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Thread: Calculating the distance between the Earth and the Sun by the fine-structure constant

  1. #1

    Calculating the distance between the Earth and the Sun by the fine-structure constant

    Hello, All.

    During developing my model of electron, I have accidentally derived a math formula with help of which the average distance between the Earth and the Sun can be calculated:

    Click image for larger version. 

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    where µ ≈ 3.986∙1014 m^3/s^2 is the Earth's gravitational parameter, α ≈ 1/137.04 is the fine-structure constant, C = 1 m/s is a matching coefficient.

    So I have a question for space flight specialists. If we replace "8" with "1" in the obtained formula, we can get a value close to the radius of the sphere of influence of the Earth (not the Hill sphere!):

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    where a is the semi-major axis of the Earth's orbit, mE is the Earth's mass, MS is the Sun's mass.

    I do not know exactly how to calculate trajectories of interplanetary spacecrafts, but I read that the radius of the sphere of influence may be more correct for "gluing" trajectories of spacecrafts. Maybe now these trajectories are calculated empirically? If so, can the obtained value "2.3584 10^9 m" (if it is not a coincidence) be used to accurately calculate the trajectories of spacecrafts?

    Thank you.

    Source: Calculation of the distance from the Earth to the Sun
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    Last edited by Nikolay Sukhorukov; 2018-Nov-22 at 05:27 PM.

  2. #2
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    Quote Originally Posted by Nikolay Sukhorukov View Post
    Hello, All.

    During developing my model of electron, I have accidentally derived a math formula with help of which the average distance between the Earth and the Sun can be calculated:...
    Hi Nikolay Sukhorukov.
    A problem is that it is possible to take a number such as an AU and construct many formula that return that number. That does not mean that the formula has any significance in physics. An AU has nothing to do with the fine structure constant. A model of the electron will not give the distance between the Sun and Earth.

    Look at the formula for the sphere of influence of the Earth and other bodies. You have a formula for an AU, the sphere of influence involves the semi-major axis (close to an AU) so maybe it is not a surprise that you can get the sphere of influence by altering your formula.

  3. #3
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    Quote Originally Posted by Nikolay Sukhorukov View Post
    Hello, All.

    During developing my model of electron, I have accidentally derived a math formula with help of which the average distance between the Earth and the Sun can be calculated:

    Click image for larger version. 

Name:	formula12.GIF 
Views:	70 
Size:	2.8 KB 
ID:	23758

    where µ ≈ 3.986∙1014 m^3/s^2 is the Earth's gravitational parameter, α ≈ 1/137.04 is the fine-structure constant, C = 1 m/s is a matching coefficient.

    So I have a question for space flight specialists. If we replace "8" with "1" in the obtained formula, we can get a value close to the radius of the sphere of influence of the Earth (not the Hill sphere!):

    Click image for larger version. 

Name:	formula14.GIF 
Views:	60 
Size:	4.2 KB 
ID:	23757

    where a is the semi-major axis of the Earth's orbit, mE is the Earth's mass, MS is the Sun's mass.

    I do not know exactly how to calculate trajectories of interplanetary spacecrafts, but I read that the radius of the sphere of influence may be more correct for "gluing" trajectories of spacecrafts. Maybe now these trajectories are calculated empirically? If so, can the obtained value "2.3584 10^9 m" (if it is not a coincidence) be used to accurately calculate the trajectories of spacecrafts?

    Thank you.

    Source: Calculation of the distance from the Earth to the Sun
    Where does the number 3/8 come from?

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    Quote Originally Posted by Nikolay Sukhorukov View Post
    Hello, All.

    During developing my model of electron, I have accidentally derived a math formula with help of which the average distance between the Earth and the Sun can be calculated:

    Click image for larger version. 

Name:	formula12.GIF 
Views:	70 
Size:	2.8 KB 
ID:	23758

    where µ ≈ 3.986∙1014 m^3/s^2 is the Earth's gravitational parameter, α ≈ 1/137.04 is the fine-structure constant, C = 1 m/s is a matching coefficient.
    Have you tried plugging in the numbers with other bodies (other planets, asteroids, etc. and their distance from the sun)? I don't feel like doing the math, but I would very strongly suspect it doesn't generalize. And of course, it's only approximate for the AU.

    I do not know exactly how to calculate trajectories of interplanetary spacecrafts, but I read that the radius of the sphere of influence may be more correct for "gluing" trajectories of spacecrafts.
    You're referring to patched conics. That's an approximation method useful for trajectories because it can simplify calculation, but sometimes it isn't accurate enough, and there are ways to do calculations more accurately. Essentially it focuses on the gravitationally dominant object (planet, sun, moon) depending on where the spacecraft is, but sometimes even small perturbations from other objects can have very significant effects on trajectory. As for how trajectories are calculated, look up "orbital mechanics."

