Satellite tracking - Dynamics, Angular velocity, Circular motion

[blueparukia]

Been sick for a bit so missed a fair bit of "advanced" maths including locuses, particularly parabolas, but what's worrying me more at the moment is dynamics. So got a moderately difficult question here to help catch me up. I don't really comprehend any of the rules and this question is tricky because it seems I'm going to need trig (albeit basic trig) to find distances and stuff as well as dealing with time.

Homework Statement

A satellite traveliling in a circular orbit 1600km above the Earth is due to pass over a satellite tracking station at noon. Assume that the satellite takes two hours to make one complete orbit and that the radius of the Earth is 6400km.

If the tracking antenna is aimed 30 degrees above the horizon (an imaginary line, tangent to the Earth and passing through the station), at what time, to the nearest second, will the satellite pass through the beam of the antenna?

Homework Equations

w = dx/dt (differential of x/theta n respect to time)
v = wr
T (period) = 2pi/w
Theta = wt

Measure velocity in radians per hour

 

[Coto]

The hardest part about this problem is the trig, so why don't we start there.

Since we can find the speed of the satellite (given the known quantities in the problem), it suffices to find the distance between the point directly above the tracking station and the point where the satellite passes through the tracking beam.

What formula would you use to find this distance? So what do we need?

 

[blueparukia]

Yeah see that's kinda where I'm having the problem. I think I've worked out the velocity, and I could use trig to find out a side for an approximate distance, but not on the curve. I'm fairly sure I'm trying to find out how many radians it has covered rather than metres but I'm not sure.

 

[Coto]

Well, how do you find out the arc length of a circle? (Hint: http://en.wikipedia.org/wiki/Arc_(geometry))

So what angle do you need in order to find the distance the satellite will travel between those two points?

 

[blueparukia]

Arc length is not part of the syllabus and I'd get marked or not rated trying that on a test :p

The angle I would need to find is of course the one at the centre of the Earth. But now this is just getting back to trig, so I should be good. If the angle between the horizon and the tracking station beam is 30 degrees (pi/6) than the angle between the tracking beam and the position at noon would be pi/3.

However the middle triangle is not a right angle triangle whereas the triangle to the center of the Earth is. I'm sure this can be done without the sin/cosine rules and there's something in my mind saying that the angle I'm after is pi/6, but I'm not sure - been a year or two since I've done trig with multiple triangles.

Thanks for your help.

 

[Coto]

Ok.

Still, to solve the problem you will need to find the angle at the center of the earth. As you mentioned, it is just trig. I suggest looking up the sine law (http://en.wikipedia.org/wiki/Law_of_sines).

Good luck!

 

[Coto]

Yeah see that's kinda where I'm having the problem. I think I've worked out the velocity, and I could use trig to find out a side for an approximate distance, but not on the curve. I'm fairly sure I'm trying to find out how many radians it has covered rather than metres but I'm not sure.

Ah yes. Once you have found the angle at the center of the Earth (in radians), your suspicions are correct and this is the right idea!

 

[blueparukia]

Cheers, is using the sine rule necessary? I know it well enough, but something tells me this can be solved by simple ratios, but we'll see :)

Cheers

 

[Coto]

It seemed to me to be the easiest way. Like any trig/geometry problem, there are many different routes to get to the same solution.

 

[Susanne217]

It seemed to me to be the easiest way. Like any trig/geometry problem, there are many different routes to get to the same solution.

You remember the Universal law of of gravity which originates from Newton 2

$$F = m_{sat} \cdot a_c = G \cdot \frac{m_{earth} \cdot m_{sat}}{r^2}$$

$$a_{c} = \frac{G \cdot m_{earth}}{r^2} = \frac{v^2}{r}$$

Thus $$v = \sqrt{\frac{G \cdot m_{earth}}{r}}$$

How is that?

 

[blueparukia]

Yeah I think this is going a little too far :p This isn't actually a real world situation, just an excercise, that is only allowed to use the formulas I mentioned above plus trigonometry. Thanks for all your help though, it's been most informative.

 

[Coto]

You remember the Universal law of of gravity which originates from Newton 2

$$F = m_{sat} \cdot a_c = G \cdot \frac{m_{earth} \cdot m_{sat}}{r^2}$$

$$a_{c} = \frac{G \cdot m_{earth}}{r^2} = \frac{v^2}{r}$$

Thus $$v = \sqrt{\frac{G \cdot m_{earth}}{r}}$$

How is that?

Overkill :). And this only finds a velocity which can be found using the data of the problem in a more direct manner.

 

[blueparukia]

Hmm, I've just spent a good 45 minutes trying to find the angle with no luck. So it turns out I'm alright with the new angular velocity, I just can't do simple trig.

There's no way I can conceive this (sin/cosine rule possibly, but I don't understand them enough to be able to make use of them), so I tried some trig functions but they all require a side or angle I don't have, so I've tried doing the sum of all angles = 180 degrees, but again the information seems insufficient to use that to find any extra angles.

Some help would be appreciated thanks :D

 

[rock.freak667]

Hmm, I've just spent a good 45 minutes trying to find the angle with no luck. So it turns out I'm alright with the new angular velocity, I just can't do simple trig.

There's no way I can conceive this (sin/cosine rule possibly, but I don't understand them enough to be able to make use of them), so I tried some trig functions but they all require a side or angle I don't have, so I've tried doing the sum of all angles = 180 degrees, but again the information seems insufficient to use that to find any extra angles.

