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19 April 2024 19:24
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Question |
| Asked by: |
Dr. & Mrs. Hsien-Lu & Hui-Lien Peng Huang |
| Subject: |
Can Spin kinetic energy & angular momentum of Planets be heglected in orbital equations? |
| Question: |
Dear Honorable Co-Chair Dr. Cerven & Dr. Gabor:
According to your kind and wise advice, we have found Wikipedia, the free encyclopedia, Talk Specific Orbital Energy in the Websites, and communicated through the channel Talk, as shown below:
Talk:Specific orbital energy/Comments
From Wikipedia, the free encyclopedia
< Talk:Specific orbital energy
• Learn more about citing Wikipedia •
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In astrodynamics the specific orbital energy (or vis-viva energy) of an orbiting body traveling through space under standard assumptions is the sum of its potential energy () and kinetic energy () per unit mass. According to the orbital energy conservation equation (also referred to as vis-viva equation) it is the same at all points of the trajectory:
Can the spinning kinetic energy of the Earth be neglected in the calculation of its specific orbital energy which is derived to be GMc/2a in the current literature? Let's take a look of the value of GMc/2a where a, Earth semi-major axis, is 1.50E+11 meters, and GMc for the Sun is 1.33E+20 mmm/ss. Then GMc/2a is 4.44E+08 mm/ss, where the Earth spinning (or self-rotation) kinetic energy per unit mass is neglected.
For Earth its GMc/2a is 4.44E+08, its spinning kinetic energy per unit mass is 4.29E+04.
On 5/15/08, Wednesday, we continued to find planet data for other Planets to calculate and compare. Surprisingly, we found out the spinning kinetic energy per mass of Jupiter and Saturn to be in the same order of magnitude of the value of their GMc/2a.
For Jupiter its GMc/2a is 8.52E+07, its spinning kinetic energy per unit mass is 3.16E+07.
For Saturn its GMc/2a is 4.63E+07, its spinning kinetic energy per unit mass is 1.95E+07.
Therefore, it is assured that the spinning angular momentum and kinetic energy of the planets cannot not be neglected as in the existing in the derivation of their orbital equations. Praise our Lord Jesus Christ. Thank you for your wise advice.
What next steps should we take? Please further critique and advise. Thanks.
God bless you all.
Sincerely,
Elder Dr. & Mrs. Hsien-Lu & Hui-Lien Peng Huang
- Show quoted text -
On 5/1/08, William T Cerven wrote:
Dr. Huang,
Your line of research is outside of the area in which Dr. Gabor and myself are most fluent and thus are not well-equipped to give you specific guidance on your research. I would suggest you find others (via a literature search) that have published articles similar to your research and, once your work is complete, submit it to the same conferences in which those authors publish. Unfortunately, I do not have a list of any such authors to give you apriori.
Sincerely,
W. Todd Cerven
AIAA Co-Technical Chair, 2008 AIAA/AAS Astrodynamics Conference
"grandpa huang"
04/30/2008 10:20 PM
To William T Cerven , "MICHAEL GABOR"
cc grandpa.huang@gmail.com
Subject Further findings
Dear Honorable Technical Co-Chairs Dr. Cerven & Dr. Gabor:
Thank you for your advice to augment our abstract. During our
augmentation we continued to find out most recently some very
interesting results as indicated below:
EARTH MOON VEUS
At Apogee
Centripetal force 3.43E+22 1.78E+20 5.44E+22
Centrifugal force 3.38E+22 1.69E+20 5.40E+22
At Perigee
Centripetal force 3.67E+22 2.22E+20 5.65E+22
Centrifugal force 3.72E+22 2.34E+20 5.61E+22
For EARTH & MOON, at Perigee each's Centrifugal force is greater than
Centripetal force. But for Venus, its Centrifugal force is less than
Centripetal
force.
Remember that the orbital direction and the spin direction of EARTH &
MOON are all counter clockwise. But the spin direction of VENUS is
Clockwise, or retrograde,
although its orbital direction is counter clockwise as that of EARH &
MOON. Why does such difference exist? Would you please advise us
whether It is worthwhile for further study? Thanks.
Praise be to our LORD JESUS CHRIST.
