SPACE GEOMETRY

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Space-Math of Visibility[edit]

Can we see the ISS with an amateur telescope?
You can calculate the size of an object by measuring its angle if you know the distance of that object.
Take a look at this triangle:

1 - 1.png


A: is the visible full-lenght of the object
a: is the half-lenght of the object
b: is the distance of the object
α: is the half-angle of its angle

For calculation of the full-lenghts we do need theese equations:

1 - 2.png


2 - 1.png


and we get the exact formular with this:

2 - 2.png


For example:
according to NASA the MOON has a distance of approx 350000 km and its diameter is about 3000 km.

Calculation:
PHI = 2 * arctan (3000000 [meter] / (2*350000000) [meter])
PHI = 0,5 deg

It is not important how far the MOON really is.
In all observations the MOON appears with it's constant Angle of 0,5 deg. Therefore the angle of the MOON is our Reference.

Now:
The ISS: Lenght 100 meter, Distance 400 km.

PHI = 2 * arctan (100 [meter] / (2 * 400000) [meter])
PHI = 0,0143 deg


That means: If there is an "ISS" it is visible and the angle of the MOON is 35 times bigger then the angle of ISS.
If the ISS is going around in 90 minutes,
than it could be visible max 7,5 seconds in front of the MOON.
7,5 [sec] = 0,5 deg(moon) * ((90 * 60 [sec]) / 360 deg)