    Maybe now these trajectories are calculated empirically? If so, can the obtained value "2.3584 10^9 m" (if it is not a coincidence) be used to accurately calculate the trajectories of spacecrafts?
    Could you explain how you think it could be used to calculate trajectories? Thanks.

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  5. #5
    Quote Originally Posted by Reality Check View Post
    A problem is that it is possible to take a number such as an AU and construct many formula that return that number. That does not mean that the formula has any significance in physics.
    Thank you. Of course, I thought about it. That is why I asked an additional question about the radius of the sphere of influence of the Earth. I find it difficult to navigate the sea of specialized space flight literature. If optimal trajectories of spacecrafts are calculated mainly empirically relative to some radius of the sphere of influence, and if this radius suddenly coincides with the obtained value of "2.3584 10^9 m", then this can hardly be called a coincidental coincidence.

    Quote Originally Posted by Hornblower View Post
    Where does the number 3/8 come from?
    In general, it happened accidentally. Quantum physics likes beautiful integer coefficients in formulas. Agree that the numbers "3/8" and "3" are more beautiful than "34/21.02", "17", or "45.454". You would better ask the question: where did the coefficient "C = 1 m/s" come from? The answer will be more interesting.

    Quote Originally Posted by Van Rijn View Post
    Have you tried plugging in the numbers with other bodies (other planets, asteroids, etc. and their distance from the sun)?
    Yes, of course. No coincidences. Thus, either the obtained formula is a coincidental coincidence or our planet is once again becoming a unique space object, may even be the center of the Universe.

    Quote Originally Posted by Van Rijn View Post
    As for how trajectories are calculated, look up "orbital mechanics."
    ...
    Could you explain how you think it could be used to calculate trajectories? Thanks.
    Thank you. I read a lot of such literature. The information on the calculation of the trajectories is contradictory. They may be calculated relative to the sphere of influence - "9.28 10^8 m" or another sphere of influence - "2.48 10^9 m" or the Hill sphere - "1.50 10^9 m". That is why I would like to ask the question to space flight specialists: whether they empirically use the orbit with the radius close to "2.3584 10^9 m" to calculate interplanetary trajectories, if this is not secret information that NASA hides.
    Last edited by Nikolay Sukhorukov; 2018-Nov-24 at 08:24 AM.

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    Nicolay, your presentation is not convincing as an exercise in physics. First, I will digress a bit and present your equation in my preferred notation, not relying on Greek letters for ease of typing.

    A = u(8a/3)2/C2
    where A = Earth orbital radius
    u = Earth gravitational parameter
    C = 1m/s
    a = fine structure constant

    It makes more sense to me to put a in the numerator than its reciprocal in the denominator, but maybe that's just me. Anyway, cutting to the chase. You are multiplying the gravitational parameter by the square of the product a dimensionless number that emerges from elementary particle physics and another dimensionless number that is "beautiful" in the eyes of you, the beholder. Then you are dividing the product by the square of what appears to be an arbitrary unit of velocity to put the result in units of distance. This result happens to be within about 1% of Earth's orbital radius.

    The gravitational parameter is proportional to the planet's mass. If we take this seriously as something other than a coincidence, it suggests that the orbital radius is somehow a function of the planet's mass. Trying the equation for the other planets gives results that are all over the place. This reinforces my inference that the close match for Earth is merely a coincidence, which I cannot reject from first principles. I would conclude that this is merely a number-crunching curiosity.

    By the way, what is the physical significance of your C = 1m/s rather than some other value?

    I do not know exactly how to calculate trajectories of interplanetary spacecrafts, but I read that the radius of the sphere of influence may be more correct for "gluing" trajectories of spacecrafts.
    My bold. I would say useful rather than correct. See the following Wiki articles. As always, Wiki is not the last word, but they are usually pretty good on these topics.
    https://en.wikipedia.org/wiki/Sphere...astrodynamics)
    https://en.wikipedia.org/wiki/Patche..._approximation

    It appears that the orbital mechanics experts have found from experience that choosing the sphere of influence radius as the transition point in this patched conic approximation gives the least error in this first estimate of a trajectory. By the way, your number is not even close. Try again.

