numeth-ans2prob

Page 1

Problem: • The time equation for elliptical orbits has the form M = φ - E sin φ ; where M is known as the mean anomaly, φ is known as the eccentric anomaly and E is the eccentricity (between 0 and 1) of the elliptic orbit. Obtain a solution that satisfies the εra < 0.001. Use Simple Fixed Point Iteration if E = 1.


Solution: • φ - E sin φ - M = 0 • εra < 0.001 • • • •

E=1 M = mean anomaly φ = eccentric anomaly E is the eccentricity (between 0 and 1) of the elliptic orbit.


• f(φ) = φ - E sin φ -M • M = (0+1)/2 = 0.5 • f(φ) = φ - 1 sin φ - 0.5

• f(φ) = φ - sin φ - 0.5

E=1


SFPI:

f(φ) = φ - sin φ - 0.5

• φ = sin φ + 0.5

---- 1

• φ =sin -1 ( φ - 0.5 )

---- 2


φ = sin φ + 0.5 φi

F(φi+1)

φi

F(φi+1)

0.0000 0.5000 0.9794 1.3302 1.4712 1.4950 1.4971 1.4973

0.5000 0.9794 1.3302 1.4712 1.4950 1.4971 1.4973 1.4973

1 1.3415 1.4738 1.4953 1.4972 1.4973

1.3415 1.4738 1.4953 1.4972 1.4973 1.4973


NRM:

f(φ) = φ - sin φ - 0.5

• F’(φ) = 1 - cos φ if φ = 0 φi F(φi) F’(φi) φi+1 0.0000 -0.5000 0.0000 #DIV/0! #DIV/0! #DIV/0! #DIV/0! #DIV/0!


f(φ) = φ - sin φ - 0.5

NRM:

• F’(φ) = 1 - cos φ if φ = 1.0 φi

F(φi)

F’(φi)

φi+1

1 1.7428 1.5229 1.4976 1.4973

-0.3415 0.2576 0.0240 0.0003 0.0000

0.4597 1.1712 0.9521 0.9269 0.9266

1.7428 1.5229 1.4976 1.4973 1.4973


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