Calculate Area of Ellipse in Keplerian Orbit

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In a Keplerian orbit, the area swept out by an orbiting body can be calculated using specific angular momentum and the relationship between the radius and velocity. The area of the entire ellipse is given by the formula πab, where a and b are the lengths of the major and minor axes. The discussion clarifies that the focus of the ellipse is at the Sun, and the area of interest is not the whole ellipse but the area swept out by the orbiting body. The specific angular momentum, denoted as h, is crucial for determining the rate at which this area is swept out. Understanding these principles aligns with Kepler's second law, which states that the line connecting a planet to the Sun sweeps out equal areas in equal times.
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Hello everybody,

I'm trying to know, in a keplerian orbit, how to calculate the area of a swaped area; since the Sun is at one of the focus, I wish to calculate given an angle measured from focus to the orbiting body, the area swaped.
I don't know if I'm explaning this right...Hope so.

Kind regards,

CPtolemy
 
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cptolemy said:
Hello everybody,

I'm trying to know, in a keplerian orbit, how to calculate the area of a swaped area; since the Sun is at one of the focus, I wish to calculate given an angle measured from focus to the orbiting body, the area swaped.
I don't know if I'm explaning this right...Hope so.

Kind regards,

CPtolemy

swaped = ? What is this word?

Calculating the area of an ellipse is pretty straightforward. There are several formulas if you know the equation of the ellipse. See http://en.wikipedia.org/wiki/Ellipse#Area.
 
Hi,

I mean swept. Sorry for my english... :(

I don't want to know the entire area of the ellipse - just the swept area by the body.

Regards,

CPtolemy
 
I think he meant the area "swept out" by the planets motion- the area inside the elliptic orbit.

cptolemy, the area of an ellipse with major and minor axes of lengths a and b is \pi ab.
 
In Fundamentals of Astrodynamics by Bate, Mueller and White, ISBN 0-486-60061-0, I can see the following equation:

dt = 2/h dA

h is the specific angular momentum, given by h = r v sin(γ), where γ is the flight path angle, i.e. the angle between the r and v vectors. This is consistent with Kepler's second law as h is a constant for a given orbit.
 
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