0
$\begingroup$

I try to figure out if a satellite would be visible. Is it possible to calculate the theoretical line of sight having the following variables?:

Human

Location: 49.836573, 24.083819, Height: 27 meters above ground

Satellite

LATITUDE: -0.04

LONGITUDE: 100.51

ALTITUDE [km]: 35779.65

  • 0
    Have you searched for satellite visibility calculators? – 2017-01-24
  • 0
    Says its on right on horizon, 0.0 yet unable to find it on the dish. – 2017-01-24
  • 0
    Maybe it is possible with only those variables. I still suggest asking physics on maybe physics.SE. – 2017-01-24
  • 0
    Also, note that precision doesn't equal to the amount of decimal places you have. – 2017-01-24
  • 1
    If it is right on the horizon there is often stuff on the ground that scatters the signal. If you have a calculation you should show it. It is easier to check numbers than to produce them. – 2017-01-24

1 Answers 1

0

I'd convert your positions to Carstesian coordinates, and then scale these positions down to the unit sphere. This can be achieved using a function

$$f(\lambda,\phi,h)= \begin{pmatrix} \tfrac1a\cdot\bigl(N(\phi)+h\bigr)\cos\phi\cos\lambda \\ \tfrac1a\cdot\bigl(N(\phi)+h\bigr)\cos\phi\sin\lambda \\ \tfrac1b\cdot\bigl(\frac{b^2}{a^2}N(\phi)+h\bigr)\sin\phi \\ \end{pmatrix} \quad\text{with } N(\phi)=\frac{a^2}{\sqrt{a^2\cos^2\phi+b^2\sin^2\phi}}$$

where $a$ and $b$ are the equatorial resp. polar radius of the earth. So plugging in your numbers you'd get

\begin{align*} h=f(49.836573°, 24.083819°, 27\,\mathrm m)&=\begin{pmatrix} 0.58915599 \\ 0.69807533 \\ 0.40693307\end{pmatrix} \\ s=f(-0.04°, 100.51°, 35779.65\,\mathrm{km})&=\begin{pmatrix} -1.206 \\ 0.001 \\ 6.517\end{pmatrix} \end{align*}

Now you want to know whether the line connecting $h$ to $s$ is above or below the horizon. To do that, compute the dot product between $h$ and $s-h$. That dot product is proportional to the sine of the elevation.

$$e=\arcsin \frac{\langle h,s-h\rangle}{\lVert h\rVert\cdot\lVert s-h\rVert} = 8.5°$$

So yes, the sattelite should be a bit over $8°$ above the horizon. The measurement was taken out on the sphere, and correcting for the ellipsoid shape of the earth may change that slightly, but I doubt that effect will be noticable compared to other inaccuracies.