Mathematics
Undergraduate
Resolved
波線を引いた式の導出がよく分かりません。u~²を展開して整理するとたしかに1-e²/2は出てくるのですが、eφsinφなどはどうしたら出てくるのでしょうか？
To see this, recall from Appendix C that the Newtonian orbit equation
(d'u/dp?)+ u= M/L? has à =
helion at p = 0. The hypotheses imply r> L»M SO this is close to the
relativistic solution. To obtain a more refined approximation, we use ü in
(M/L")(1 + e cos φ) as solution with peri-
the relativistic correction term thus
代人
(d'u/dp?) + u = (M/L?) + 3Mz?.
A routine computation gives
3m3
(1+e cos p)+
L
e?
e?
1+
2
M
cos 2p + ep Sin p
6
u=
L?
as the solution with perihelion at φ = 0. To find the next perihelion we need
3M°ele
p+
L*
du
Me
sin
L?
sin 2p + sinφ+p cos φ
3
dp
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できました！テイラー展開かと思ったら単純に代入して解くだけでした！