Some help would be appreciated thanks :D

If you call the angle between the long orange line and the long red line as θ, and you know that the angle between the radius of the Earth and the imaginary line is 90°, what is the other angle in the triangle?

The you will have two sides and two angles. Sine rule should work to get the angle θ.

 

[blueparukia]

and you know that the angle between the radius of the Earth and the imaginary line is 90°, what is the other angle in the triangle?

No idea, that's what I was trying to work out with sum of all angles = 180 degrees. And there's no other trig I can think of that will solve it without knowing theta or the other lengths.

 

[rock.freak667]

No idea, that's what I was trying to work out with sum of all angles = 180 degrees. And there's no other trig I can think of that will solve it without knowing theta or the other lengths.

I'll redraw what I meant:

http://img291.imageshack.us/img291/8250/newaj.png

Since the object is traveling in a circle, shouldn't the effective radius i.e. the long red and orange lines be the same?

 

[blueparukia]

Huh, common sense. It strikes again. Thanks a lot.

 

[blueparukia]

OK, I can't figure this out. I'm stumped, I'm frustrated, I can't answer the question, it's been 4 days and the trig is just standing in the way of what I need to know. I am not skilled enough at trig to be able to work this out. The sides can't be equal on a right angle triangle, so if I connect all the dots I get an isoscoles triangle with insufficient data to complete the problem. All I want is the angle, I do not remember anything about trig about from the basic ratios and I feel this answer attempt has been made overally complicated.

At any rate, diagram attached. I want this question answered today and any help given will be hugely appreciated. Thank you all.

 

[rock.freak667]

http://img718.imageshack.us/img718/6299/54193918.png

I removed the confusing parts of your diagram. You are only dealing with triangle ABC, you have angle A=θ, angle B is shown (90+30 gives?) and you can write angle C in terms of θ, since you know the sum of the interior angles of a triangle = 180°.

Then just apply the sine rule.

 

[blueparukia]

Yes I think I got it! Thank you so much, I have an answer :D Thank you very, very much. Hoping the answer is ~40 degrees.

 

[blueparukia]

Hmm, double checking my answer (40.5 degrees) is most likely not right since I substituted the wrong sides. The problem now is that the sine rule to my knowledge won't work with both side a and angle A missing. So I'm thinking use my original diagram to solve for side a then subsitute that into the sin rule?

Cheers.

 

[rock.freak667]

Hmm, double checking my answer (40.5 degrees) is most likely not right since I substituted the wrong sides. The problem now is that the sine rule to my knowledge won't work with both side a and angle A missing. So I'm thinking use my original diagram to solve for side a then subsitute that into the sin rule?

Cheers.

you have side BC and angle B, you have side AB and angle C (when you get that in terms of θ)

 

[blueparukia]

angle C (when you get that in terms of θ)

No idea what that means. Does that mean I have to solve θ first? Sorry, I sound stupid, but I'm tired of this question and it's been a while since I've done trig.

 

[rock.freak667]

No idea what that means. Does that mean I have to solve θ first? Sorry, I sound stupid, but I'm tired of this question and it's been a while since I've done trig.

Don't get frustrated, look at the diagram I posted.

the angle B is just 120° right? The angle A is just θ

if A+B+C =180°, what is C ?

 

[blueparukia]

Well 180-120 is equal to 60. That would mean the remaining two angles are 60 degrees combined. So C would be...60-θ?

 

[rock.freak667]

Well 180-120 is equal to 60. That would mean the remaining two angles are 60 degrees combined. So C would be...60-θ?

Yes so now you can apply the sine rule using the sides that you know.

 

[blueparukia]

Not really sure I know how to solve it like that.

First of all, radius of satellite orbit = 6400+1600 = 8000km

(sin 120)/8000 = (sin (C-θ))/6400

Would mean:
sin (C-θ) = 0.693
And C-θ = sin^-1(0.693)
= 43.853

But I can't see how I can get either C or θ from that, since it would be C = θ + 43.853 or θ = C - 43.853.

Thanks, I can see this going somewhere.

 

[rock.freak667]

Not really sure I know how to solve it like that.

First of all, radius of satellite orbit = 6400+1600 = 8000km

(sin 120)/8000 = (sin (C-θ))/6400

Would mean:
sin (C-θ) = 0.693
And C-θ = sin^-1(0.693)
= 43.853

But I can't see how I can get either C or θ from that, since it would be C = θ + 43.853 or θ = C - 43.853.

Thanks, I can see this going somewhere.

No no C= 60-θ. So what you have is this

$$\frac{sin120}{8000}= \frac{sin(60 - \theta)}{6400}$$

 

[blueparukia]

Oops, got a bit lost there.

So 60-θ = 43.853
So -θ = 43.853-60 = -16.146
θ = 16.146? :D

Thanks a load, you've been unbelievably helpful.

 

[rock.freak667]

Oops, got a bit lost there.

So 60-θ = 43.853
So -θ = 43.853-60 = -16.146
θ = 16.146? :D

Thanks a load, you've been unbelievably helpful.

Yes that should work now.

 

[Susanne217]

Overkill :). And this only finds a velocity which can be found using the data of the problem in a more direct manner.

What can I say other than I'm a nerd and yes I'm proud of it :)