  7. #7
    Quote Originally Posted by Hornblower View Post
    Nicolay, your presentation is not convincing as an exercise in physics ...
    ...
    By the way, what is the physical significance of your C = 1m/s rather than some other value?
    Yes, this is the main. I did not start right away with this because I wanted to make sure that this topic is generally interesting to someone. This forum section - ATM - allows to publish non-standard theories. One of these theories claims that, firstly, the sphere of influence of any space object is not infinite, but limited in space, and, secondly, at least some fundamental constants may be different in the spheres of influence of other planets. These are not my original ideas but the authors of these ideas (which i read) did not give the exact formulas. I tried to develop these hypotheses. As a result, about 2 years ago I derived the formula for calculating the Earth's radius by fundamental physical constants, and now I have derived the formula for calculating the distance between the Earth and the Sun. I do not say categorically that this distance is the maximum radius of the sphere of influence of the Earth, but it is very similar to this because it is obtained using the most important constant - the fine-structure constant.
    Now about the "1 m/s". I argued as follows: if fundamental physical constants are somehow related to a planet, then they must necessarily be related to the gravitational parameter µ of the planet. Next. If the radius of the sphere of influence of the planet is limited in space, then it must also be related with the gravitational parameter µ of the planet. But the radius of an orbit in which any body (a satellite, a spacecraft etc) rotates around the planet is related with its orbital speed accordingly to the formula √(µ/Ro) (if we neglect the ellipticity of the orbit Ro). That is, the orbital speed of a body related to the maximum radius of the sphere of influence of the planet must be minimal. And I thought: could this minimum speed also exist in the micro-world of the planet, that is, in atoms, nuclei and electrons? And I got something.
    If you are interested, I can continue. I would not like to divert attention to an uninteresting topic for you.

    P.S.
    If the constant α was universal for all space objects, then it would be tempting to use it to calculate maximum spheres of influence of other planets with this formula:
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    where µp is a gravitational parameter of a planet or a space object. Now I am analyzing this possibility.

    Quote Originally Posted by Hornblower View Post
    It appears that the orbital mechanics experts have found from experience that choosing the sphere of influence radius as the transition point in this patched conic approximation gives the least error in this first estimate of a trajectory.
    May be. But it seems that the experience of the experts is still insufficient:

    Wiki: "Although this method gives a good approximation of trajectories for interplanetary spacecraft missions, there are missions for which this approximation does not provide sufficiently accurate results."

    Every interplanetary mission is a very expensive process. Perhaps the experts chose that method of calculation, which is not quite effective, but does not lead to the loss of spacecrafts. But this is just my guess.
    Last edited by Nikolay Sukhorukov; 2018-Nov-25 at 12:18 PM.

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    Quote Originally Posted by Nikolay Sukhorukov View Post
    Thank you. Of course, I thought about it. That is why I asked an additional question about the radius of the sphere of influence of the Earth.... then this can hardly be called a coincidental coincidence.
    Nikolay Sukhorukov, it is a coincidental coincidence. You created a formula that accidently gets close to an AU. You altered that formula and got "a sphere of influence". Thus: so maybe it is not a surprise that you can get the sphere of influence by altering your formula.

    Another problem is that your formula are wrong.
    1 astronomical unit = 149597870700 metres (exactly) and you have 150940000000 m. That is ~1% off.
    Sphere of influence of the Earth = 0.924 * 10^6 km. You have an even worse value of "2.3584 10^9 m". That is ~250% off.

    Quantum physics does not "like beautiful integer coefficients in formulas". Integer values are a consequence of solving quantum equations, e.g. the quantum numbers. The fine-structure constant appears in quantum physics formula and is not an integer.

    Quantum physics is still not orbital mechanics (sphere of influence) or a mostly random distance of the Earth from the Sun.

    Setting C = 1m/s possibly comes from not trying very hard to get the actual value of an AU.

  9. #9
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    Quote Originally Posted by Nikolay Sukhorukov View Post
    Yes, this is the main. I did not start right away with this because I wanted to make sure that this topic is generally interesting to someone. This forum section - ATM - allows to publish non-standard theories. One of these theories claims that, firstly, the sphere of influence of any space object is not infinite, but limited in space, and, secondly, at least some fundamental constants may be different in the spheres of influence of other planets. These are not my original ideas but the authors of these ideas (which i read) did not give the exact formulas. I tried to develop these hypotheses. As a result, about 2 years ago I derived the formula for calculating the Earth's radius by fundamental physical constants, and now I have derived the formula for calculating the distance between the Earth and the Sun. I do not say categorically that this distance is the maximum radius of the sphere of influence of the Earth, but it is very similar to this because it is obtained using the most important constant - the fine-structure constant.
    Now about the "1 m/s". I argued as follows: if fundamental physical constants are somehow related to a planet, then they must necessarily be related to the gravitational parameter µ of the planet. Next. If the radius of the sphere of influence of the planet is limited in space, then it must also be related with the gravitational parameter µ of the planet. But the radius of an orbit in which any body (a satellite, a spacecraft etc) rotates around the planet is related with its orbital speed accordingly to the formula √(µ/Ro) (if we neglect the ellipticity of the orbit Ro). That is, the orbital speed of a body related to the maximum radius of the sphere of influence of the planet must be minimal. And I thought: could this minimum speed also exist in the micro-world of the planet, that is, in atoms, nuclei and electrons? And I got something.
    If you are interested, I can continue. I would not like to divert attention to an uninteresting topic for you.

