Newton solution of Keplerian equation



(1)  M=MA*2*3.14/256                       : Radian from MA
(2)  to=M+e*sin(M)+0.5*e^2*sin(2*M)      : Initial value
(3)  mo=to-e*sin(to)                       : Keplerian equation
(4)  dto=(M-mo)/(1-e*cos(to))              : e=Eccentricity
(5)  t1=to+dto                             : 1st approximate value
(6)  m1=t1-e*sin(t1)                       : calculation
(7)  dt1=(M-m1)/(1-e*cos(t1))              : calculation
(8)  t2=t1+dt1                             : 2nd approximate value
(9)  m2=t2-e*sin(t2)                       : calculation
(10) dt2=(M-m2)/(1-e*cos(t2))              : calculation
(11) t3=t2+dt2                             : 3rd approximate value
(12) t4=tan(t3/2)                          : calculation
(13) u=root((1+e)/(1-e))*t4                : calculation
(14) s=2*arctan(u)                         : Real Anomaly
(15) p=24*3600/n                           : Period (total seconds)
(16) 4*3.14^2*(a^3/p^2)=G*Q                : A law of the universal graviation
(17) a=(((G*Q*p^2)/(4*3.14^2))^(1/3))*10^9 : Semi Major Axis
(18) r0=a(1-e^2)/(1+e*cos(s))              : Altitude from the center of the earth
(19) r=r0-637814000                        : Altitude from the face of the earth


Supplement of (16),(17):

  a = (((G*Q*p^2)/(4*3.14^2))^(1/3))*10^9
  a = (6.6732*10^(-8)*5.9732*10^27*(p*p)/(4*3.14159*3.14159))^(1/3)

Then, G is the constant of the universal gravitation,
      Q is the mass of the earth.


  [N] = [(kg*m)/s^2] = 10^3 * 10^2 [(g*cm)/s^2]
                     = 10^5        [(g*cm)/s^2]

  [m^2] = 10^4 [cm^2],  [kg^2] = 10^6 [g^2]

  G = 6.6732 * 10^(-11)                       [N*m^2/kg^2]
    = 6.6732 * 10^(-11) * 10^5 * 10^4 / 10^6  [cm^3/(g*(s^2))]
    = 6.6732 * 10^(-8)                        [cm^3/(g*(s^2))]

  Q = 5.9732 * 10^24  [kg]
    = 5.9732 * 10^27  [g]

  R : Mass of the satellite
  a : Semi Major Axis
  p : Period


Then, R * a * (2π/p)^2 = (G * Q * R) / a^2

      i.e.  p^2 = (4π^2 / (G * Q)) * a^3

      so,   a = ((G * Q) * p^2 / 4π^2 )^(1/3)

The unit is [(cm^3/(g*(s^2)) * g * s^2)^(1/3)] = [cm].


--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
           Mean Anomaly       Eccentri.  Ini.val. Kep.equation      1st appro.value            2nd appro.value            3rd appro.value          Real anomaly      Mean motion   Period  (h)   (m)    SMA           Dist.C        Dist.S

 MA        MA'     M          e          to       Mo       dto      t1       m1       dt1      t2       m2       dt2      t3          t4           u        s        n             p        p1    p2    a             r             rh
 
(sample1 : 16 Jan 1994, AO-13)
 22        30.9    0.53969    0.72099    1.139    0.484    0.079    1.218    0.541   -0.002    1.216    0.540    0.000    1.216       0.696        1.729    2.093    2.09721276    41198    11    27    2578169623    1933173674    12953.6
 
(sample2 : 28 May 2001, AO-40)
0.01        0.0    0.00025    0.81492    0.001    0.000    0.000    0.001    0.000    0.001    0.002    0.000    0.001    0.003       0.002        0.006    0.012    1.27026844    68017    18    54    3601408456     666581751      287.7
128       180.0    3.14000    0.81492    3.140    3.139    0.001    3.141    3.141   -0.001    3.140    3.139    0.001    3.141    3374.653    10567.542    3.141    1.27026844    68017    18    54    3601408456    6536251656    58984.4
255.99    360.0    6.27975    0.81492    6.275    6.282   -0.010    6.265    6.280   -0.001    6.264    6.280   -0.001    6.263      -0.010       -0.031   -0.062    1.27026844    68017    18    54    3601408456     667135754      293.2
 
