Consider Kepler\'s equation: E = M + esin E. A first-order approximation for the

Question

Consider Kepler's equation: E = M + esin E. A first-order approximation for the solution is given by assuming that e = 0, so E M. Substituting this into Kepler's equation gives the first-order approximation E M + esin M. Your goal is to continue this approach to derive a second-order approximation for E. Begin by substituting E M + esin M back into Kepler's equation. Then assume that e is small. Your final form should look like E M + esin M + c(e, M), i.e., your goal is to determine c(e, M) which is a function of e and M.

Explanation / Answer

E = M + esinE

In the limit of zero eccentricity, we just have

E = M

a simple iterative scheme for improving the starting approximation

Ek = M + esin*Ek-1

E1 = M+esinM

Eo = M

E2 = M+esinE1

E2 = M + esin(M+esinM)

E2 = M + esinM + e^2sinM*cosM

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.