- #1

Spanky Deluxe

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I'm guessing I need to calculate the eccentricity too. I really suck at conversions like this. :(

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In summary, to calculate the semi major axis of a body in an elliptical orbit, you need to know the body's gravitational constant, G*, the mass, and the position. You also need to know the eccentricity.

- #1

Spanky Deluxe

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I'm guessing I need to calculate the eccentricity too. I really suck at conversions like this. :(

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- #2

ideasrule

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Take a look at http://en.wikipedia.org/wiki/Orbit#Analysis_of_orbital_motion, especially the equation at the very end. Provided you know the mass & position of the body being orbited, the semi-major axis (a) can easily be calculated.

- #3

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You better have at least one other thing. You need to know the body's gravitational constant, G*M, which are often combined as a single parameter μ.Spanky Deluxe said:

I'm guessing I need to calculate the eccentricity too. I really suck at conversions like this. :(

Hint: How is the specific angular momentum related to the semi-major axis?

Hint: You need to calculate the eccentricity.

No, they can't. He is given non-zero values. Ignoring them *will* yield the wrong answer.ideasrule said:Keplerian orbits are all two-dimensional, so you don't need to worry about z or vz; they can be defined as 0.

- #4

ideasrule

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- #5

Spanky Deluxe

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D H said:You better have at least one other thing. You need to know the body's gravitational constant, G*M, which are often combined as a single parameter μ.

Hint: How is the specific angular momentum related to the semi-major axis?

Hint: You need to calculate the eccentricity.

No, they can't. He is given non-zero values. Ignoring them *will* yield the wrong answer.

aha, so I can use:

[tex]\textbf{h}=\textbf{r}\times\textbf{v}[/tex]

to find the specific angular momentum and then

[tex]a\left(1-e\right)=\frac{h^{2}}{GM}[/tex]

But then how can I separate the eccentricity and semi-major axis?

- #6

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Spanky Deluxe said:But then how can I separate the eccentricity and semi-major axis?

Simple: Compute the eccentricity vector and take its magnitude. So what's this eccentricity

[tex]\vec e = \frac{\vec v \times \vec h}{GM} - \frac{\vec r}{||\vec r||}[/tex]

- #7

Spanky Deluxe

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D H said:Simple: Compute the eccentricity vector and take its magnitude. So what's this eccentricityvectorthing?

[tex]\vec e = \frac{\vec v \times \vec h}{GM} - \frac{\vec r}{||\vec r||}[/tex]

Ok, I've done that now.

One thing though, I also found the following equations which I found http://microsat.sm.bmstu.ru/e-library/Ballistics/kepler.pdf :

[tex]a=\frac{GMr}{2GM-rv^{2}}[/tex]

and

[tex]e=\sqrt{1-\frac{h^{2}}{GMa}}[/tex]

I've tested both sets of equations and I get the same numbers for the eccentricity, which is good. However, I'm getting different results for the semi major axis. :(

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- #8

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http://photos-d.ak.fbcdn.net/hphotos-ak-snc1/hs017.snc1/4224_95107990559_95107705559_3010885_4034015_s.jpgSpanky Deluxe said:[tex]a=\frac{GMr}{2GM-rv^{2}}[/tex]

The vis-viva equation, doh!

[tex]v^2=GM\left(\frac 2 r - \frac 1 a\right)[/tex]

I was thinking to much of your problem in term of converting cartesian position and velocity to orbital elements (six of them). The specific angular momentum and the eccentricity vector are the keys to unlocking that puzzle. However, all you wanted was the semi-major axis. The vis-viva is the key to answering that particular problem.

That's because you have the relation between angular momentum wrong, here:I've tested both sets of equations and I get the same numbers for the eccentricity, which is good. However, I'm getting different results for the semi major axis. :(

Spanky Deluxe said:[tex]a\left(1-e\right)=\frac{h^{2}}{GM}[/tex]

That should be

[tex]a(1-e^2) = \frac {h^2}{GM}[/tex]

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- #9

Spanky Deluxe

- 19

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DH, you are an absolute star and a lifesaver! Thanks so much for the help!

A semi-major axis is a measurement in an ellipse that extends from the center to the farthest point on the ellipse. In cartesian coordinates, it is one of the two axes that define the shape and orientation of the ellipse.

The semi-major axis can be calculated using the distance formula, which is the square root of the sum of the squared x and y coordinates. It can also be calculated by finding the distance between the center of the ellipse and one of its foci.

The semi-major axis is half the length of the major axis, and the major axis is the longest diameter of an ellipse. Therefore, the semi-major axis is a crucial factor in determining the size and shape of an ellipse.

The semi-major axis is directly related to the orbital period of an object. The larger the semi-major axis, the longer the orbital period will be. This relationship is described by Kepler's Third Law of Planetary Motion.

No, the semi-major axis cannot be negative in cartesian coordinates. The x and y coordinates are always positive, and the distance formula only produces positive values. Therefore, the semi-major axis is always a positive value in cartesian coordinates.

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