An Earth orbit has a semi-major axis a of 23, 500 km, and an eccentricity e of 0
ID: 2078884 • Letter: A
Question
An Earth orbit has a semi-major axis a of 23, 500 km, and an eccentricity e of 0.65. The time of periapsis passage T_0 is 0 sec. If you find you need to iterate for this problem, use an initial guess equal to the mean anomaly M, and only check that the absolute tolerance meets the condition Delta_A = 10^-3 (i.e. there is no need to also check a relative tolerance Delta_R). a. Find the value of the eccentric anomaly E at time t = 19, 760 sec. b. At a different future time, the eccentric anomaly E is 5.75 rad. Calculate the position and velocity vectors in perifocal coordinates at this different future time.Explanation / Answer
A) a = 23500 km
e = 0.65
time of periapsis passage = 0
mean anamoly M =n(t-tp)
µ=a3n2
at t = 19760 sec, M= 0.67556 { since, µ=3.986 x 1014}
by iterative method, eccentric anamoly is
E1=M + e sin M = 0.68322
E2=M + e sin E1 = 0.68331
E3=M + e sin E2 = 0.68331
When E2= E3 iteration can be stopped
Therefore E = 0.68331
B) position vectors
x= a (cos E-e) = 8160.760
y=a sqrt(1-e2) sin E = 1789.203
= x + y = 8106.760 + 1789.760
velocity components are
V = x + y
R = a(1-e cos E) = 8301.855
= an/r
= 9.675*10-6
x = -0.0227
y = 0.1719
V = -0.0227 + 0.1719