Orbital elements - what's wrong with my calculation?

• dominicfhk
In summary: Hi dominicfhk,Well, first of all, there are several errors in your calculations, but you didn't post the original equations correctly anyway. I don't really know how you could even get some of those results.In any case, your pdf is wrong too.By following it, you would get n as a null vector. For the eccentricity vector, check the sign of the j component and the calculated magnitude; how can a magnitude be negative :confused:For the velocity vector, try with these values: R=(DU√2/2)(-I+K)V=DU/2TU Kthis gives

Homework Statement

Ok so this satellite is orbiting the earth. The position vector is r = -0.707i + 0.707j + 0k and velocity vector is v = 0i + 0.5j + 0k. To find Vo (the angle between vector n and e), my book said the formula to use is arccos{(e dot r)/(|e||r|)}. However, I got 98 degree by using this formula while the correct answer is 173 degree.

Homework Equations

arccos{(e dot r)/(|e||r|)}

The Attempt at a Solution

I got the e vector as -.5304i-0.707j+0k, and therefore |e|=-.3535
r vector was given as -0.707i+0.707j+0k, and so |r|=0.9998
"e dot r" = -0.125 and "|e||r|" = 0.884
so arccos{(e dot r)/(|e||r|)} = 98 degree, which does not match the correct answer.

This is from the textbook "Fundamentals of Astrodynamics" problem 2.2 on P.114
http://aeroden.files.wordpress.com/2011/12/fundamentals_of_astrodynamics.pdf [Broken]

Thank you so much.

Last edited by a moderator:
Hi dominicfhk,

Well, first of all, there are several errors in your calculations, but you didn't post the original equations correctly anyway. I don't really know how you could even get some of those results.
In any case, your pdf is wrong too.
By following it, you would get n as a null vector.

I fiddled around and found the correct values for the exercise that give me the same answers as in the book.
Well, almost, I don't get the exact same numerical answers because of the precision involved, this is because I write √2/2 instead of .707, it's much cleaner for working with the equations and I am quite certain that this is where .707 comes from.

So try with those values instead:
R=(DU√2/2)(-I+K)
V=DU/2TU K

this gives
ρ=DU/8 (as expected)
e=5√2/8≈.883 (vs .885 in the book)
v0=arcos(-7√2/10)≈171.87° (vs 173° in the book)

Cheers...

dominicfhk said:

Homework Statement

Ok so this satellite is orbiting the earth. The position vector is r = -0.707i + 0.707j + 0k and velocity vector is v = 0i + 0.5j + 0k. To find Vo (the angle between vector n and e), my book said the formula to use is arccos{(e dot r)/(|e||r|)}. However, I got 98 degree by using this formula while the correct answer is 173 degree.

Homework Equations

arccos{(e dot r)/(|e||r|)}

The Attempt at a Solution

I got the e vector as -.5304i-0.707j+0k, and therefore |e|=-.3535
For the eccentricity vector, check the sign of the j component and the calculated magnitude; how can a magnitude be negative
r vector was given as -0.707i+0.707j+0k, and so |r|=0.9998
As pointed out by Oli4, the 0.707's are probably meant to be ##\sqrt(2)/2##, else a radius of 0.9998 implies that the satellite is orbiting below the Earth's surface! Thus you can reasonably assume the radius magnitude to be 1.0 .
"e dot r" = -0.125 and "|e||r|" = 0.884
The value for |e||r| looks okay, but check your calculation for e dot r. I'm not sure how you managed to get the value for |e||r| correct when your magnitude of e above is wonky.
so arccos{(e dot r)/(|e||r|)} = 98 degree, which does not match the correct answer.
Once you straighten out the details for the eccentricity vector and its magnitude, all should be well. Don't worry overly much about being bang on in the third decimal place; it's possible that some of the problem results were calculated by slide rule!
This is from the textbook "Fundamentals of Astrodynamics" problem 2.2 on P.114
http://aeroden.files.wordpress.com/2011/12/fundamentals_of_astrodynamics.pdf [Broken]

Thank you so much.

Last edited by a moderator:

1. Why are my orbital elements not matching the expected values?

There are several potential reasons for this. One possibility is that there may be an error in your calculation. Double check your equations and input values to ensure accuracy. Another possibility is that your data may be outdated or incorrect. Make sure you are using the most current and accurate data available. Additionally, certain factors such as perturbations from other celestial bodies or atmospheric drag can also affect the accuracy of orbital element calculations.

2. What units should I use for my orbital element calculations?

The units used for orbital elements can vary depending on the specific equations and software being used. However, some commonly used units include astronomical units (AU) for distance, seconds or days for time, degrees or radians for angles, and kilograms for mass. It is important to be consistent with the units used throughout your calculations.

3. Can I calculate orbital elements for objects other than planets?

Yes, orbital elements can be calculated for any object orbiting another body, including moons, asteroids, and artificial satellites. However, the specific equations and parameters used may vary depending on the object's characteristics and the desired accuracy of the calculation.

4. How do I account for the effects of relativity in my orbital element calculations?

In most cases, the effects of relativity on orbital elements are negligible and can be ignored. However, for highly precise calculations, such as those used for space missions, relativity may need to be taken into account. This can be done through the use of specialized equations and software that incorporate relativistic effects into the calculations.

5. Can I use orbital elements to predict future positions of celestial objects?

Yes, orbital elements can be used to predict the future positions of objects in orbit. However, this requires a precise and accurate calculation of the orbital elements, as well as accounting for any external influences on the object's orbit. Minor errors in the calculation or unaccounted for perturbations can result in significant differences in predicted positions.