Integrating the inverse square law

In summary, the small planet moves in response to the gravitational force of the Sun, but its position function changes over time.
  • #1
haaj86
17
0
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 doesn't 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 textbook, please if the calculation is long and you can't type it then tell me where to find it.
 
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  • #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
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
Cantab Morgan said:
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]

jtbell said:
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
haaj86 said:
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
haaj86 said:
[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
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
Hootenanny said:
However, I suspect that missing off the dx was just a typo.

You are absolutely right, sorry about that.


Cyosis said:
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
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
Yes it works perfectly, thank you for all your help.


haaj86
 
  • #10
You're welcome.
 
  • #11
Hootenanny said:
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
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
Cantab Morgan said:
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:
Cyosis said:
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.
 

1. What is the inverse square law?

The inverse square law is a physical principle that states that the intensity of a physical quantity decreases in proportion to the square of the distance from the source. This law is commonly applied to the fields of physics, astronomy, and engineering.

2. How is the inverse square law used in science?

The inverse square law is used to describe the relationship between the intensity of a physical quantity, such as light or gravitational force, and the distance from its source. It is commonly used to calculate the brightness of light sources, the strength of electromagnetic fields, and the gravitational pull between two objects.

3. What is the formula for the inverse square law?

The formula for the inverse square law is I = k/d^2, where I is the intensity of the physical quantity, k is a constant, and d is the distance from the source. This means that as the distance from the source increases, the intensity decreases exponentially.

4. How does the inverse square law affect light and sound?

The inverse square law has a significant impact on the behavior of light and sound. As distance from the source increases, the intensity of both light and sound decreases. This means that objects appear dimmer and sounds become quieter the further away they are from the source.

5. What are some real-life examples of the inverse square law?

One common example of the inverse square law is the brightness of a light bulb. As you move further away from the light source, the light appears dimmer due to the decrease in intensity. Another example is the gravitational pull between two objects, such as the Earth and the Moon. The gravitational force decreases as the distance between the two objects increases.

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