(sample3 : 23 Jun 2001, AO-40 before arcjet)
0.01        0.0    0.00025    0.81508    0.001    0.000    0.000    0.001    0.000    0.001    0.002    0.000    0.001    0.003       0.002        0.006    0.012    1.2711484     67970    18    53    3599749208     665697946      278.8
 86       121.4    2.10969    0.81508    2.517    2.040    0.042    2.559    2.111   -0.001    2.558    2.109    0.000    2.558       3.329       10.430    2.950    1.2711484     67970    18    53    3599749208    6046194040    54083.8
128       180.0    3.14000    0.81508    3.140    3.139    0.001    3.141    3.141   -0.001    3.140    3.139    0.001    3.141    3374.653    10572.585    3.141    1.2711484     67970    18    53    3599749208    6533816936    58960.0
 
(sample4 : 23 Jun 2001, AO-40 after arcjet)
0.01        0.0    0.00025    0.81501    0.001    0.000    0.000    0.001    0.000    0.001    0.002    0.000    0.001    0.003       0.002        0.006    0.012    1.2710861     67973    18    53    3599855129     665944691      281.3
100       140.5    2.45313    0.81501    2.645    2.257    0.114    2.759    2.455   -0.001    2.758    2.453    0.000    2.758       5.150       16.132    3.018    1.2710861     67973    18    53    3599855129    6321346299    56835.3
128       180.0    3.14000    0.81501    3.140    3.139    0.001    3.141    3.141   -0.001    3.140    3.139    0.001    3.141    3374.653    10570.598    3.141    1.2710861     67973    18    53    3599855129    6533782042    58959.7
--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------


The solution of this algorithm corrected by W3IWI and JE9PEL :

                    Before      After
  Perigee height :   278.8        281.3  ....... +2.5 km
  Apogee height  : 58960.0      58959.7  ....... -0.3 km


The report of Peter Guelzow, DB2OS :

                    Before      After
  Perigee height :   279.754      282.2427  .... +2.5 km
  Apogee height  : 58971.166    58971.024   .... -0.1 km


Excel sheet of the Satellite Altitude analysis,

http://www.ne.jp/asahi/hamradio/je9pel/satrange.zip


> Date: Sat, 23 Jun 2001 10:08:15 +0000
> From: Peter Guelzow
> To: amsat-bb, amsat-dl
> Subject: AO-40: more successful ARCJET operations
> 
> Dear All,
> 
> The arc-jet thruster has been invoked on Orbit 296, MA 118-135,
> which gave about 1 hour of thrusting gas only. 
> 
> The S2 TX is OFF from MA 100-180 to spare power for the ATOS. 
> The gas generator for the ammonia draws about 120 - 130 W of power
> when cycled on by the thermostat. 
> 
> The IHU-2 is running and logging telemetry into a circular buffer
> and hold about 2.5 days worth of data. Downloaded telemetry from
> this 1h burn indicated positive power budget and everything looks
> nominally. 
> 
> The thrust on orbit 296 started at MA 121.4 and lasted for 3618s.
> The acceleration is guestimated to be 54E-6 m/s^2 ,
> and the direction of acceleration is towards alon 274, alat -2 
> (the current attitude).
> 
> Give or take the unknowns, the expected outcome of this was:
> 
>                      Before            After
>   --------------------------------------------
>   Epoch year            2001              2001
>   Epoch time       173.12145         173.16312
>   Inclination         5.2833         5.2833592
>   R.A.A.N          180.71591         180.70361
>   Eccentricity      0.815077         0.8150139
>   Arg perigee      288.69088         288.71333
>   Mean Anomaly         121.4         140.46836
>   Mean motion MM   1.2711484         1.2710861
>   Revolution             296               296
>   SMA                36003.6         36004.773
>       
>   Perigee height     279.754          282.2427   ...  +2.5 km
>   Apogee height    58971.166         58971.024
>   --------------------------------------------
> 
> So the perigee was raised by this 1h burn by about +2.5 km, but only
> a thrust rate of 50% was used. 
> 
> Since everything went so well, the computer onboard AO-40 was 
> commanded to initiate 2h "burns" starting around apogee on orbit 297
> for the next three orbits. 
> 
> The first 2-hour burn stopped at Orbit 297 MA 142, which is 2001 
> Jun 23 0154 utc. Again, all telemetry is looking good and we may soon
> expand to 4h "burns" and possibly increase the thrust level as well.
> 
> While we haven't yet seen the latest NORAD data, some effects of the
> changing orbit should be soon noticeable. 
> 
> Indeed, we all are very happy with the successful results of the ATOS
> (Arcjet Thruster on OSCAR Satellite) so far.
> 
> Some more information about the ATOS system is available here:
> 
> http://www.irs.uni-stuttgart.de/RESEARCH/EL_PROP/PROJ/e_atos.html
> 
> best 73s
>  Peter DB2OS for the AO-40 team


