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【第二坑】有关天体运行轨道的研究,如果直接这样……

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为了简便,一切写在字母后面的非零数字表示乘方


IP属地:美国1楼2013-01-29 17:31回复
    题目还是这个:
    平面内两质点O和P,距离为r0,O固定,质量为M,P初速度v0,方向垂直两点连线。P仅受万有引力作用。仅考虑牛顿力学(相对论一边去),求P轨迹
    转换成数学语言
    P''=-GM*P/|P|3
    P(0)=(r0,0)
    P'(0)=(0,v0)
    P(t)


    IP属地:美国本楼含有高级字体2楼2013-01-29 17:33
    回复
      P=(rcosθ,rsinθ)
      则|P|=r
      P''=((rcosθ)'',(rsinθ)'')
      =(r''cosθ-2r'θ'sinθ-rθ''sinθ-rθ'2cosθ,
      r''sinθ+2r'θ'cosθ+rθ''cosθ-rθ'2sinθ)
      原方程变为:
      r''cosθ-2r'θ'sinθ-rθ''sinθ-rθ'2cosθ=-GMcosθ/r2
      r''sinθ+2r'θ'cosθ+rθ''cosθ-rθ'2sinθ=-GMsinθ/r2

      r''-2r'θ'tanθ-rθ''tanθ-rθ'2=-GM/r2……①
      r''+2r'θ'cotθ+rθ''cotθ-rθ'2=-GM/r2……②
      ②-①:
      (2r'θ'+rθ'')(tanθ+cotθ)=0
      ∵tanθ+cotθ≠0
      ∴2r'θ'+rθ''=0……③
      ③代入①:
      r''-rθ'2=-GM/r2


      IP属地:美国本楼含有高级字体4楼2013-01-29 17:41
      回复
        r''-rθ'2=-GM/r2……④
        ④整理可得
        θ'=±√(GM/r3+r''/r)⑤
        θ''=±[-3GMr'/r4+(r'''r-r''r')/r2]/√(GM/r3+r''/r)⑥
        ⑤⑥代入③
        ±r'√(GM/r3+r''/r)±r[-3GMr'/r4+(r'''r-r''r')/r2]/√(GM/r3+r''/r)=0
        4r'(GM/r3+r''/r)=3GMr'/r3-r'''+r''r'/r
        GMr'/r3+r'''+3r''r'/r=0
        GMr'+r3r'''+3r2r'r''=0
        r3r'''+3r2r'r''=-GMr'
        (r3r'')'=(-GMr)'
        r3r''=-GMr + constA


        IP属地:美国5楼2013-01-29 17:48
        回复
          r3r''=-GMr + constA……⑦
          考虑constA:
          constA=r(0)3r''(0)+GMr(0)=r03r''(0)+GMr0
          至于r''(0),考虑④式
          r''(0)=-GM/r(0)2+r(0)θ'(0)2=-GM/r02+v02/r0
          ∴constA=-GMr0+v02r02+GMr0=v02r02
          因此⑦式为
          r3r''=-GMr+v02r02


          IP属地:美国6楼2013-01-29 17:51
          收起回复
            r3r''=-GMr+v02r02
            r''=-GM/r2+v02r02/r3
            dr/dt d(dr/dt) = dr(-GM/r2+v02r02/r3)
            r'2/2=GM/r-v02r02/(2r2)+ constB


            IP属地:美国7楼2013-01-29 17:57
            回复
              r'2/2=GM/r-v02r02/(2r2)+ constB ……⑧
              考虑constB:
              constB=r'(0)2/2-GM/r(0)+v02r02/(2r(0)2)=-GM/r0+v02/2
              ∴⑧式为
              r'2/2=GM/r-v02r02/(2r2)-GM/r0+v02/2
              r'2=2GM(1/r-1/r0)+v02(1-r02/r2)


