Astrodynamics - radius of the Earth (Q/A)

• Astrodynamics - radius of the Earth (Q/A) by aero on 29 Aug, 2014 22:50
• I've been playing with formulas for circular orbits and characteristic orbital energy hoping to ovoid the need to remember the value of the gravitational parameter, mu. It turns out that the equations have been and can be simplified to some surprising results, (to me). But the accepted value of the radius of the Earth seems to be in error. Please follow my logic and let me know where I erred, or not.
  
  Start by finding velocity of a circular orbit, V_o around Earth.
  eqn. 1) V_o² = mu/r , but I don't want to use mu, so I go the long way by setting gravitational acceleration equal to centrifugal acceleration and solving for V_o².
  eqn. 2) g_h = a_c
  eqn. 3) g = mu/r²
  eqn. 4) a_c = V_o²/r where
      r is the Earth centered distance of the circular orbit, equals r_e + h , that is, Earth's radius plus orbit altitude.
  Write eqn. 3) twice and divide equations. After substituting values gives
  eqn. 5) g_h = g_surface * r_e²/(r_e +h)²
  where g_surface = 9.80665 m/s² by definition
  Substitute eqn. 4) and eqn. 5) into eqn. 2), rearranging and using r = r_e + h, gives
  V_o² = g_surface*r_e²/(r_e + h)
  
  So I now have the equation I was after, circular orbital velocity without resorting to the use of mu.
  Now I looked at characteristic energy, C₃ of the orbit. From Wikipedia I find
  eqn. 6) V_o/2 - mu/r = C₃/2 where r is again equal to r_e + h.
  Just out of curiosity and to avoid using mu, I substitute eqn. 1) into eqn. 6) and multiply through by 2. The result
  eqn. 7) C₃ = - V_o²
  
  Characteristic energy equals the square of circular orbital velocity. This did surprise me so I checked numerically over a range of heights using Excel. With
  mu = 398600.4418 km^3/s^2 and
  r_e = 6,371,000 meters
  I got good agreement, but not perfect agreement between the C₃ values calculated using eqn. 6) and eqn. 7) . The difference was a constant 0.2766% over the altitude range from 0 to 55,000 km. So what is wrong? The only common input parameter in both calculations for C₃ is Earth radius. A bit of variation on r_e results in 6,375,417 meters giving 0.0000% difference to four decimal places in the values of C₃ calculated by eqn. 6) and eqn. 7).
  
  I will need to read your comments before I'll know if they are welcome.
  
  
  add: Looking at http://en.wikipedia.org/wiki/Earth_radius I see a lot of different values for the radius of the Earth, ranging from 6,353 km to 6,384 km.
  I will bravely conjecture that the Astrodynamic radius of the earth = 6,375.417 km.
• #1 by Proponent on 30 Aug, 2014 05:10
• Earth is not perfectly spherical, so I think agreement to within 0.2766% is pretty good and I wouldn't worry about it.
  
  By the way, my own trick for not having to remember mu is to use "natural" units.  Take a convenient orbit as a reference, say a circular parking orbit around Earth at an altitude of 400 km.  Use the radius and speed of that orbit as your units of distance and velocity, respectively.  Take the time unit to be the orbital period divided by 2*pi (in other words, it's the distance unit divided by the velocity unit).  In these units, mu = 1.
• #2 by aero on 30 Aug, 2014 05:50
• Quote from: Proponent on 30 Aug, 2014 05:10

  Earth is not perfectly spherical, so I think agreement to within 0.2766% is pretty good and I wouldn't worry about it.
  
  By the way, my own trick for not having to remember mu is to use "natural" units.  Take a convenient orbit as a reference, say a circular parking orbit around Earth at an altitude of 400 km.  Use the radius and speed of that orbit as your units of distance and velocity, respectively.  Take the time unit to be the orbital period divided by 2*pi (in other words, it's the distance unit divided by the velocity unit).  In these units, mu = 1.
  

  
  Thanks - That seems like a neat trick.
  
  As for where a 4.5 km difference in r_e matters, I don't know. Maybe in outer planet navigation but as far as any effect on calculated delta-V goes, well, again, I don't know. And once escaped from Earth, it couldn't matter much for the spacecraft trajectory. It might have a small effect on the ranging signal but DSN most likely uses GPS coordinates for transmitter/receiver position and not r_e at all.
• #3 by mdo on 30 Aug, 2014 06:33
• Quote from: aero on 29 Aug, 2014 22:50

  ... hoping to ovoid the need to remember the value of the gravitational parameter, mu.
  

