1. ## dynamics

*theta=(-)
A particle of mass m moves under an attractive central force mk/r^a where r is the radial distance from the force centre O.
assuming that the radial and transverse components of accleration in polar coordinated (r, (-)) are (r''-r*(-)'^2) and (2r'*(-)' + r*(-)'') respectively, show that the differential equation for the orbit is,
d^2u/d(-)^2 = k/h^2 * u^(a-2)
where u=1/r and h=r^2 * (-)

sorry about the notations, i wasn't sure how to present the equation on this.

I am having trouble with this and a similar question. I am not sure how to derive the differential equation for the radial and transverse components.
cheers
p.s i hope this is the right sub-forum to post in

2. Hi

The 2 equations are

(1) $m \left(\ddot{r}-r \dot{\theta}^2\right) = -\frac{mk}{r^a}$

(2) $2 \dot{r} \dot{\theta} + r \ddot{\theta} = 0$

Then

$\frac{du}{d\theta} = \frac{du}{dt} \cdot \frac{dt}{d\theta} = \frac{d \left(\frac{1}{r}\right)}{dt} \cdot \frac{1}{\dot{\theta}} = -\frac{\dot{r}}{r^2} \cdot \frac{1}{\dot{\theta}} = -\frac{\dot{r}}{r^2 \dot{\theta}}$

$\frac{d^2u}{d\theta^2} = \frac{d}{d\theta} \left(\frac{du}{d\theta}\right) = \frac{d}{d\theta} \left(-\frac{\dot{r}}{r^2 \dot{\theta}}\right) = \frac{d}{dt} \left(-\frac{\dot{r}}{r^2 \dot{\theta}}\right) \cdot \frac{dt}{d\theta} = -\frac{\ddot{r}r^2 \dot{\theta}-\dot{r} \left(2r\dot{r} \dot{\theta}+r^2 \ddot{\theta}\right)}{r^4 \dot{\theta}^2} \cdot \frac{1}{\dot{\theta}}$

From equation (2) we know that the parenthesis is equal to 0

From equation (1) we can substitute $\ddot{r} = r \dot{\theta}^2 - \frac{k}{r^a}$

Therefore

$\frac{d^2u}{d\theta^2} = -\frac{\ddot{r}r^2 \dot{\theta}}{r^4 \dot{\theta}^3} = -\frac{\ddot{r}}{r^2 \dot{\theta}^2} = -\frac{r \dot{\theta}^2 - \frac{k}{r^a}}{r^2 \dot{\theta}^2} = -\frac{1}{r}+\frac{k}{r^{a+2} \dot{\theta}^2} = -u + \frac{k}{r^4 \dot{\theta}^2} \cdot u^{a-2}$

This is not exactly what you need to find so I may have missed something ...