              IP属地:美国来自Android客户端10楼2013-01-30 08:04
              回复
                好吧,好像写成这样更好:
                r'2=2GM/r-v02r02/r2-2GM/r0+v02
                这个方程的解:
                Ⅰ.当v02<2GM/r0时
                引进变参φ∈[0,∞)
                r与t的关系满足参数方程
                t=[GMφ+(GM-v02r0)sinφ]/(2GM/r0-v02)^(3/2)
                r=[GM+(GM-v02r0)cosφ]/(2GM/r0-v02)
                Ⅱ.当v02>2GM/r0时
                引进变参ψ∈[0,∞)
                r与t的关系满足参数方程
                t=[GMψ+(GM-v02r0)sinhψ]/(v02-2GM/r0)^(3/2)
                r=[GM+(GM-v02r0)coshψ]/(2GM/r0-v02)
                事实上ⅠⅡ两种情况的结果可用φ=iψ相互转化
                Ⅲ.当v02=2GM/r0时
                r与t的关系满足方程
                2/3 * (r-r0)^(3/2) + 2r0*(r-r0)^(1/2) = t√(2GM)


                IP属地:美国15楼2013-01-30 20:43
                回复
                  在ⅠⅡ两种情况下继续推导
                  (之所以一块写是因为两个很相似,且对比起来很有意思)
                  Ⅰ的用黑色表示
                  Ⅱ的用红色表示
                  r'
                  =(dr/dφ)/(dt/dφ)
                  =(dr/dψ)/(dt/dψ)
                  ={-(GM-v02r0)sinφ/(2GM/r0-v02)} / {[GM+(GM-v02r0)cosφ]/(2GM/r0-v02)^(3/2)}
                  ={(GM-v02r0)sinhψ/(2GM/r0-v02)} / [GM+(GM-v02r0)coshψ]/(v02-2GM/r0)^(3/2)}
                  =-(GM-v02r0)sinφ/[GM+(GM-v02r0)cosφ]√(2GM/r0-v02)
                  =-(GM-v02r0)sinhψ/[GM+(GM-v02r0)coshψ]√(v02-2GM/r0)


                  IP属地:美国本楼含有高级字体16楼2013-01-30 20:49
                  回复
                    r''
                    =(d(r')/dφ)/(dt/dφ)
                    =(d(r')/dψ)/(dt/dψ)
                    ={-(GM-v02r0)cosφ*[GM+(GM-v02r0)cosφ]-(GM-v02r0)sinφ*(GM-v02r0)sinφ}/[GM+(GM-v02r0)cosφ]^2*√(2GM/r0-v02) / {[GM+(GM-v02r0)cosφ]/(2GM/r0-v02)^(3/2)}
                    ={-(GM-v02r0)coshψ*[GM+(GM-v02r0)coshψ]+(GM-v02r0)sinhψ*(GM-v02r0)sinhψ}/[GM+(GM-v02r0)coshψ]^2*√(v02-2GM/r0) / {[GM+(GM-v02r0)coshψ]/(v02-2GM/r0)^(3/2)}
                    =(2GM/r0-v02)^2[-GM(GM-v02r0)cosφ-(GM-v02r0)^2]/[GM+(GM-v02r0)cosφ]^3
                    =(2GM/r0-v02)^2[-GM(GM-v02r0)coshψ-(GM-v02r0)^2]/[GM+(GM-v02r0)coshψ]^3


