Author Topic: Astrodynamics - radius of the Earth (Q/A)  (Read 4053 times)

Offline aero

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Astrodynamics - radius of the Earth (Q/A)
« on: 08/29/2014 10:50 PM »
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, Vo around Earth.
eqn. 1) Vo2 = 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 Vo2.
eqn. 2) gh = ac
eqn. 3) g = mu/r2
eqn. 4) ac = Vo2/r where
    r is the Earth centered distance of the circular orbit, equals re + h , that is, Earth's radius plus orbit altitude.
Write eqn. 3) twice and divide equations. After substituting values gives
eqn. 5) gh = gsurface * re2/(re +h)2
where gsurface = 9.80665 m/s2 by definition
Substitute eqn. 4) and eqn. 5) into eqn. 2), rearranging and using r = re + h, gives
Vo2 = gsurface*re2/(re + 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) Vo/2 - mu/r = C3/2 where r is again equal to re + 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 = - Vo2

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
re = 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 re 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.
« Last Edit: 08/29/2014 11:56 PM by aero »
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Offline Proponent

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Re: Astrodynamics - radius of the Earth (Q/A)
« Reply #1 on: 08/30/2014 05:10 AM »
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.

Offline aero

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Re: Astrodynamics - radius of the Earth (Q/A)
« Reply #2 on: 08/30/2014 05:50 AM »
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 re 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 re at all.
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Offline mdo

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Re: Astrodynamics - radius of the Earth (Q/A)
« Reply #3 on: 08/30/2014 06:33 AM »
... 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.

Offline ddunham

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Re: Astrodynamics - radius of the Earth (Q/A)
« Reply #4 on: 08/30/2014 07:02 AM »
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.

Offline aero

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Re: Astrodynamics - radius of the Earth (Q/A)
« Reply #5 on: 08/30/2014 07:20 AM »
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.
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Offline R7

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Re: Astrodynamics - radius of the Earth (Q/A)
« Reply #6 on: 08/30/2014 08:54 AM »
g = mu/r2

ge = mu/re2 = 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  :P ;)
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Offline aero

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Re: Astrodynamics - radius of the Earth (Q/A)
« Reply #7 on: 08/30/2014 04:31 PM »
g = mu/r2

ge = mu/re2 = 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?"
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Offline aero

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Re: Astrodynamics - radius of the Earth (Q/A)
« Reply #8 on: 08/31/2014 07:14 PM »
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?
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Offline pericynthion

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Re: Astrodynamics - radius of the Earth (Q/A)
« Reply #9 on: 09/01/2014 11:08 PM »
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.

Offline aero

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Re: Astrodynamics - radius of the Earth (Q/A)
« Reply #10 on: 09/03/2014 06:09 PM »
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?
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Offline pericynthion

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Re: Astrodynamics - radius of the Earth (Q/A)
« Reply #11 on: 09/03/2014 06:19 PM »
What field of study would concern itself with this and related questions about the physical characteristics of earth?

Geodesy.

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