Celestial Mechanics Problems

Abstract

The equations of motion of central force motion of a point mass are given by

EquationSource% MathType!MTEF!2!1!+- % feaagCart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyBamaabm % aabaGabmOCayaadaGaeyOeI0IaamOCaiqbew9aMzaacaWaaWbaaSqa % beaacaaIYaaaaaGccaGLOaGaayzkaaGaeyypa0JaamOzamaabmaaba % GaamOCaaGaayjkaiaawMcaaiaacYcacaaMc8UaamOCaiqbew9aMzaa % daGaey4kaSIaaGOmaiqadkhagaGaaiqbew9aMzaacaGaeyypa0Zaae % WaaeaacaaIXaGaai4laiaadkhaaiaawIcacaGLPaaadaqadaqaaiaa % dsgacaGGVaGaamizaiaadkhaaiaawIcacaGLPaaadaqadaqaaiaadk % hadaahaaWcbeqaaiaaikdaaaGccuaHvpGzgaGaaaGaayjkaiaawMca % aiabg2da9iaaicdaaaa!5CF9!]]</EquationSource><EquationSource Format="TEX"><![CDATA[$$ m\left( {\ddot r - r{{\dot \phi }^2}} \right) = f\left( r \right),\,r\ddot \phi + 2\dot r\dot \phi = \left( {1/r} \right)\left( {d/dr} \right)\left( {{r^2}\dot \phi } \right) = 0 $$

(2.2.1)

If the attracting force directed to the center satisfies Newton’s laws, we have

EquationSource% MathType!MTEF!2!1!+- % feaagCart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzamaabm % aabaGaamOCaaGaayjkaiaawMcaaiabg2da9maabmaabaGaamyBaiaa % d2eacqaHZoWzcaGGVaGaamOCamaaCaaaleqabaGaaGOmaaaaaOGaay % jkaiaawMcaaiabg2da9iabgkHiTmaabmaabaGaamyBaiaadUeacaGG % VaGabmOCayaacaWaaWbaaSqabeaacaaIYaaaaaGccaGLOaGaayzkaa % GaaiOlaaaa!4A8C!]]</EquationSource><EquationSource Format="TEX"><![CDATA[$$ f\left( r \right) = \left( {mM\gamma /{r^2}} \right) = - \left( {mK/{{\dot r}^2}} \right). $$

(2.1.2)

r and γ are the polar coordinates of motion, m and M are the masses of the two bodies, and γ and K are certain constants.