    P.S.
    If the constant α was universal for all space objects, then it would be tempting to use it to calculate maximum spheres of influence of other planets with this formula:
    Click image for larger version. 

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    where µp is a gravitational parameter of a planet or a space object. Now I am analyzing this possibility.



    May be. But it seems that the experience of the experts is still insufficient:
    Says who? There have been numerous spectacularly successful interplanetary missions over the past half century, and the one I know of that failed because of a math error involved a mixup of metric and English units, not a failure to use your line of thought.

    Wiki: "Although this method gives a good approximation of trajectories for interplanetary spacecraft missions, there are missions for which this approximation does not provide sufficiently accurate results."

    Every interplanetary mission is a very expensive process. Perhaps the experts chose that method of calculation, which is not quite effective, but does not lead to the loss of spacecrafts. But this is just my guess.
    My bold. Yes, you appear to be guessing. They would be foolhardy to use the patched conic approximation as the final choice. I will make the educated guess that they used this approximation to get close for a start, and thus minimize the computer time needed to do the multibody numerical integration number-crunching that is needed to find the trajectory that will hit the target. This is just an example of using elegant simplifying techniques to reduce the grunt work and leave more time and money available for the spacecraft.

  10. #10
    Quote Originally Posted by Reality Check View Post
    Nikolay Sukhorukov, it is a coincidental coincidence. You created a formula that accidently gets close to an AU.
    ...
    Another problem is that your formula are wrong.
    1 astronomical unit = 149597870700 metres (exactly) and you have 150940000000 m. That is ~1% off.
    Thank you, but, may be, AU accidently gets close to the obtained value "1.51∙1011 m"? It is too self-confident to say that we know everything about the creation of the Solar system. Note that AU is only the average value between aphelion and perihelion. Uncertainty is inevitable.

    Quote Originally Posted by Reality Check View Post
    Sphere of influence of the Earth = 0.924 * 10^6 km. You have an even worse value of "2.3584 10^9 m". That is ~250% off.
    Please look at the figure below:

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    The translation is: "... The sphere of influence of the planet P1 relative to the Sun P0 is a sphere whose center coincides with the center of the planet and with the radius

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    [r1 is AU, m1 is the Earth's mass, m0 is the Sun's mass.]

    ... M.D. Kislik has shown that building trajectories of spacecrafts using the “gluing” method is more beneficial if, instead of spheres of action, we consider spheres of influence. In this case, errors in the parameters of the trajectory in the transition from one attracting center to another are on average minimal ..."

    (Of course, I rely on the competence of the author.)

    The obtained value "2.3584∙109 m" is close to the value obtained using the above formula.

    Quote Originally Posted by Reality Check View Post
    Quantum physics does not "like beautiful integer coefficients in formulas". Integer values are a consequence of solving quantum equations, e.g. the quantum numbers. The fine-structure constant appears in quantum physics formula and is not an integer.
    I am sorry, but I meant integer coefficients and not fundamental constants. For example, in the formula of the Rydberg constant

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    we use "8" but not "8.12".

    Quote Originally Posted by Reality Check View Post
    Setting C = 1m/s possibly comes from not trying very hard to get the actual value of an AU.
    I swear it's not. I just have not reached this point yet, as it seems that you are not very interested in it.

