a, e, i, aW, w, M

 Element Symbol Value Epoch Time tepoch 23:05:28.22 UTC January 3, 2012 Second Geocentric Range r2 7,194.4110 km Earth's Equatorial Radius RE 6,378.137 km Gravitational Parameter m 398600.44 km3/s2 x Position x 5052.4587 km y Position y 1056.2713 km z Position z 5011.6366 km x Velocity ẋ 3.8589872 km/s y Velocity ẏ 4.2763114 km/s z Velocity ż -4.8070493 km/s

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:

v2 = ẋ2 + ẏ2 + ż2

v2 = 56.286344 (km/s)2

Now we can find the Semi-major Axis (a) of the orbit by using the following equation:

a = [ 2/r2 - v2/m ]-1

a = 7,310.8163 km

Remember that the Semi-major 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 Semi-major Axis, just subtract the Earth's equatorial radius:

aalt = a - RE

aalt = 932.6793 km

Since we now know the Semi-major Axis of the orbit, we also know the orbit period through Kepler's Third Law:

T = 2p (a3/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 Semi-major 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:

ei = (1/r2 - 1/a) x - (xẋ + yẏ + zż) ẋ / m
ej = (1/r2 - 1/a) y - (xẋ + yẏ + zż) ẏ / m
ek = (1/r2 - 1/a) z - (xẋ + yẏ +zż) ż / m

ei = 0.011926
ej = 0.0031623
ek = 0.0101645

e = [ (ei)2 + (ej)2 + (ek)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 cross-product of the position and velocity vectors:

hi = yż - zẏ
hj = zẋ - xż
hk = xẏ - yẋ

hi = -26,508.867 km2/s
hj = 43,627.260 km2/s
hk = 17,529.749 km2/s

We will also need the magnitude of the angular momentum vector, which is determined by the following:

h = [ (hi)2 + (hj)2 + (hk)2 ]1/2

h = 53,975.457 km2/s

The Inclination of this orbit can now be determined by using the following equation:

i = cos-1 (hk / 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 (aW). This requires a few additional quantities:

Wi = -hj
Wj = hi
W = [ (Wi)2 + (Wj)2 ]1/2

Wi = -43,627.260 km2/s
Wj = -26,508.867 km2/s
W = 51,049.563 km2/s

Now we can find the location of the ascending node using the following conditional equation:

If Wj > 0 Then:
aW
= cos-1 (
Wi / W )

If Wj < 0 Then:
aW
= 360° - cos-1 (
Wi / W )

aW = 211°.28377

Next is the Argument of Perigee (w). Use the following conditional equation:

If ek > 0 Then:
w
= cos-1 [ (
Wiei + Wjej) / We ]

If ek < 0 Then:
w
= 360° - cos-1 [ (
Wiei + Wjej) / We ]

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:
n
= cos-1 [ (xei + yej + zek) / er2 ]

If (xẋ + yẏ + zż) < 0 Then:
n
= 360° - cos-1 [ (xei + yej + zek) / er2 ]

n = 354°.80860

The Eccentric Anomaly (E) is found from the True Anomaly by using the following equation:

If 0 < n < p Then
E = cos-1 [ (e + cos
n) / (1 + ecosn) ]

If p < n < 2p Then
E = 360° - cos-1 [ (e + cos
n) / (1 + ecosn) ]

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:

 Keplerian Orbit Element Symbol Value Epoch Time tepoch 23:05:28.22 UTC January 3, 2012 Semi-major Axis a 7,310.8163 km Period T 103.68323 min Mean Motion n 13.888456 orbits/day Eccentricity e 0.0159858 Inclination i 71°.048202 Right Ascension of the Ascending Node aW 211°.28377 Argument of Perigee w 137°.75619 Mean Anomaly M 354°.97240

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