God bless you, your families, and your careers,
Sincerely yours,
Elder Dr. & Mrs. Hsien-Lu & Hui-Lien Peng Huang
---------- Forwarded message ----------
From: grandpa huang
Date: Apr 18, 2008 2:46 PM
Subject: Re: Regarding your abstract submission
To: William T Cerven , Michael.Gabor@ngc.com
Cc: grandpa.huang@gmail.com
Dear Honorable Technical Co-Chairs Dr. Cerven & Dr. Gabor:
Sorry for the email which is difficult to read. Please see the attached file.
God bless you all.
Sincerely,
Elder Dr. & Mrs. Hsien-Lu & Hui-Lien Peng Huang
On 4/18/08, grandpa huang wrote:
> Dear Honorable Technical Co-Chairs Dr. Cerven & Dr. Gabor:
>
> Thank you very much for your kind encouragement. You encouraged us to
> either augment our abstract with additional appropriate content for
> the next joint AAS/AIAA meeting or, alternatively, resubmit the
> abstract to another conference more relevant to our topic.
>
> During augmenting our abstract at your advice, we found a bonus from
> the derivation of the radial acceleration. Here we present the result
> to you for your further comments, critiques, advice, and direction.
> Below is our derivation.
>
> nConservation of energy gives
> E=(1/2) m(dr/dt)(dr/dt) + (½) IrWrWr + (1/2) Is WsWs - GmMc/r =
> constantA
> (1)
> where G is the gravitational constant, m is the planet mass, Mc, mass
> of the central body e.g. Sun for Earth or Earth for Moon, r, distance
> between the orbiting Planet and Sun, Wr, Planet orbital angular
> velocity, Ws, Planet Spin angular velocity, Ir, moment of inertia of
> orbiting Planet, and Is, moment of inertia of spinning Planet.
>
> nConservation of angular momentum for the motion of the orbiting Planet gives
> L = IrWr + IsWs = constantB
>
> (2)
>
> From Eq.(1) we have
> (dr/dt) = SQRT [ (2/m)E – (1/m)IrWrWr – (1/m)IsWsWs + 2GMc / r ]
>
> (3)
>
> Radial Acceleration = d(dr/dt)/dt = [d(dr/dt)/dr](dr/dt) = {d(SQRT[
> (2/m)E – (1/m)IrWrWr – (1/m)IsWsWs + 2GMc/r]/dr}
>
> {SQRT[ (2/m)E – (1/m)IrWrWr – (1/m)IsWsWs + 2GMc/r]}
> =
> (1/2){1 / SQRT[ (2/m)E – (1/m)IrWrWr – (1/m)IsWsWs + 2GMc/r]}
> {d[
> (2/m)E – (1/m)IrWrWr – (1/m)IsWsWs + 2GMc/r]/dr}
>
> {SQRT[ (2/m)E – (1/m)IrWrWr – (1/m)IsWsWs + 2GMc/r]}
> =
> (1/2) {d[ (2/m)E – (1/m)IrWrWr – (1/m)IsWsWs + 2GMc/r]/dr}
> =
> (1/2){ 0 - d[(1/m)IrWrWr ]/dr – 0 + d[2GMc / r]/dr}
> = -
> (1/m) [(1/2)d(IrWrWr)/dr] + GMc [d(1/r)/dr]
> = -
> (1/m) [(1/2)WrWr (dIr/dr) +Ir(1/2) d(WrWr)/dr] + [GMc (-1/rr)]
> =
> -(1/m)[(1/2)WrWr dIr/dr + Ir(1/2) 2Wr dWr/dr] - GMc/rr
> (4)
> To find out dIr/dr and dWr/dr, we have,
> for the orbiting Planet, Ir = mrr
> dIr / dr= 2mr
>
> (5)
> equation (2) gives IrWr = L – IsWs
>
> (6)
> d(IrWr)/dr = d(L-IsWs) = 0
> and Ir and Wr are varying with respect to r, so
> d(IrWr)/dr = Ir
> (dWr/dr) + Wr(dIr/dr) = 0
> (7)
> Equation (7) gives dWr/dr = - (Wr / Ir) (dIr/dr)
>
> (8)
> Equations (5) and (8) give dWr/dr = - (Wr/mrr)(2mr) = - 2Wr / r
>
> (9)
> Equations (4), (5), and (9) give
> Radial Acceleration = d(dr/dt)/dt = - r WrWr + 2rWrWr – GMc/rr =
> r WrWr – GMc/rr
> (10)
> Where GMc/rr is due to the attraction between the Planet and the
> center mass Mc, i.e. the Centripetal Acceleration, and
> r WrWr is in the direction of r, pointing outwardly
> radially from the center mass Mc, i.e. the Centrifugal Acceleration.