    Quote Originally Posted by Hornblower View Post
    ... They would be foolhardy to use the patched conic approximation as the final choice. I will make the educated guess that they used this approximation to get close for a start, and thus minimize the computer time needed to do the multibody numerical integration number-crunching that is needed to find the trajectory that will hit the target. This is just an example of using elegant simplifying techniques to reduce the grunt work and leave more time and money available for the spacecraft.
    May be, but what if for optimal calculations with using the patched conic approximation it is useful to use theory, which presupposes that the spheres of influence of planets and the Sun are limited in space, as shown on the figure:

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    where "Radius of an intermediate sphere of influence" is "2.3584∙109 m" ≈ AU/64 . I can not do such calculations myself. I just suggest that experts think about this possibility.
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  11. #11
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    Quote Originally Posted by Nicolay Sukhorukov
    May be, but what if for optimal calculations with using the patched conic approximation it is useful to use theory, which presupposes that the spheres of influence of planets and the Sun are limited in space, as shown on the figure:
    They do use theory, specifically Newton's theory, or Einstein's theory for supercritical stuff.
    where "Radius of an intermediate sphere of influence" is "2.3584∙109 m" ≈ AU/64 . I can not do such calculations myself. I just suggest that experts think about this possibility.
    My bold. The burden is on you to convince them that you have a good idea. Judging by the hodgepodge of conjectures I have seen so far I would not have any false hopes of making a favorable impression on them.

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    Quote Originally Posted by Nikolay Sukhorukov View Post
    Thank you, but, may be, AU accidently gets close to the obtained value...
    Close is not good enough. Your formula gives the wrong number.

    Sceince does not say it knows everything about anything. We know enough about the formation of all solar systems to be confident that it is a chaotic, almost random process. Thus, an AU is not a value embedded inside whatever you are doing for an electron.
    Quote Originally Posted by Reality Check View Post
    1 astronomical unit = 149597870700 metres (exactly) and you have 150940000000 m. That is ~1% off.
    Sphere of influence of the Earth = 0.924 * 10^6 km. You have an even worse value of "2.3584 10^9 m". That is ~250% off.
    An AU is defined by the IAU as an exact value as I highlight. This was set to match the measured 2009 value of 149597870700±3 m. That is an uncertainty of 3 meters. You are wrong by 1342129300 meters!

    Your Sphere of influence of the Earth value us stull wrong. My link is to the Sphere of influence of the Earth Wikipedia article in English and with numbers. The OP that you wrote is wrong. You do not get a "value close to the radius of the sphere of influence of the Earth".

    This is the radius of the sphere of influence of the Earth: 0.924 * 10^6 km. This is your value for the radius of the sphere of influence of the Earth: 2.3584 10^9 m. These numbers are not the same. Your number is wrong by a factor of ~2.5.

    You are correct. I am not interest in a formula that is not based in working physics and gives wrong numbers.
    Last edited by Reality Check; 2018-Nov-26 at 08:10 PM.

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    Quote Originally Posted by Nikolay Sukhorukov View Post
    Please look at the figure below:

    Click image for larger version. 

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    The translation is: "... The sphere of influence of the planet P1 relative to the Sun P0 is a sphere whose center coincides with the center of the planet and with the radius
    Please read: Hill sphere and the equation after "When eccentricity is negligible (the most favourable case for orbital stability), this becomes": r ~ a * cube root (m/3M) (for any body m orbiting a body M)
    Your image has: r ~ 1.13 AU cube root (m/M) for the Earth alone.

    The [URL=https://en.wikipedia.org/wiki/Sphere_of_influence_(astrodynamics)sphere of influence has an equation[/URL]: r ~ a * 2/5 root (m/M).

    The translation of the image, the image itself or the context of the image is wrong. It is the Hill sphere that is proportional to the cube root of masses.

  14. #14
    Quote Originally Posted by Hornblower View Post
    ... The burden is on you to convince them that you have a good idea. Judging by the hodgepodge of conjectures I have seen so far I would not have any false hopes of making a favorable impression on them.
    Experience is the best judge. Why not NASA do one interplanetary launch relative to the orbit "2.3584∙109 m"? If the launch will be more effective than previous launches, it can save hundreds of millions of dollars on rocket fuel.

    Quote Originally Posted by Reality Check View Post
    Please read: Hill sphere and the equation after "When eccentricity is negligible (the most favourable case for orbital stability), this becomes": r ~ a * cube root (m/3M) (for any body m orbiting a body M) ...
    I know it. The definition of Hill's sphere from my source:

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    The translation is: "Hill’s sphere radius is determined by the formula"

    Quote Originally Posted by Reality Check View Post
    The [URL=https://en.wikipedia.org/wiki/Sphere_of_influence_(astrodynamics)sphere of influence has an equation[/URL]: r ~ a * 2/5 root (m/M).
    My source defines it the same way:

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    The translation is: "The approximate value of the radius of the sphere of action of a planet is determined by the formula"

    Quote Originally Posted by Reality Check View Post
    The translation of the image, the image itself or the context of the image is wrong. It is the Hill sphere that is proportional to the cube root of masses.
    It is very strange that formula

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    is not in Wiki. Maybe it is a secret formula that is hidden by space powers?