> Date: Sun, 24 Jun 2001 18:08:22 -0400
> From: Tom Clark
> To: Mineo Wakita, amsat-bb
> Subject: Re: AO-40: more successful ARCJET operations
> 
> Mineo -- Nice work. The perigee difference is probably due to your
> definition of altitude, i.e. distance from the surface of the earth.
> You said:
> 
> > (18) r0=a(1-e^2)/(1+e*cos(s)) : Height from the center of the earth
> > (19) r=r0-6371                : Height from the face of the earth
> 
> The equatorial radius of the oblate earth is 6378.160 km, larger than
> the 6371 km you use. Since the inclination is only 5.3 deg, it is
> appropriate to use the equatorial value.
> 
> Therefore, using your formulation, I come up with:
> 
> > The solution of this algorithm [posted by JE9PEL]
> >                     Before      After
> >   Perigee height :   286.0        288.4  ....... +2.4 km
> >
> > The report of Peter Guelzow, DB2OS
> >                     Before      After
> >   Perigee height :   279.754      282.2427  .... +2.5 km
> 
>   The JE9PEL perigee corrected by W3IWI:
>                       Before        After
>     Perigee height :   278.8        281.3  ....... +2.5 km
> 
> This agrees to ~1 km with the results that Peter reported. Since the
> object of the "burn" was to raise perigee (with very little change
> in apogee), and since a post-burn mean motion is not yet available,
> I haven't bothered to examine the apogee differences.
> 
> 73 de Tom, W3IWI


> Date: Mon, 25 Jun 2001 22:24:29 +0900
> From: Mineo Wakita
> To: tac
> Subject: Re: AO-40: more successful ARCJET operations
> 
> Thank you, Tom Clark / W3IWI.
> 
> I confirmed the equatorial radius of the oblate earth is 6378.140 km
> (6378140 m = 637814000 cm) by a astronomical journal of Japanese.
> As a result, this perigee height agrees to ~1 km with the results 
> that Peter reported.
> 
> I corrected as the following (18),(19) and I reposted the Excel sheet.
> http://www.ne.jp/asahi/hamradio/je9pel/satrange.zip
> 
> 
> (18) r0=a(1-e^2)/(1+e*cos(s)) : Distance from the center of the earth
> (19) r=r0-637814000           : Distance from the surface of the earth
> 
> 
> The solution of this algorithm corrected by W3IWI and JE9PEL :
> 
>                     Before      After
>   Perigee height :   278.8        281.3  ....... +2.5 km
>   Apogee height  : 58960.0      58959.7  ....... -0.3 km
> 
> 
> The report of Peter Guelzow, DB2OS :
> 
>                     Before      After
>   Perigee height :   279.754      282.2427  .... +2.5 km
>   Apogee height  : 58971.166    58971.024   .... -0.1 km


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