# Calculating the Velocity of an object far far away

I saw a satellite the other day, it moved very fast through the night sky.

Now I know it took around 110 seconds for it to fully travel out of the night sky(out of my visual bounds assuming, that I'm observing from a flat area), how can I calculate an approximation of the velocity that this object was moving at?

Altitude is going to be a factor in this calculation so let's assume that the altitude is 40.000 feet just like an airplane. How would I go about solving this problem?

I also found this image which illustrates exactly what I'm trying to calculate: The Star in the illustration. Any help is appreciated, Thanks.

• Anything in orbit that low would instantly burn up, so unless you were watching the last seconds of a satellite's life... Also, you should note the altitude in the diagram is astronomical altitude, not to be confused with the aviation altitude you talk about.
– user10851
May 10, 2016 at 2:48
• i know, i also said "let's assume" May 10, 2016 at 2:58

The formula is

$$v = \varphi\cdot r/t$$

Where v is the velocity, φ the traversed angle in radians, r the distance and t the time, if the object is moving in a transversal direction. If it is approaching or receding you have to split the radial and transversal components with Pythagoras. In your example I get around 1140 ft/sec. • First of all, thanks for the answer and the formula really helped :) what was your angle? I estimated the angle to be around 110 degrees 1.1919 rads -> v = 1.919*40000/110 = 697 ft/sec May 10, 2016 at 4:31
• I thought 110 was the time in seconds, I estimated 180° with that time but that was just aimed over the thumb. May 10, 2016 at 4:35
• Oh I see my bad, apparently chose an angle close to 110 which might have caused confusion with the time, well the formula works perfectly, calculated the ISS: 1.919*400/110 = 6.9 km/sec, google says its 7.6 km/sec which is not far off. And the UFO's that I saw: if we assume we saw an UFO at 10000-20000 ft with a time distance of 3-4 seconds: 1.919*20000/3 = 12800 ft/sec hmm not bad but seems a bit slow, maybe it was just two birds crossing :/ May 10, 2016 at 4:44
• All you can calculate is its angular speed, $\frac{\partial \theta}{\partial t}$, to determine its actual speed you need to know how far away the object is, given by $r \frac{\partial \theta}{\partial t}$.
– jim
May 10, 2016 at 12:46