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    Quote Originally Posted by Nikolay Sukhorukov View Post
    Maybe it is a secret formula that is hidden by space powers?
    You mean:
    Russia, Ukraine, Byelorussia, Armenia, Georgia, Azerbaijan, Lithuania, Latvia, Estonia, Moldova, Kazhakstan, Kyrgyzstan, Uzbekistan, Turkmenistan, Tajikistan, United States, France, Japan,
    China, United Kingdom, Austria, Belgium, Czech Republic, Denmark, Finland, Germany, Greece, Hungary, Ireland, Italy, Luxemborg, Netherlands, Norway, Poland, Portugal, Romania, Spain, Sweden, Switzerland, Canada, Australia, New Zealand, India, Israel, Iran, North Korea

    Scenario a: Every country on the above list managed to keep the formula secret. As have the commercial space providers, of which there are a few now.
    Scenario b: Your version of the formula is wrong or being quoted/used out of context
    Which one seems more likely?

  16. #16
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    Quote Originally Posted by Nikolay Sukhorukov View Post
    I know it. The definition of Hill's sphere from my source:
    That is correct. Hill sphere. The radius of the Hull sphere has a cube root. This new image has that cube root and other terms. The sphere of influence has an equation: r ~ a * 2/5 root (m/M). Your first image has a cube root. It is not the sphere of influence.

    The Wikipedia page is a reliable source. It states the equation. It derives the equation. It references textbooks. Your value for the sphere of influence is wrong.

    This new image, which is not your first image, does have the same sphere of influence equation and thus the same value for the Earth. You have cited your source showing that your value is wrong.

    It is a "secret" formula which is hidden by not looking at the full context of the formula. It is probably a Hill radius because it has a cube root of m1/m0 as in Hill sphere. It may be a formula derived using different approximations than "D.P. Hamilton & J.A. Burns (1992). "Orbital stability zones about asteroids. II - The destabilizing effects of eccentric orbits and of solar radiation". Icarus. 96 (1): 43–64. Bibcode:1992Icar...96...43H. doi:10.1016/0019-1035(92)90005-R. "
    Last edited by Reality Check; 2018-Nov-29 at 08:23 PM.

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    Quote Originally Posted by Nikolay Sukhorukov View Post
    Experience is the best judge. Why not NASA do one interplanetary launch relative to the orbit "2.3584∙109 m"?
    Leaving aside the cost issue, what exactly would you expect such a mission to do? What do you expect NASA (or any other space agency) to measure that couldn't be determined from the data of the many interplanetary flights that have already occurred?

    If the launch will be more effective than previous launches, it can save hundreds of millions of dollars on rocket fuel.
    How did you determine that? I have the impression that you aren't aware of the extreme accuracy that's possible when determining spacecraft trajectory, or the key issues involved. You might want to look up the Pioneer anomaly, which was a case where both spacecraft had a very small unpredicted acceleration. It turned out it was because the radiators for the radioisotope thermal generators emitted a bit more in one direction. In order to determine this miniscule effect, the researchers needed to know what the trajectory should be to extreme accuracy.

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  18. #18
    Quote Originally Posted by Shaula View Post
    ...Your version of the formula is wrong or being quoted/used out of context
    Which one seems more likely?
    Now I do not know what to say. This is not my formula. The source is: "G.N. Duboshin. Reference guide to celestial mechanics and astrodynamics. 1976." The link is (in russian): http://sci.sernam.ru/book_dsm.php?id=215
    Even if space specialists cannot understand each other, then all the more I don’t understand where they got this formula from.

    Quote Originally Posted by Reality Check View Post
    That is correct. Hill sphere. The radius of the Hull sphere has a cube root.
    This is a cube root (in red rectangle).

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    Just this formula is given in a more expanded form than in the Wiki.

    Quote Originally Posted by Reality Check View Post
    This new image has that cube root and other terms. The sphere of influence has an equation: r ~ a * 2/5 root (m/M). Your first image has a cube root. It is not the sphere of influence.

    The Wikipedia page is a reliable source. It states the equation. It derives the equation. It references textbooks. Your value for the sphere of influence is wrong.

    This new image, which is not your first image, does have the same sphere of influence equation and thus the same value for the Earth. You have cited your source showing that your value is wrong.
    It is not about terminology, which may be different in other languages. Please see my previous answer and link to Shaula.

