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Satellite Orbit Characteristics Problem

The Italian satellite SIRIO had the following orbital parameters on March 26, 1979 at 00:00:00UT.

Semimajor Axis: 42,167.911km

Eccentricity: 0.00033

Mean Anomaly: 28.3866 degrees



Given the provided orbital measurements above, determine the following:

a) The orbital period in hours, minuets, and seconds

b) The mean orbital angular velocity in radians per second

c) The maximum and minimum distances of the spacecraft from the center of the earth during each orbital revolution

d) The time (expressed as a date, hour, minute, and second) of the next perigee passage after 00:00:00UT on March 26, 1979.


Solution:

a) The orbital period time (T):

T = 2π sqrt [ a3 / μ ]

T = 2π sqrt [ (42,167.911km)3 / 3.9861352E5 km3/s2 ]

T = 23.9373 hours = 23hrs, 56min, 14.15sec


b) The orbital angular velocity eta (η):

η = 2π / T = 2π / 23.9373 hours = 7.291E-5 rad/s


c) The min and max distance at each orbit revolution (rmin, rmax)

rmin = a - ae = a ( 1 - e ) = 42,167.911km ( 1 - 0.00033 ) = 42,153.9956km

rmax = a + ae = a ( 1 + e ) = 42,167.911km ( 1 + 0.00033 ) = 42,181.826km

Note: daily range (r) is equal to 27.83km


d) Perigee passage is calculated as:

t = tP = M / η = (28.3866 deg / 7.291E-5 rad/s) (2π rad / 360 deg) = 6795sec = 1.888hrs

Hence, the previous perigee passage was 1.888hrs prior to the given measurement. Therefore, the next passage is:

T - 1.888hrs = 23.9373hrs - 1.888hrs = 22.0493hrs after 00:00:00 on March 26th.

So the next perigee passage will occur at 22:02:57.48 on March 26, 1979


Satellite key words: orbit measurement, period, angular velocity, mean anomaly, spacecraft, radians, eccentricity, semi major axis, date, UT, revolution, location, point source, GEO, LEO, MEO, maximum distance, minimum distance, perigee, apogee, speed.


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