may call me a stickler, but... what is the differential on the right hand side integral?
 
  • #3
jtbell
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It's much easier to use conservation of energy. The potential energy in this case is

[tex]U(x) = - \frac {GMm}{x}[/tex]

(where we define U = 0 at x = infinity). The total of potential plus kinetic energy is constant.
 
  • #4
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I'm not sure I followed your very first step. You may call me a stickler, but... what is the differential on the right hand side integral?
It is simply F=ma, where [tex]F=-\frac{GMm}{x^{2}}[/tex] and [tex]ma=m\frac{dv}{dt}[/tex]

It's much easier to use conservation of energy. The potential energy in this case is

[tex]U(x) = - \frac {GMm}{x}[/tex]

(where we define U = 0 at x = infinity). The total of potential plus kinetic energy is constant.
You are definitely right, I was going to get to equation three by using the fact that the total energy is constant, but I decided to start by integrating the e.o.m to show exactly what I mean by integrating the inverse square law. I am really stuck with the last integral and I am shocked to not being able to find it solved. The closest solution I found was to assume that [tex]x_{0}[/tex] is so large such that we can ignore the [tex]1/x_{0}[/tex] term and approximate the integral. But I really want to find the exact solution as x(t)= ....
 
  • #5
Hootenanny
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It is simply F=ma, where [tex]F=-\frac{GMm}{x^{2}}[/tex] and [tex]ma=m\frac{dv}{dt}[/tex]
I believe he was referring to the right hand side of
[tex]m\frac{dv}{dt}=-\frac{GMm}{x^{2}}[/tex]
which does not appear to have a variable of integration. However, I suspect that missing off the dx was just a typo.
 
  • #6
Cyosis
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Defining a=1/x0 the integral can be written as:

[tex] \int \frac{1}{\sqrt{\frac{1}{x}-a}}\, dx =\int \frac{\sqrt{x}}{\sqrt{1-ax}}\, dx[/itex]

Now use the substitution [itex]u=\sqrt{1-ax}[/itex]. Ignoring constants you should arrive at an integral of the form [tex]\int \sqrt{1-u^2}\,du[/tex] which is easily solved by using another substitution, u=sin(z).

Edit: After some work I got to the following expression. Barring any mistakes it doesn't look like you can solve this for x.
[tex]
t=-\frac{1}{\sqrt{2GM}}\int^{x}_{x_{0}}{(\frac{1}{x}-\frac{1}{x_{0}})^{-\frac{1}{2}}}dx= \frac{x_0 \sqrt{x_0}}{\sqrt{2GM}} \left(\sqrt{\frac{x}{x_0}(1-\frac{x}{x_0})}+\arcsin(\sqrt{1-\frac{x}{x_0}})\right)
[/tex]
 
Last edited:
  • #7
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However, I suspect that missing off the dx was just a typo.
You are absolutely right, sorry about that.


Defining a=1/x0 the integral can be written as:

[tex] \int \frac{1}{\sqrt{\frac{1}{x}-a}}\, dx =\int \frac{\sqrt{x}}{\sqrt{1-ax}}\, dx[/itex]

Now use the substitution [itex]u=\sqrt{1-ax}[/itex]. Ignoring constants you should arrive at an integral of the form [tex]\int \sqrt{1-u^2}\,du[/tex] which is easily solved by using another substitution, u=sin(z).

Edit: After some work I got to the following expression. Barring any mistakes it doesn't look like you can solve this for x.
[tex]
t=-\frac{1}{\sqrt{2GM}}\int^{x}_{x_{0}}{(\frac{1}{x}-\frac{1}{x_{0}})^{-\frac{1}{2}}}dx= \frac{x_0 \sqrt{x_0}}{\sqrt{2GM}} \left(\sqrt{\frac{x}{x_0}(1-\frac{x}{x_0})}+\arcsin(\sqrt{1-\frac{x}{x_0}})\right)
[/tex]

Thank you so much, I worked through it following your steps and I got a very similar answer which is

[tex]t=-\frac{1}{\sqrt{2GM}}\int^{x}_{x_{0}}{(\frac{1}{x}-\frac{1}{x_{0}})^{-\frac{1}{2}}}dx= \frac{x_0 \sqrt{x_0}}{\sqrt{2GM}} \left(\frac{1}{2}
sin(2arcsin(\sqrt{1-\frac{x}{x_{0}}})+\arcsin(\sqrt{1-\frac{x}{x_0}})\right)[/tex]

I am not sure if this is what you basically got and you managed to reduce the first term to the form you have in your solution or that I have basically made a mistake somewhere. If you are willing to help me further to get it to the form of your solution then please let me know so that I can type my full solution and you can help me to spot the mistake I made. But for now, regardless whether I got the first term right or not the fact will remain that I will not be able solve it for x, so you have answered my main question, thank you for your help.
 
  • #8
Cyosis
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Your answer is exactly the same. Putting [tex]u=\arcsin(\sqrt{1-\frac{x}{x_0}})[/tex] for convenience. [tex]\frac{1}{2} \sin(2u)=\cos(u) \sin(u)=\sqrt{1-\sin^2 u} \sin u[/tex]. If you fill in u now you will see that the arcsin cancels and you will get the same answer as me.
 
  • #9
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Yes it works perfectly, thank you for all your help.


haaj86
 
  • #10
Cyosis
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You're welcome.
 
  • #11
However, I suspect that missing off the dx was just a typo.
Please look again, it's not a typo. Can you see what I was gently driving at? The differential is not dx. It is actually dt. This attempt to solve the differential equation fails, and produces such an intractable integral, because of this error in the very first step.
 
  • #12
Cyosis
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You missed the chain rule [tex] \frac{dv}{dt}=\frac{dv}{dx} \frac{dx}{dt}=v \frac{dv}{dx}[/tex]. So the differential is dx not dt.
 
  • #13
Hootenanny
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Please look again, it's not a typo. Can you see what I was gently driving at? The differential is not dx. It is actually dt. This attempt to solve the differential equation fails, and produces such an intractable integral, because of this error in the very first step.
If it is not a typo, then it is just plain wrong for the reason that Cyosis points out:
You missed the chain rule [tex] \frac{dv}{dt}=\frac{dv}{dx} \frac{dx}{dt}=v \frac{dv}{dx}[/tex]. So the differential is dx not dt.
 
  • #14
Aha. I am wrong. I was going to suggest multiplying both sides by v, but I now see more carefully that you have already done that.
 

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