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Can anyone help me with the following mechanics question:

Information:

A Planet moves under the gravitational influence of a massive star, so that (ignoring the centre of mass) its motion is restricted to a plane, and its position vector and velocity vector have the form

$r = r$ $\Biggl(cos\theta,sin\theta,0 \Biggr)$, $\dot{r} = \dot{r} \Biggl(cos\theta,sin\theta,0 \Biggr) + r \dot{\theta}\Biggl(-sin\theta,cos\theta,0 \Biggr)$.(All column vectors)

respectively where $r$ is the radial distance and $\theta$ is the polar angle.

If $r(0) = (1,8,0)$ (Column vector) and $\frac{dr}{dt} = (4,2,0)$ (Column vector) are the initial position and velocity of the planet, calculate $r(0), \frac {dr}{dt}(0), \theta (0)$ and $\frac{d\theta}{dt}(0)$.

All help will be appreciated.

Thanks,

Euden

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    Even though @Robjohn already answered, next time this may be better suited for Physics.SE.2012-03-28

1 Answers 1

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I assume that you mean $\vec{r}(0)=(1,8,0)$ and $\frac{\mathrm{d}\vec{r}}{\mathrm{d}t}=(4,2,0)$ and that $r=|\vec{r}|$, that is, $ r=\sqrt{65}\tag{1} $ Note that $r^2=\vec{r}\cdot\vec{r}$ so that $r\frac{\mathrm{d}r}{\mathrm{d}t}=\vec{r}\cdot\frac{\mathrm{d}\vec{r}}{\mathrm{d}t}$. Thus, $\frac{\mathrm{d}r}{\mathrm{d}t}=\frac{1}{\sqrt{65}}(1,8,0)\cdot(4,2,0)$. That is, $ \frac{\mathrm{d}r}{\mathrm{d}t}=\frac{4}{13}\sqrt{65}\tag{2} $ We also have $r^2\frac{\mathrm{d}\theta}{\mathrm{d}t}\vec{n}=\vec{r}\times\frac{\mathrm{d}\vec{r}}{\mathrm{d}t}$, where $\vec{n}$ is the unit normal to the plane of motion. Thus, $\frac{\mathrm{d}\theta}{\mathrm{d}t}\vec{n}=\frac{1}{65}(1,8,0)\times(4,2,0)=-\frac{6}{13}(0,0,1)$. That is, $ \frac{\mathrm{d}\theta}{\mathrm{d}t}=-\frac{6}{13}\tag{3} $ I further assume that $\theta(1,0,0)=0$. Then, $\theta(1,8,0)=\arctan(8)$. That is, $ \theta=\arctan(8)\tag{4} $

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    The angle from periapsis cannot be computed simply from the position and velocity of the orbiting body. The shape of the orbit depends on the strength of the gravitational field.2012-03-29