Help with a math problem

July 29, 2018 8:31 PM

I'm trying to help a friend study for an exam using example problems from past exams, which were provided by their teacher as a study aid. This is strictly for test prep, provided by the instructor, and is not graded homework. Unfortunately answers were not provided. One question has us stuck though, can you help? Details inside.

The question is about the distance between two points on the surface of the earth observing a satellite in orbit. We have what we believe are the correct answers for the first two parts but are stuck on the third part.

Here is the question and a link to images of my work:

A satellite A is directly above the point B on the Earth's surface and can just be seen (on the horizon) from another point C on the Earth's surface. The radius of the Earth is R and the height of the satellite above the Earth is h. The point O denotes the center of the Earth.

(a) By considering the angle Ɵ = ∠BOC, or otherwise, find the distance s between B and C along the Earth's surface in terms of R and H.

(b) If the satellite is in low orbit, so that h is small compared with R, show that
s ≈ k(RH)^(1/2)
where the value of k is to be found.

(c) If the satellite is very disant from the Earth so that R is small compared with h, show that
s ≈ aR + b(R^2)/h
where the values of a and b are to be found.

Solution for (a) & (b)
Setup for (c)

Part a is just basic geometry and trig, part b depends on the simplification that h^2 is small and can be discarded and also that s is approximately AC.

The "trick" for part c is eluding us. A full solution would be ideal, but even a hint towards the proper "trick "would be appreciated.

posted by Reverend John to Education (14 answers total)

Best answer: h large compared to R implies that R/(R+h) can be approximated as R/h by way of 1/(1+h/R). Then look at the Taylor series for arccos().
posted by solitary dancer at 9:02 PM on July 29, 2018 [1 favorite]

The trick may depend on your definition of "very distant." For stellar-scale values of h, isn't the object visible all the way until theta is 90 degrees? Which would lead to values of a = pi/2 and b=0. Which of course is why I despise problems formed in this manner, unless I suppose you are explicitly told to avoid degenerate cases like b=0, which it appears would imply a definition of "very large" that was in fact medium-large.
posted by range at 9:03 PM on July 29, 2018 [3 favorites]

Or to put it another way, my physicist brain is having conniptions at the idea that "much greater than" might mean anything other than "all second order terms and beyond have coefficient zero" while the problem seems to want a second order term. I'd say b=0 because history shows that I am an asshole on tests; if I were feeling like playing along I also would head to the Taylor expansion of arc cosine. (It is however a little tricky since I believe you'd rather have the expansion about pi/2 since that's where you're going to evaluate it.)
posted by range at 9:09 PM on July 29, 2018 [1 favorite]

(So as not to abuse the edit window I'll quickly mention that you need the expansion about cos(pi/2) which is indeed the expansion about zero. Don't drink and post to math AskMe, kids.)
posted by range at 9:14 PM on July 29, 2018 [1 favorite]

For stellar-scale values of h, isn't the object visible all the way until theta is 90 degrees? Which would lead to values of a = pi/2

I think you mean a=R*pi/2
posted by w0mbat at 9:38 PM on July 29, 2018

(it's just a definitional thing from the problem -- s = aR, the arc length would be R* pi/2 so a = pi/2.)

Anyway the Taylor expansion more or less gives something in the expected form though I don't immediately see the rationale for reducing R^2/(R+h) to R^2/h instead of reducing it all the way to zero for large h. But the problem-setter may have gotten there via some other approximation on a different solution path - unfortunately I think this part of the problem hinges a bit on the dreaded "guess how I solved the problem" quiz-writing style.
posted by range at 10:02 PM on July 29, 2018

If a = pi/2, then that is the maximum case when h is at infinity. For h less than infinity, then b must be negative. The little sliver that you must subtract is approximated by the side of a tiny triangle similar to the large right triangle with ratio R over h. The long side of the similar tiny triangle is R so the short side is R times the ratio R over h, so R * R/h. So b = - (R^2)/h
posted by JackFlash at 11:11 PM on July 29, 2018 [1 favorite]

More accurately according to the problem, b = minus 1.
(R^2)/h is the approximation to the small sliver you must subtract from the 90 degree arc length, using similar triangles.
posted by JackFlash at 11:21 PM on July 29, 2018

And one more clarification, the large right triangle has the ratio R over h+R which is equal to R over h for h>>R.
posted by JackFlash at 11:42 PM on July 29, 2018

Response by poster: Checks out, thanks you guise!
posted by Reverend John at 8:24 AM on July 30, 2018

But as range pointed out, there is no good rationale for approximating R/(R+h) by R/h, as suggested by JackFlash and used in the last line of your posted answer. I think it would be worth pressing the instructor on this point. If you run the calculation for h>>R, you'll find that the error from using R/(R+h) is significantly less than the error from using R/h (by an order of magnitude at R/h=100 and diverging further for larger ratios), which indicates that this approximation is indeed suspect.
posted by doctord at 8:54 AM on July 30, 2018

Doctord, what you are saying is simply that an approximation is less accurate than a non-approximation which isn't particularly profound. But put that in the context of the entire problem. What is the final error when starting with the first half of the solution that the arc length is R*pi/2. The approximation for the b component is less important. It is by definition small to begin with.
posted by JackFlash at 9:26 AM on July 30, 2018

Yeah, I suspect JackFlash found the question-author's solution path and the similar triangles argument definitely motivates the approximation used - I haven't completely run down the path but it looks like it makes more sense in terms of a small-angle approximation to use the short side of that similar triangle as a proxy for arc length, rather than h >> R as an explicit limit. The Taylor approximation of arc cosine is for sure what I would go to first but that's just because I like algebra more than geometry; working it out with similar triangles would seem to be more in the spirit of the implied genre of the question.
posted by range at 5:21 PM on July 30, 2018 [1 favorite]

doctord: If we let t = R/h, we can write R/(R+h) = t/(1+t) = t(1–t+O(t²)). Then we can give Reverend John's answer an explicit error bound.

(Edit: Actually, just t/(1+t) = t(1+O(t)) is enough for the level of precision requested here!)
posted by aws17576 at 10:19 AM on July 31, 2018