  
  Have a look at the definition for canonical units http://en.wikipedia.org/wiki/Canonical_units.
• #4 by ddunham on 30 Aug, 2014 07:02
• Quote

  where gsurface = 9.80665 m/s2 by definition

  
  This is the definition of a standard value for 'g', but it does not necessarily reflect the actual value of 'g' at any particular point on the earth's surface.  (Supposed to be value at 45 degree latitude, but it's off a bit, and the earth isn't a perfect spheroid anyway)    I don't know if it could be responsible for all of the 0.2% error, but I imagine it's part of it.
  
• #5 by aero on 30 Aug, 2014 07:20
• Quote from: ddunham on 30 Aug, 2014 07:02

  Quote

  where gsurface = 9.80665 m/s2 by definition

  
  This is the definition of a standard value for 'g', but it does not necessarily reflect the actual value of 'g' at any particular point on the earth's surface.  (Supposed to be value at 45 degree latitude, but it's off a bit, and the earth isn't a perfect spheroid anyway)    I don't know if it could be responsible for all of the 0.2% error, but I imagine it's part of it.
  

  
  I think that's probably right and I imagine it is all of it. The radius I calculated to remove the 0.2% is probably just the radius of a sphere where acceleration of gravity is 9.80665. Some places it's above the surface and some places it's below. It's just a little larger than the sphere with r = 6371 km. It would be an iso-something surface.
• #6 by R7 on 30 Aug, 2014 08:54
• g = mu/r²
  
  g_e = mu/r_e² = 398600.4418 km^3/s^2 / (6371 km)² = 0.00982025 km/s² = 9.82025m/s² which is not g₀. Mean radius and standard gravity are separate agreements.
  
  Anyhow, hiding from mu in astrodynamics is like hiding from pi in geometry 
• #7 by aero on 30 Aug, 2014 16:31
• Quote from: R7 on 30 Aug, 2014 08:54

  g = mu/r²
  
  g_e = mu/r_e² = 398600.4418 km^3/s^2 / (6371 km)² = 0.00982025 km/s² = 9.82025m/s² which is not g₀. Mean radius and standard gravity are separate agreements.
  
  

  
  My point exactly- 398600.4418 km^3/s^2 / (6,375.417 km)² = 0.0098665 km/s² = 9.80665m/s² which is g₀.  It boils down to, "What difference does it make?"
• #8 by aero on 31 Aug, 2014 19:14
• A related question: When a payload is launched into a 185 km or 200 km or any designated LEO altitude, what zero altitude reference do they use? Radius of the launch site, Volumetric mean radius (km), 6371.0 or something else?
  
• #9 by pericynthion on 01 Sep, 2014 23:08
• Quote from: aero on 31 Aug, 2014 19:14

  A related question: When a payload is launched into a 185 km or 200 km or any designated LEO altitude, what zero altitude reference do they use? Radius of the launch site, Volumetric mean radius (km), 6371.0 or something else?
  

  
  There's no universally-accepted convention, and this can cause confusion.  For engineering purposes, usually the issue will by bypassed by using the center of the earth as reference, e.g. with a Cartesian state vector or osculating orbital elements.
• #10 by aero on 03 Sep, 2014 18:09
• Quote from: pericynthion on 01 Sep, 2014 23:08

  Quote from: aero on 31 Aug, 2014 19:14

  A related question: When a payload is launched into a 185 km or 200 km or any designated LEO altitude, what zero altitude reference do they use? Radius of the launch site, Volumetric mean radius (km), 6371.0 or something else?
  

  
  There's no universally-accepted convention, and this can cause confusion.  For engineering purposes, usually the issue will by bypassed by using the center of the earth as reference, e.g. with a Cartesian state vector or osculating orbital elements.
  

  
  Thank you. Now I have another question. Of the various physical characteristics of earth, which ones are known more accurately than the gravitational parameter? For example, Mean density = 5.514 g/cm^3[3] (Wikipedia) but that is based on the volumetric mean radius. Mu is known much more precisely than Mass of earth so wouldn't Mass of earth be more accurately calculated using mu, which would change the mean density value?
  
  Another question. Since g = mu/r^2, can we determine r without resorting to earth parameters? I'd guess the answer is "Yes," using an accurate clock and Newton's laws. That is, g' = mu/r^2 = v^2/r, or mu/r = v^2 from centrifugal acceleration of a circular orbit. One accurate value of r plus the inverse square law gives g at any radius. All formulas well known to us.
  
  What field of study would concern itself with this and related questions about the physical characteristics of earth?
• #11 by pericynthion on 03 Sep, 2014 18:19
• Quote from: aero on 03 Sep, 2014 18:09

  What field of study would concern itself with this and related questions about the physical characteristics of earth?
  

  
  Geodesy.