                    IP属地:美国本楼含有高级字体17楼2013-01-30 21:04
                    回复
                      根据前面的⑤式
                      θ'2=GM/r3+r''/r
                      =GM(2GM/r0-v02)^3/[GM+(GM-v02r0)cosφ]^3 + (2GM/r0-v02)^3[-GM(GM-v02r0)cosφ-(GM-v02r0)^2]/[GM+(GM-v02r0)cosφ]^4
                      =GM(2GM/r0-v02)^3/[GM+(GM-v02r0)coshψ]^3 + (2GM/r0-v02)^3[-GM(GM-v02r0)cohψ-(GM-v02r0)^2]/[GM+(GM-v02r0)cohψ]^4
                      =(2GM/r0-v02)^3/[GM+(GM-v02r0)cosφ]^4 *{GM[GM+(GM-v02r0)cosφ] -GM(GM-v02r0)cosφ-(GM-v02r0)^2}
                      =(2GM/r0-v02)^3/[GM+(GM-v02r0)coshψ]^4 *{GM[GM+(GM-v02r0)coshψ] -GM(GM-v02r0)coshψ-(GM-v02r0)^2}
                      =(2GM/r0-v02)^3/[GM+(GM-v02r0)cosφ]^4 *(2GMv02r0-v04r02)
                      =(2GM/r0-v02)^3/[GM+(GM-v02r0)coshψ]^4 *(2GMv02r0-v04r02)
                      =(2GM/r0-v02)^4/[GM+(GM-v02r0)cosφ]^4 * v02r02
                      =(2GM/r0-v02)^4/[GM+(GM-v02r0)coshψ]^4 * v02r02
                      ∴θ'
                      =v0r0(2GM/r0-v02)^2/[GM+(GM-v02r0)cosφ]^2
                      =v0r0(2GM/r0-v02)^2/[GM+(GM-v02r0)coshψ]^2


                      IP属地:美国本楼含有高级字体18楼2013-01-31 18:08
                      回复
                        θ=∫θ'dt
                        =∫θ'dφ*dt/dφ
                        =∫θ'dψ*dt/dψ
                        =∫v0r0(2GM/r0-v02)^2/[GM+(GM-v02r0)cosφ]^2 * {[GM+(GM-v02r0)cosφ]/(2GM/r0-v02)^(3/2)} *dφ
                        =v0r0(2GM/r0-v02)^2/[GM+(GM-v02r0)coshψ]^2 * {[GM+(GM-v02r0)coshψ]/(v02-2GM/r0)^(3/2)} *dψ
                        =∫√(2GMv02r0-v04r02)/[GM+(GM-v02r0)cosφ] *dφ
                        =∫√(v04r02-2GMv02r0)/[GM+(GM-v02r0)coshψ] *dψ


                        IP属地:美国本楼含有高级字体19楼2013-01-31 18:26
                        回复
                          θ
                          =∫√(2GMv02r0-v04r02)/[GM+(GM-v02r0)cosφ] *dφ
                          =∫√(v04r02-2GMv02r0)/[GM+(GM-v02r0)coshψ] *dψ
                          =√(2GMv02r0-v04r02)∫dφ[cos²(φ/2)+sin²(φ/2)]/[(2GM-v02r0)cos²(φ/2)+v02r0sin²(φ/2)]
                          =√(v04r02-2GMv02r0)∫dψ[cosh²(ψ/2)-sinh²(ψ/2)]/[(2GM-v02r0)cosh²(ψ/2)-v02r0sinh²(ψ/2)]
                          =2√(2GMv02r0-v04r02)∫d(φ/2)[1+tan²(φ/2)]/[2GM-v02r0+v02r0tan²(φ/2)]
                          =2√(v04r02-2GMv02r0)∫d(ψ/2)[1-tanh²(ψ/2)]/[2GM-v02r0-v02r0tanh²(ψ/2)]
                          =2/(2GM-v02r0)*√(2GMv02r0-v04r02)∫d(φ/2)sec²(φ/2)/[1+v02r0/(2GM-v02r0)*tan²(φ/2)]
                          =2/(2GM-v02r0)*√(v04r02-2GMv02r0)∫d(ψ/2)sech²(ψ/2)/[1-v02r0/(2GM-v02r0)*tanh²(ψ/2)]
                          =2/√(2GMv02r0-v04r2)*√(2GMv02r0-v04r02)∫d(φ/2)√[v02r0/(2GM-v02r0)]sec²(φ/2)/[1+v02r0/(2GM-v02r0)*tan²(φ/2)]
                          =2/√(v04r2-2GMv02r0)*√(v04r02-2GMv02r0)∫d(ψ/2)√[v02r0/(v02r0-2GM)]sech²(ψ/2)/[1+v02r0/(v02r0-2GM)*tanh²(ψ/2)]
                          =2∫d{√[v02r0/(2GM-v02r0)]tan(φ/2)}/{1+[√[v02r0/(2GM-v02r0)]tan(φ/2)]^2}
                          =2∫d{√[v02r0/(v02r0-2GM)]tanh(ψ/2)}/{1+[√[v02r0/(v02r0-2GM)]tanh(ψ/2)]^2}
                          =2arctan{√[v02r0/(2GM-v02r0)]tanφ/2} + constC
                          =2arctan{√[v02r0/(v02r0-2GM)]tanhψ/2} + constC
                          ∵θ(0)=0,φ(0)=0,ψ(0)=0,constC=constD=0
                          tan(θ/2)=
                          √[v02r0/(2GM-v02r0)]tan(φ/2)
                          √[v02r0/(v02r0-2GM)]tanh(ψ/2)


