If I have a particle orbiting a central force $$F=-k/r^3$$ what is the shape of the orbit (the radius as a function of the angle)?

# [Physics] Shape of an inverse cube orbit

classical-mechanicshomework-and-exercisesorbital-motion

#### Related Solutions

Here is a snapshot from Fundamentals of Astrodynamics by Bate, Mueller, and White.

You do have to assume that you are working with the earth, while it is not mentioned in the problem.

You start by calculating the specific energy (ε) of the orbit given the velocities:

$$\epsilon=\frac{V^2}{2}-\frac{\mu}{r}$$

From there you calculate the semi-major axis (a) of the orbit:

$$\epsilon = -\frac{\mu}{2a} \therefore a = -\frac{\mu}{2\epsilon}$$

Next, you find the parameter (p) of the orbit:

$$r = \frac{p}{1+e\cos{v}}=\frac{p}{1+e\cos{90°}} = p$$

Finally, you can throw it all together into an equation from the definition of a conic section:

$$p = a(1-e^2) \therefore e = \sqrt{1-\frac{p}{a}}$$

When numbers are plugged in from the initial problem and from the characterization of earth, I got the final answer.

Your equation is an attempt to apply Newton's 2nd Law $F=ma$. The force is central so the acceleration is purely radial. In plane polar co-ordinates the radial component of acceleration has two terms: $\ddot r$ and $-r\dot\theta^2$. The 1st term is zero if the particle is constrained to move in a circle. The 2nd term is the centripetal acceleration.

In your equation, the $\ddot r$ term is missing. You clearly expect elliptical orbits, so $\ddot r \ne 0$. You also have the wrong sign for the $r\dot\theta^2$ term. The centripetal acceleration $r\dot\theta^2$ and the gravitational force $G\frac{Mm}{r^2}$ are both directed inwards, so they should have the same sign.

So the equation of motion which allows for elliptical motion should be

$m(\ddot r - r\dot\theta^2) = -\frac{GMm}{r^2}$.

The variable $\dot\theta$ can be eliminated using the fact that angular momentum is conserved $(mr^2\dot\theta=constant)$ since there is no tangential component of force. You then have a differential equation involving only $r$ and $t$. It does not have a simple solution. Alternatively you can remove the time-dependence and solve for $r$ as a function of $\theta$ using the substitution $u=\frac{1}{r}$. This does have a simple solution, which is the equation of an ellipse. You can find the details of the calculation on many websites, eg Farside Physics.

If you only want to find out how the speed $v$ varies with radius $r$, the solution is much simpler. You apply Conservation of Energy. Kinetic energy is $\frac12mv^2$ and gravitational potential energy is $-\frac{GMm}{r}$, so

$\frac12mv_0^2-\frac{GMm}{r_0}=\frac12mv^2-\frac{GMm}{r}$

Given the speed at any particular radius, you can calculate the speed at any other radius. However, this equation does not tell you the shape of the orbit.

## Best Answer

Using Lagrange's equation

$$\frac{d^2}{dt^2}\left(\frac{1}{r}\right)+\frac{1}{r}=\frac{-m r^2}{l^2}F\left(r\right)$$

And plugging in $$F\left(r\right)=\frac{-k}{r^3}$$

We get $$\frac{d^2}{dt^2}\left(\frac{1}{r}\right)+\frac{1}{r}=\frac{k m}{l^2 r}$$

Which simplifies to $$\frac{d^2}{dt^2}\left(\frac{1}{r}\right)+\frac{1}{r}\left(1-\frac{k m}{l^2}\right)=0$$

Assuming $$r\left(0\right)=r_p, r'\left(0\right)=0$$

We can solve equation (3) for r which gives us $$r\left(\theta\right)=r_p sec\left(\theta\sqrt{1-\frac{km}{l^2}}\right)$$