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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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Have you given us the whole of the question? What is $\theta$? Are there any forces acting on this planet? –  Chris Eagle Mar 28 '12 at 14:29
    
I looks like i missed out a vital part, my apologies I will update the question now. –  Euden Mar 28 '12 at 15:01
    
Even though @Robjohn already answered, next time this may be better suited for Physics.SE. –  Joe Mar 28 '12 at 16:14
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1 Answer 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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Thankyou for this. However, i have a query. The second line of your computation after r^2=r.r, so that ........ Thus. I don't understand where the differential equations before the word thus come from? –  Euden Mar 29 '12 at 15:51
    
Take the derivative of each side of the previous equation with respect to $t$ and divide by $2$. Note that $\vec{r}$ is a vector and $r$ is a scalar ($r=|\vec{r}|$). –  robjohn Mar 29 '12 at 16:43
    
Okay thank you, One further question if you will: is $\theta(1,0,0) =0$ a fact? –  Euden Mar 29 '12 at 17:03
    
No, but since the plane of the motion is perpendicular to $(0,0,1)$ and the $0$-point for $\theta$ was not specified, I chose what seemed a reasonable direction for $\theta=0$ and made my assumption explicit. –  robjohn Mar 29 '12 at 17:11
    
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. –  robjohn Mar 29 '12 at 20:32
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