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Elliptical Orbits... Unfortunately.

by Nghi
Tags: elliptical, orbits
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Nghi
#1
Nov5-07, 02:47 PM
P: 18
1. The problem statement, all variables and given/known data

A particular comet has an elliptical orbit. When closest to the Sun (perihelion) it is at a distance of 8.823 x 10^10 m and moves with a speed of 54.6 km/s. The greatest distance between this comet and the Sun (aphelion) is 5.902 x 10^12 m.

Calculate its speed at aphelion.

2. Relevant equations

Initial angular momentum = final angular momentum, since Kepler's law states that equal area is swept out at equal times.

3. The attempt at a solution

Li = Lf
Iw = Iw
(mr^2)w = (mr^2)w
(mr^2)(v/r) = (mr^2)(v/r)

Simplifying it, I get the result:

rv = rv

Plugging in the numbers, I get the following answer:

(8.823e10 m)(54,600 m/s) = (5.902e12)(v)
v = 816.225 m/s
= 0.816225 km/s

That's not the right answer, though, and it bothers me like whoa. It's because my professor sent a mass email out yesterday, explaining how a similar problem on the homework had been wrong. So he posted this question up instead, and it's the exact same problem, only with different numbers. When I did the original problem, I got the right answer following this method. Now, it's wrong, and it's confusing me. Any ideas?
1. The problem statement, all variables and given/known data



2. Relevant equations



3. The attempt at a solution
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dynamicsolo
#2
Nov5-07, 03:03 PM
HW Helper
P: 1,662
Quote Quote by Nghi View Post
(8.823e10 m)(54,600 m/s) = (5.902e12)(v)
v = 816.225 m/s
= 0.816225 km/s

That's not the right answer, though, and it bothers me like whoa.
What is the answer claimed to be? The comet obeys conservation of angular momentum. The aphelion distance is about 67 times greater than the perihelion distance, so the aphelion speed should be about 1/67 of the perihelion speed (these are the only two points on the orbit where you can do the calculation this simply), or around 54.6/67 km/sec = 0.82 km/sec. I'm not seeing anything wrong with your method given the information in the problem.
Nghi
#3
Nov5-07, 03:31 PM
P: 18
It's an online homework problem, so the answers won't be shown until tomorrow.

I'm just frustrated that I'm going to get deducted for a problem that I know that I'm doing right. :( I just emailed my professor, asking him for clarification, so hopefully he'll reply soon.

But thank you for verifying what I did. I wasn't sure whether I was wrong, or whether the homework was trying to crush my self-estem into little itty, bitty pieces, ha ha. :'(

I do have one relevant question, though: can you use conservation of energy to answer this question? Because I was browsing through the threads, and I found one where the guy used 1/2mv^2 - GMm/r = 1/2mv^2 - GMm/r to try and find his answer. I tried doing that for this problem, and I got a radically different (but still wrong) answer.

lakersgrl786
#4
Nov5-07, 04:07 PM
P: 2
Elliptical Orbits... Unfortunately.

I'm having the same issues with that problem -- can you post if you find an answer from the professor?
dynamicsolo
#5
Nov5-07, 06:37 PM
HW Helper
P: 1,662
Quote Quote by Nghi View Post

I do have one relevant question, though: can you use conservation of energy to answer this question? Because I was browsing through the threads, and I found one where the guy used 1/2mv^2 - GMm/r = 1/2mv^2 - GMm/r to try and find his answer. I tried doing that for this problem, and I got a radically different (but still wrong) answer.
OK, this took some time to track down the problem. Yes, in principle, you should get the same result using the conservation of mechanical energy equation.

We ran into a similar difficulty a few weeks ago on another celestial mechanics problem of this sort. What it illustrates is the danger of providing data for a highly elliptical orbit to insufficient precision. If you use conservation of mechanical energy and make the calculations for energy per unit mass (just drop the little m) with the data you are given, you get this (GM = 1.3273310^20 for the Sun):

using perihelion data --

r_p = 8.82310^10 m
v = 54,600 m/sec

E/m = -13,817,600 J/m
2a (major axis) = 9.60610^12 m
r_a (aphelion distance) = 9.51810^12 m


using aphelion data --

r_a = 5.90210^12 m
v = 816.2 m/sec

E/m = -22,156,404 J/m
2a = 5.99110^12 m
r_p (perihelion distance) = 8.8710^10 m


We get the major axis of the elliptical orbit using the vis-viva equation (I don't know if you had it in your course -- it's either in your book or look on Wiki, for instance). The value of E/m is equal to -GM/2a , where 2a is the distance from the perihelion point to the aphelion point (the "length" of the ellipse) and [tex]2a = r_{p} + r_{a}[/tex].

So from this, at first glance, it looks like two different orbits and that your instructor goofed, with the results using the aphelion information being consistent with the problem's distances and the perihelion velocity not being consistent. But if you use the conservation of angular momentum as you did for the problem and apply it to your aphelion velocity (which looks reliable at present), you get a perihelion velocity of

(5.90210^12 / 8.82310^10) x 816.2 m/sec = 66.893 816.2 m/sec = 54,598 m/sec ,

which is awfully close to what he used.

So what gives? The answer is that for a highly elliptical orbit, the values of the aphelion distance and aphelion speed are extremely sensitive to the value of the perihelion speed. When you make up a problem like this, you must either avoid using an orbit with a huge difference between the perihelion and aphelion distances or else given the values to at least four or five significant figures. (This is just the trouble the other student ran into on their problem last month. Depending on how they rounded off the numbers they used, they got considerable discrepancies in their answers.)

You're welcome to pass these calculations on to your instructor, if they haven't discovered the difficulty already. Obviously, they're going to have to allow a fair amount of leeway in grading everyone's work...


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