    Quote Originally Posted by Van Rijn View Post
    How did you determine that? I have the impression that you aren't aware of the extreme accuracy that's possible when determining spacecraft trajectory, or the key issues involved.
    Of course, I do not aware of this. That is why I asked space specialists for help. After all, can anyone say whether an interplanar launch from the orbit close to "2.3584∙109 m" has ever been realized? Thank's. Maybe there is an open NASA database in which the detailed parameters of interplanetary missions are stored?
    Last edited by Nikolay Sukhorukov; 2018-Nov-30 at 10:34 AM.

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    Quote Originally Posted by Nikolay Sukhorukov View Post
    Now I do not know what to say...
    Even if space specialists cannot understand each other, then all the more I don’t understand where they got this formula from.
    I don't think space specialists have a problem understanding each other. I can't find a version of this in English so I cannot be sure but it is more likely that this equation is being used out of context.

    After all, can anyone say whether an interplanar launch from the orbit close to "2.3584∙109 m" has ever been realized?
    Pretty sure it is safe to say one hasn't. Because there is no compelling reason to do so. Why on earth would anyone shift their material to a point nearly ten times further than the Moon and then perform a large delta-v? You haven't presented any good evidence that this would be beneficial. So - direct questions. Please show the benefits of this maneuvre over a convetional lauch. Please explain what you would expect to see at this point in space.

  20. #20
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    Quote Originally Posted by Shaula View Post
    Pretty sure it is safe to say one hasn't. Because there is no compelling reason to do so. Why on earth would anyone shift their material to a point nearly ten times further than the Moon and then perform a large delta-v? You haven't presented any good evidence that this would be beneficial. So - direct questions. Please show the benefits of this maneuvre over a convetional lauch. Please explain what you would expect to see at this point in space.
    That's similar to my unanswered question below:

    Quote Originally Posted by Van Rijn View Post
    Leaving aside the cost issue, what exactly would you expect such a mission to do? What do you expect NASA (or any other space agency) to measure that couldn't be determined from the data of the many interplanetary flights that have already occurred?
    Nikolay Sukhorukov, you haven't yet made clear what you expect or why. I was guessing you might be thinking there might be some kind of physical change at that distance affecting trajectory, but if so, that should be detectable from the paths of existing interplanetary missions that go farther than that. I'd really like to see some clarification on what you are arguing. Do you have an ATM gravity hypothesis, with gravity working differently than what is normally understood? Do you think there is some unknown effect at that distance that affects spacecraft differently based on what they are doing? Something else?
    Last edited by Van Rijn; 2018-Dec-01 at 12:17 AM. Reason: spelling

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  21. #21
    Quote Originally Posted by Shaula View Post
    Pretty sure it is safe to say one hasn't. Because there is no compelling reason to do so. Why on earth would anyone shift their material to a point nearly ten times further than the Moon and then perform a large delta-v? You haven't presented any good evidence that this would be beneficial. So - direct questions. Please show the benefits of this maneuvre over a convetional lauch. Please explain what you would expect to see at this point in space.
    Well, I intuitively suppose that along the radius of "2.3584∙109 m" there is a certain force gravity line. If so, we can use it somehow, but I don’t know how to do it. Let me remind you that the topic of my post is calculating the average distance (not even AU) from the Earth to the Sun, but not optimizing the launch of spacecrafts. I am not competent to discuss such subject. Nevertheless, I suppose that when an orbit close to "2.3584∙109 m" is crossed by a spacecraft, some effects should occur, for example, interruptions in radio communications. It is interesting to know whether such communication interruptions exist when a spacecraft leaves the sphere of influence of the Earth.

    Quote Originally Posted by Van Rijn View Post
    Nikolay Sukhorukov, you haven't yet made clear what you expect or why. I was guessing you might be thinking there might be some kind of physical change at that distance affecting trajectory, but if so, that should be detectable from the paths of existing interplanetary missions that go farther than that.
    ... Do you think there is some unknown effect at that distance that affects spacecraft differently based on what they are doing?
    Exactly! Therefore, I asked about problems with radio communications when leaving the sphere of influence of the Earth. Do you can guarantee that such problems do not exist? Personally, I think that the leaders of interplanetary missions, for various reasons, not all tell the public.

    Quote Originally Posted by Van Rijn View Post
    Do you have an ATM gravity hypothesis, with gravity working differently than what is normally understood?
    An important difference in my understanding of gravity is that the sphere of influence of gravity is not infinite, but limited in space. Are you ready to admit such a hypothesis?

