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Suppose that the gravitational force is not given by the inverse-square law, and instead is $$ F_{grav}=(\frac{A}{r^{2}}+\frac{B}{r^{4}})\hat{r}, $$ where A and B are real constants. Derive the equations of motion and determine whether angular momentum is conserved. By using the substitution r=1/u, transform the radial component of the equations of motion into a differential equation for u as a function of the angular coordinate thetha. Determine for what values of A, B and integration constants stationary solutions exist and whether they are stable.

I have been having trouble with this question for a few days now and not sure if what I did is correct. Would really appreciate any help, thanks!

Samad Miah
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3 Answers3

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It is rather easy to show that angular momentum is conserved. The rate of change of angular momentum $\mathbf{L}$ is the torque ($\mathbf{\tau}$):

$$ \frac{d \mathbf{L}}{dt} = \frac{d \mathbf{}}{dt} (\mathbf{r} \times \mathbf{p}) = \frac{d \mathbf{\mathbf{r}}}{dt} \times \mathbf{p} + \mathbf{r} \times \frac{d \mathbf{\mathbf{p}}}{dt} = \mathbf{r} \times \mathbf{F} = \mathbf{\tau} $$

since $\mathbf{F}$ is along $\hat{r}$, this is zero. Note also that $\frac{d \mathbf{\mathbf{r}}}{dt} \times \mathbf{p}$ is zero because velocity $\frac{d \mathbf{\mathbf{r}}}{dt} $ is parallel to momentum $\mathbf{p}$.

This is actually true for all central forces (in case one is not familiar with central forces, any force which its magnitude depends only on the distance of the object to a center is central force).

The other way is to write down equations of motion in polar coordinates (which is the preferred coordinates system in case of central forces due to symmetry). You just need to take time derivatives from $\mathbf{r} = r \hat{r}$ to find:

$$ \frac{\mathbf{F}}{m} = \mathbf{a} = (\ddot{r}-r\dot{\theta}^2) \hat{r} + (2\dot{r}\dot{\theta}+r^2\ddot{\theta}) \hat{\theta} $$

since $\theta-$component of a central force is zero, we have:

$$ 2\dot{r}\dot{\theta}+r^2\ddot{\theta} = \frac{1}{r}\frac{d}{dt}(r^2\dot{\theta}) = 0 \Rightarrow m r^2\dot{\theta} = \textrm{constant} = mr (r\dot\theta) = mr\omega = L $$

To write down equation of motion in terms of $u$ ($u=r^{-1}$), just replace $r$ by $u$ in the $r-$component of $\mathbf{F}$:

$$ \ddot{r}-r\dot{\theta}^2 = \frac{A}{r^{2}}+\frac{B}{r^{4}} $$

In case you wonder how to deal with $\dot{\theta}$, just rewrite it in terms of constant $L$ (physics note: this gives you an effective potential).

Mahdi
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    To add more detail: therefore the formula for the force is irrelevant, we only need to know that the force parallel to the distance vector. +1! – Rory Daulton Aug 04 '15 at 15:44
  • If you can, could you show me exactly how to do the substitution and then forming the differential equation. Still finding it confusing, thanks! – Samad Miah Aug 13 '15 at 20:10
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To get the equation of motion, you use acceleration components in polar form.

Along the radius vector, $F=m(\ddot{r}-r\dot{\theta}^2)$

Tangentially, there is no force so $2\dot{r}\dot{\theta}+r^2\ddot{\theta}=0$ $$\Rightarrow\frac 1r\frac{d}{dt}(r^2\dot{\theta})=0$$ $$\Rightarrow r^2\dot{\theta}=h, \text {a constant}$$ Which also shows angular momentum per unit mass is conserved.

This enables you to do the substitution, so that to start with, $$\frac{dr}{dt}==-\frac{1}{u^2}\frac{du}{d\theta}\frac{d\theta}{dt}=-h\frac{du}{d\theta}$$

And so on...

David Quinn
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To David,

In the question when you make the substitution it says to derive a differential equation, and find the constants A and B. Could you please show how you can do that?

Samad could you get in touch with me

Keplerhimself
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