Believe it or not, a state vector adequately describes the entire orbit of a satellite even though it gives the position and velocity of a satellite at just one epoch time. We can convert the state vector to the Keplerian (classical) orbit elements to show what this initial orbit looks like. First, we will determine the square of the total velocity of the satellite by taking the sum of the squares of the velocity components: v^{2} = ẋ^{2} + ẏ^{2} + ż^{2} v^{2} = 56.286344 (km/s)^{2} Now we can find the Semimajor Axis (a) of the orbit by using the following equation: a = [ 2/r_{2}  v^{2}/m ]^{1} a = 7,310.8163 km Remember that the Semimajor Axis of the satellite's orbit is measured from the center of the Earth and not from the surface. If you want to find the average altitude from the Semimajor Axis, just subtract the Earth's equatorial radius: a_{alt} = a  R_{E} a_{alt} = 932.6793 km Since we now know the Semimajor Axis of the orbit, we also know the orbit period through Kepler's Third Law: T = 2p (a^{3}/m)^{1/2} T = 6220.9941 seconds = 103.68323 minutes This orbit period is consistent with a low earth orbit (LEO) satellite. We can also find the Mean Motion of this satellite by taking the inverse of the period: n = 1 / T n = 0.0001607 orbits/s = 13.888456 orbits/day The Semimajor Axis just tells us the average distance of the satellite in its orbit. We will now need to find the Eccentricity of the orbit in order to know the orbit's shape. To do this, we will need to determine the satellite's eccentricity vector and find its magnitude: e_{i} = (1/r_{2
} 1/a) x  (xẋ + yẏ + zż) ẋ / m e_{i} = 0.011926 e = [ (e_{i})^{2} + (e_{j})^{2} + (e_{k})^{2} ]^{1/2} e = 0.0159858 The Eccentricity of this orbit is estimated to be very low, which is expected for a low earth orbit satellite. Next, we will extract the orbit Inclination from the state vector. To accomplish this, we will need to determine the satellite's angular momentum vector. This is done by taking the crossproduct of the position and velocity vectors: h_{i} = yż  zẏ h_{i} = 26,508.867
km^{2}/s We will also need the magnitude of the angular momentum vector, which is determined by the following: h = [ (h_{i})^{2} + (h_{j})^{2} + (h_{k})^{2} ]^{1/2} h = 53,975.457 km^{2}/s The Inclination of this orbit can now be determined by using the following equation: i = cos^{1} (h_{k} / h) i = 71°.048202 We only have three more elements to go to complete this step. The next element to determine is the Right Ascension of the Ascending Node (a_{W}). This requires a few additional quantities:
W_{i}
= h_{j}
W_{i}
= 43,627.260 km^{2}/s Now we can find the location of the ascending node using the following conditional equation: If
W_{j}
> 0 Then: If
W_{j}
< 0 Then: a_{W} = 211°.28377 Next is the Argument of Perigee (w). Use the following conditional equation: If e_{k} >
0 Then: If e_{k} <
0 Then: w = 137°.75619 Finally, we will determine the Mean Anomaly (M) at the epoch time. We will first find the True Anomaly (n), translate this to the Eccentric Anomaly (E), then finally find the Mean Anomaly (M). The True Anomaly is found using the following conditional equation: If (xẋ + yẏ + zż)
> 0 Then: If (xẋ + yẏ + zż)
< 0 Then: n = 354°.80860 The Eccentric Anomaly (E) is found from the True Anomaly by using the following equation: If 0
< n < p
Then If
p
< n < 2p
Then E = 354°.89083 = 6.1940135 radians Finally, the Mean Anomaly (M) can be found from the Eccentric Anomaly using Kepler's Equation: M = E  esinE M = 6.1954371 radians = 354°.97240 Finally, we can gather the determined orbit elements into one table:


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Step 14: Keplerian Elements Was Last Modified On September 23, 2013 