                          IP属地:美国本楼含有高级字体20楼2013-01-31 21:21
                          回复
                            x=rcosθ
                            =[GM+(GM-v02r0)cosφ]/(2GM/r0-v02) *cosθ
                            =[GM+(GM-v02r0)coshψ]/(2GM/r0-v02) *cosθ
                            =[GM+(GM-v02r0)cosφ]/(2GM/r0-v02) *[1-tan²(θ/2)]/[1+tan²(θ/2)]
                            =[GM+(GM-v02r0)coshψ]/(2GM/r0-v02) *[1-tan²(θ/2)]/[1+tan²(θ/2)]
                            =[GM+(GM-v02r0)cosφ]/(2GM/r0-v02) *[(2GM-v02r0)-v02r0tan²(φ/2)]/[(2GM-v02r0)+v02r0tan²(φ/2)]
                            =[GM+(GM-v02r0)coshψ]/(2GM/r0-v02) *[(v02r0-2GM)-v02r0tanh²(ψ/2)]/[(v02r0-2GM)+v02r0tanh²(ψ/2)]
                            =[GM+(GM-v02r0)cosφ]/(2GM/r0-v02) *[(2GM-v02r0)(1+cosφ)-v02r0(1-cosφ)]/[(2GM-v02r0)(1+cosφ)+v02r0(1-cosφ)]
                            =[GM+(GM-v02r0)coshψ]/(2GM/r0-v02) *[(v02r0-2GM)(-1-coshψ)-v02r0(1-coshψ)]/[(v02r0-2GM)(-1-coshψ)+v02r0(1-coshψ)]
                            =[GM+(GM-v02r0)cosφ]/(2GM/r0-v02) *[2GM-2v02r0+2GMcosφ]/[2GM+(2GM-2v02r0)cosφ]
                            =[GM+(GM-v02r0)coshψ]/(2GM/r0-v02) *[2GM-2v02r0+2GMcoshψ]/[2GM+(2GM-2v02r0)coshψ]
                            =(GM-v02r0+GMcosφ)/(2GM/r0-v02)
                            =(GM-v02r0+GMcoshψ)/(2GM/r0-v02)


                            IP属地:美国本楼含有高级字体24楼2013-02-06 09:25
                            回复
                              y2=r2-x2
                              ={[GM+(GM-v02r0)cosφ]^2-(GM-v02r0+GMcosφ)^2}/(2GM/r0-v02)^2
                              ={[GM+(GM-v02r0)coshψ]^2-(GM-v02r0+GMcoshψ)^2}/(2GM/r0-v02)^2
                              =[(GM)^2-(GM-v02r0)^2](1-cos²φ)/(2GM/r0-v02)^2
                              =[(GM)^2-(GM-v02r0)^2](1-cosh²ψ)/(2GM/r0-v02)^2
                              =(2GMv02r0-v04r02)(1-cos²φ)/(2GM/r0-v02)^2
                              =(2GMv02r0-v04r02)(1-cosh²ψ)/(2GM/r0-v02)^2
                              =v02r02sin²φ/(2GM/r0-v02)
                              =v02r02sinh²ψ/(v02-2GM/r0)
                              y=v0r0sinφ/√(2GM/r0-v02)
                              y=v0r0sinhψ/√(v02-2GM/r0)


                              IP属地:美国本楼含有高级字体25楼2013-02-06 09:32
                              回复