Ok, so I've got this capsule in an orbit, and I want to know when I'll be in contact with it, and for how long.

[Dman979]

It's in a 28.4887 degree orbit, 178x178 km orbit. Right now, it's at 28.4887° N, 80.5728° W. If I did the math right, it has an orbital period of 87.91 minutes, or 5274.6 seconds.

I have a radio station at 28^oN, 80.5728° W. It has a magical antenna at ground level, with no atmosphere to worry about. We don't have to worry about signal strength, or anything like that- only how long the capsule has line of sight to that magical antenna.

What is the formula to figure out how long I'll be in radio contact? What other information would i need to figure this out?

[insert_name]

you have something in orbit of earth irl, or are you playing a modded ksp install

[cubinator]

45 minutes ago, insert_name said:

you have something in orbit of earth irl, or are you playing a modded ksp install

Based on the orbital period and altitude, I would assume it's in orbit of a hypothetical Earth with no atmosphere.

Since the orbit is circular (whew!) we don't have to worry about how fast the capsule is going. Your ground antenna has a half-sphere where it can see the sky. You could figure it out by drawing a tangential plane that crosses Earth at your antenna, seeing where the satellite's orbit intercepts the plane and where it exits, finding the central angle, and figuring out what fraction of the total orbit it is. Unfortunately, I could maybe figure that out in 2D but not in 3D. Sorry.

BTW, I think you need the longitude of ascending node or whatever it's called to be able to know at what longitude it crosses the equator.

Edited March 29, 2016 by cubinator

[Dman979]

16 minutes ago, cubinator said:

Based on the orbital period and altitude, I would assume it's in orbit of a hypothetical Earth with no atmosphere.

Since the orbit is circular (whew!) we don't have to worry about how fast the capsule is going. Your ground antenna has a half-sphere where it can see the sky. You could figure it out by drawing a tangential plane that crosses Earth at your antenna, seeing where the satellite's orbit intercepts the plane and where it exits, finding the central angle, and figuring out what fraction of the total orbit it is. Unfortunately, I could maybe figure that out in 2D but not in 3D. Sorry.

BTW, I think you need the longitude of ascending node or whatever it's called to be able to know at what longitude it crosses the equator.

I'd imagine that the longitude would be changing, based on a 0^oN0^oW frame of reference. Based on an inertial frame of reference (say, the stars, roughly speaking), it wouldn't matter, I think.

Let's assume that for the little amount of time it's in Line of sight, that the earth isn't moving.
To transcribe 2D to 3D, couldn't you use the relative height above the horizon? For a station right below the orbit, that would be 90 degrees above the horizon, and 180km. For a station 30^o away from the orbit, couldn't you use some variant of the pythagoram theorem to find out how far above the horizon it is?

[PB666]

87.91 minutes means he is rougly a few 100 miles up because of that you cannot see 180 degrees.

If we assume up is 0 degrees, then when the satellite is at 90 degrees he is signalling through the earth at 45 down.

So it has to do with altitude.

So lets assume its the earth. mu for earth is 5.9722E²⁴ * 6.67408E-11 = 3.985E14

period = 2pi * SQRT(a^3/mu)

5274.6 = 2pi * SQRT(a^3) /SQRT(mu)

839.47 = SQRT(a^3)/SQRT(mu)

16757890306 = SQRT(a^3)

280826887538650000000 = a^3

a = 6548567 the radius of the earth is 6,371000 ( 6,353 km to 6,384 km ) so this means elevation is about 177560 meters. 

Thus the parameters are in earth orbit (clarified) his velocity is now calculable = SQRT(mu/r) = 7800 m/s. (how damn convinient)

The view of his satellite is obstructed. While this seems difficult its actually not. There is a theta that is where cos theta = 6,371000 / 6548567 therefore the

Arccos (0.9728) = 13.37 degrees. THis assumes the reciever is at ground level, and the ground is absolutely flat. This means he can recieve that he can recieve a satellite in 26.74 of 360 degrees. If the period is 5274.6 seconds he can send and receive for 391.8 seconds.

You need to know how to calculate the Specific gravitational constant for point masses. See above,

You need to know how to convert fractions into sin or cosine using an ArcCos or ArcSin (Acos or Asin) function.

You need to know how many degrees are in a period.

meh, if you cant figure it out with what I gave your doomed, :^).

