# Creating a classroom tool for satellite location

I would like to create a sun-dial type project for seventh graders which they can use to locate their favorite geostationary satellite. These are the kids who will probably hear about satellites for the first time and I want a way to engage them. I am hoping to make something like the following:

Here, the leaning side of the triangle should point to the satellite.

Let's say my target satellite is Bangabandhu-1. The parameters are as follows.

Let's say the base has a diameter of 6 inch. I am doing it assuming I am located at Dhaka, Bangladesh. The distance between Dhaka and Jakarta is 5345.8 km. The altitude is about 35789 km. So, the angle of the leaning side will be 81.505 degree according to this calculation.

Am I correct?

• What kind of distance do you mean with the distance between Dhaka and Jakarta is 5345.8 km? Google maps tell me 3781 km. But we are on the surface of Earth. So do you think of the straight line distance or the great circle distance? Both cities are not at the equator. So where is the right triangle between the cities and the satellite?
– Uwe
Aug 17, 2022 at 23:00
• What do you mean by the 'leaning side' - the angle of the metal pointer or the angle of the shadow? Aug 18, 2022 at 15:12
• @Andrew, angle of the metal pointer Aug 19, 2022 at 17:00
• Visibly locating satellites in GEO without a small telescope is challenging. They sometimes flare but usually are probably to dim. See answers and comments on Are any geosynchronous satellites visible with the naked eye? and perhaps also Has there ever been a deliberate "Find A Satellite" challenge? Actually, could there be? Oh, and in Astronomy SE: How bright are geostationary satellites due to reflected sunlight?
– uhoh
Sep 18, 2022 at 22:27
• JPL Horizons will compute the Alt/Az coordinates for you. After that, it's just a matter of creating a pointer at the right angle and direction. Jan 18 at 17:17

You'll need to calculate two angles: azimuth and elevation. And you have to keep in mind that the Earth is a sphere (an ellipsoid to be precise but I think for our purposes "sphere" is close enough as an approximation)

Since you only want geostationary satellites (which is a smart decision because it's making the whole thing a whole lot easier) we only need latitude and longitude of the observer as well as the satellite.

The azimuth angle is the direction in which we have to look. Geometrically speaking it's the angle in the horizontal plane between north and the plane that runs through the center of Earth, the observer and the satellite (here, an image would be helpful but I can't create nice illustrations).

We need the difference in longitude between the observer and the satellite, let's call it $$G$$

And we need the difference in latitude. Since the latitude of a satellite in GEO is always close to zero, we can just take the latitude of the observer. So latitude $$L$$

Azimuth: $$\alpha = 180° + \arctan\left(\frac{\tan(G)}{\tan(L)}\right)$$

Now to the elevation. That's basically "how steep do we have to look up".

$$\beta = \arctan\left( \frac{\cos(G) \cdot \cos(L) - 0.15}{\sqrt{1-\cos^2(G) \cdot \cos^2(L)}}\right)$$

With those two angles you can build your satellite pointer...

• I get the feeling I messed up the formula somehow (eitzher the math or while putting it in Mathjax). I'd be happy if a second or third pair of eyes could check if they are correct. Aug 19, 2022 at 6:29
• Why not provide the results of a few sample calculations for a given sat from selected global cities? That would show both you and readers that this works. Aug 19, 2022 at 14:07