>
> Bonus:
>
> For a circular orbital motion such as that approximated by the Moon
> around Earth, r is constant in magnitude, dr/dt =0, and d(dr/dr)/dt
> =0 in magnitude, and from Eq. (10), rWrWr is equal to GMc/rr in
> magnitude. Therefore, Equation (10) gives that
> Radial Acceleration of a Planet orbiting in a circular orbit is zero
> in magnitude and rWrWr is equal to GMc/rr which is the Centripetal
> Acceleration in magnitude, but opposite in direction to the
> Centripetal Acceleration, thus rWrWr is the Centrifugal Acceleration.
> This implies that the Centrifugal force applied to the orbiting Planet
> is not a Fictitious Force as traditionally considered in the Basic
> Physics. What will be the impact? Please advice! Thanks.
>
> Furthermore, this implication confirms what we had derived many years
> ago by applying the Bernoulli's theorem to a segment of a circular
> flow of material in a uniform speed. The total force experienced by
> the segment of the material, from the difference of the larger
> pressure due to the slower speed in the inner ring side and the
> smaller pressure due to the faster speed in the outer ring side, is
> exactly equal to the mass of the circulating segment of material
> multiplied by the equivalent radius of the center of mass of the
> material inside the circulating segment and multiplied by the square
> of the angular velocity of the circulating segment of material. Since
> the direction of the total force experienced by the mass of the
> segment is from the inner ring toward the outer ring, the total force
> is a centrifugal force. It is derived by applying the Conservation of
> Energy in terms of the Bernoulli's equation to the circulating mass.
> This total force should not be considered as a Fictitious force as we
> pointed out previously.
>
> God bless you all.
>
> Sincerely,
>
> Elder Dr. & Mrs. Hsien-Lu & Hui-Lien Peng Huang
>
>
>
> On 4/4/08, William T Cerven wrote:
> >
> > Dear Dr. Huang,
> >
> > Thank you for your submission, "Planet's Spin Energy & Angular Momentum
> > Included In Derivation of Orbit Equation," to the 2008 AIAA/AAS
> > Astrodynamics Specialists Conference in Honolulu, Hawaii. Due to the large
> > number of submissions this year, we could not accept every abstract
> > submitted. Faced with the daunting task of having fewer presentation slots
> > than we had suitable submissions, we looked at abstract quality, number of
> > abstracts submitted by the same authors, and total numbers of abstracts
> > submitted on a single topic. After multiple reviews, we determined that
> > your abstract did not contain enough content appropriate to the scope of the
> > conference relative to its peer submissions to qualify for acceptance. We
> > encourage you to consider augmenting your abstract with additional
> > appropriate content for the next joint AAS/AIAA meeting or, alternatively,
> > resubmit the abstract to another conference more relevant to your topic.
> >
> > Sincerely,
> > W. Todd Cerven and Michael J. Gabor
> > Technical Co-Chairs
> > AIAA/AAS Astrodynamics Specialists Conference
> > William.T.Cerven@aero.org
> > Michael.Gabor@ngc.com
> >
>
|
| Date: |
17 May 2008
|
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Answers (Ordered by Date)
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| Answer: |
Elder Dr. & Mrs. Hsien-Lu & Hui-Lien Peng Huang - 11/03/2012 23:37:02
| | | In our original document shown above there was a typographical error in our expression right after Equation (6), i.e.
d(IrWr)/dr = d(L-IsWs) = 0.
It should have been written as
d(IrWr)/dr = d(L-IsWs)/dr = 0.
Since we published here in 2008, we have not received any comments from our beloved viewers yet. We would appreciate very much if some honorable experts can give us some advice. Thanks in Christ Jesus.
|
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| Answer: |
Elder Dr. & Mrs. Hsien-Lu & Hui-Lien Peng Huang - 11/03/2012 23:38:10
| | | In our original document shown above there was a typographical error in our expression right after Equation (6), i.e.
d(IrWr)/dr = d(L-IsWs) = 0.
It should have been written as
d(IrWr)/dr = d(L-IsWs)/dr = 0.
Since we published here in 2008, we have not received any comments from our beloved viewers yet. We would appreciate very much if some honorable experts can give us some advice. Thanks in Christ Jesus.
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