Since a degree is roughly 69 miles or about 100 kilometers, that satellite is like right over your head, so you will only be in contact for another ~200 seconds. As it moves to the edge of the range the, Here is how you know. 7.8 km sec * 196 seconds is roughly 1600 miles, the difference from its current N position is about 30 km due north of your position and the satellite is moving along a rougly east-west trajectory. Therefore it will disappear about the same time whether it passed directly over head or 30 km north or south, however due to is inclination and the difference in position you may have lost or gained some fraction of seconds. But BTW don't forget the earth surface is also moving about 1000 miles per hour at the equator and cos of 28.' x 1000 at your position. So if the sat is moving west to east across the sky, it not moving as fast away.

 

 

 

 

Edited March 29, 2016 by PB666

[Dman979]

29 minutes ago, PB666 said:

87.91 minutes means he is rougly a few 100 miles up because of that you cannot see 180 degrees.

If we assume up is 0 degrees, then when the satellite is at 90 degrees he is signalling through the earth at 45 down.

So it has to do with altitude.

So lets assume its the earth. mu for earth is 5.9722E²⁴ * 6.67408E-11 = 3.985E14

period = 2pi * SQRT(a^3/mu)

5274.6 = 2pi * SQRT(a^3) /SQRT(mu)

839.47 = SQRT(a^3)/SQRT(mu)

16757890306 = SQRT(a^3)

280826887538650000000 = a^3

a = 6548567 the radius of the earth is 6,371000 ( 6,353 km to 6,384 km ) so this means elevation is about 177560 meters. 

Thus the parameters are in earth orbit (clarified) his velocity is now calculable = SQRT(mu/r) = 7800 m/s. (how damn convinient)

The view of his satellite is obstructed. While this seems difficult its actually not. There is a theta that is where cos theta = 6,371000 / 6548567 therefore the

Arccos (0.9728) = 13.37 degrees. THis assumes the reciever is at ground level, and the ground is absolutely flat. This means he can recieve that he can recieve a satellite in 26.74 of 360 degrees. If the period is 5274.6 seconds he can send and receive for 391.8 seconds.

You need to know how to calculate the Specific gravitational constant for point masses. See above,

You need to know how to convert fractions into sin or cosine using an ArcCos or ArcSin (Acos or Asin) function.

You need to know how many degrees are in a period.

meh, if you cant figure it out with what I gave your doomed, :^).

Since a degree is roughly 69 miles or about 100 kilometers, that satellite is like right over your head, so you will only be in contact for another ~200 seconds. As it moves to the edge of the range the, Here is how you know. 7.8 km sec * 196 seconds is roughly 1600 miles, the difference from its current N position is about 30 km due north of your position and the satellite is moving along a rougly east-west trajectory. Therefore it will disappear about the same time whether it passed directly over head or 30 km north or south, however due to is inclination and the difference in position you may have lost or gained some fraction of seconds. But BTW don't forget the earth surface is also moving about 1000 miles per hour at the equator and cos of 28.' x 1000 at your position. So if the sat is moving west to east across the sky, it not moving as fast away.

All of that sounds good. Is there a formula you can point me to?

What about if it's not right overhead?

[YNM]

You'll need to know the difference between its orbital period and its synodic period with Earth's rotation. It'll represent the shift in the ground track, from the perspective of the satellite. Use spherical trigonometry to determine the shift in ground track by longitudes. I have made some calculations like this for ISS from a hypothetical observer at the equator, assuming that the satellite has passed the zenith of the observer. Yours will be harder.

EDIT : Anyway, I wanted to ask : Are you just needing the time it'll be in contact from the instance now ? Or the next pass as well ? If it's the current instance, what PB666 did (half a degree isn't a big thing to worry actually). If not, maybe I can help a bit, given the sat would then not pass your zenith.

Edited March 30, 2016 by YNM
some questions for OP

[PB666]

4 hours ago, Dman979 said:

All of that sounds good. Is there a formula you can point me to?

What about if it's not right overhead?

 

you need your givens first, what is the exact radial velocity (omega) of the earth and what is the exact radial velocity of the satellite.

w (omega) = v/r     the earth rotates 2pi radians over 86163 seconds so omega is  7.2922 E-5 radians per second. Sat is 1.193E-3 rad/sec which means that if it is traveling west to east it will linger for about 207 seconds.

The last element we don't know is that we know is north coordinate now, but we don't know what it will be after it has traveled 13.3 (+ correction) degrees east south eastward. Thats a bit more complicated, we know it crosses the equator it is 90 degrees from its current theta, We should be able to apply the cosine function and it maximum northern position to guess that it will be 27.63 N, which means it is about 30km south and ~1,600 km. If you are answering this for school the probably want 3 significant digits, the reality is the answer is only good for about 2 digits.