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    Okay, all this speculative stuff needs to end now. We cannot discuss these ideas unless we know where these equations come from.
    Nicolay Sukhorukov, you will need to present your ideas much better, starting with how you derived your equations. Just dumping them here on the board is useless unless we know how you got to your equations.
    And also NOT at an external link, like in the first post, where you also start with "here is the equation". And also not by pictures from Russian books, where most here on the board cannot read what is stated there.
    Please present a complete and understandable model here. You can use LaTeX here to write your equations.
    All comments made in red are moderator comments. Please, read the rules of the forum here and read the additional rules for ATM, and for conspiracy theories. If you think a post is inappropriate, don't comment on it in thread but report it using the /!\ button in the lower left corner of each message. But most of all, have fun!

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    Quote Originally Posted by Nikolay Sukhorukov View Post
    Well, I intuitively suppose that along the radius of "2.3584∙109 m" there is a certain force gravity line. If so, we can use it somehow, but I don’t know how to do it. Let me remind you that the topic of my post is calculating the average distance (not even AU) from the Earth to the Sun, but not optimizing the launch of spacecrafts.
    I intuitively suppose you are completely wrong. See why that isn't a good scientific argument? If you don't want to talk about this aspect of your claims then I suggest you just say "It is out of scope" and stop discussing it. As is you have dedicated another post to it, making it fair game.

    Quote Originally Posted by Nikolay Sukhorukov View Post
    Exactly! Therefore, I asked about problems with radio communications when leaving the sphere of influence of the Earth. Do you can guarantee that such problems do not exist? Personally, I think that the leaders of interplanetary missions, for various reasons, not all tell the public.
    So you ask if any problems have been seen (no) and then simultaneously add the caveat that if any were not the answer is "it's a conspiracy". This makes your claim pretty pointless to discuss as the only answer you will accept is that you are right, anything else is covered by the unprovable, untestable conspiracy theory. Turning your argument around - I think your ideas are completely wrong and any proof that exists is part of a disinformation campaign by <insert organisation here>. See why this kind of argument makes the whole discussion pointless?

    In short you have demonstrated very neatly in this one post that you are probably not willing to have your ideas subjected to any kind of scientific scrutiny and that it is pointless to discuss them. Either focus on the topic and provide testable predictions (and accept the data that is out there rather than resorting to unlikely global conspiracy theories - unless you can prove their malign influence exists) or just be quite clear that this whole topic is unscientific because you will not accept any evidence against it.

    Edit to add: Whoops, cross post with a moderator. Happy to drop this side conversation if you are going to focus back on the main topic.
    Last edited by Shaula; 2018-Dec-01 at 11:03 AM. Reason: Caveat due to cross post with mod

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    Quote Originally Posted by Nikolay Sukhorukov View Post
    Exactly! Therefore, I asked about problems with radio communications when leaving the sphere of influence of the Earth. Do you can guarantee that such problems do not exist? Personally, I think that the leaders of interplanetary missions, for various reasons, not all tell the public.
    Can you prove there isn't an invisible elf in my back yard? It isn't my burden of proof. If you want to argue for such a radio communication effect, you need to present a valid reason such an issue should exist in the first place, and it would be up to you to present supporting evidence. So far, I have seen no evidence, seen no reason why such an issue should exist, and no reason to think there is a cover up.

    An important difference in my understanding of gravity is that the sphere of influence of gravity is not infinite, but limited in space. Are you ready to admit such a hypothesis?
    The term "sphere of influence" as conventionally used is about a relationship between one object orbiting another, and essentially is something that is handy for some approximation methods. It does not imply a set limit for gravitational influence itself. Accurate trajectory calculations of spacecraft and natural objects often requires consideration of the gravitational perturbations of objects beyond their "sphere of influence" (the range where the object's gravity is dominant).

    Anyway, you're welcome to argue any hypothesis you would like, but if you're arguing limitations to gravity's range, there's abundant evidence already available against it.

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    If the spheres of influence were hard limits, the planets would not perturb each other.

  26. #26
    Quote Originally Posted by tusenfem View Post
    Okay, all this speculative stuff needs to end now. We cannot discuss these ideas unless we know where these equations come from.
    Nicolay Sukhorukov, you will need to present your ideas much better, starting with how you derived your equations.
    I tried to do it:

    Quote Originally Posted by Nikolay Sukhorukov View Post
    One of these theories claims that, firstly, the sphere of influence of any space object is not infinite, but limited in space, and, secondly, at least some fundamental constants may be different in the spheres of influence of other planets.
    ...
    I argued as follows: if fundamental physical constants are somehow related to a planet, then they must necessarily be related to the gravitational parameter µ of the planet. Next. If the radius of the sphere of influence of the planet is limited in space, then it must also be related with the gravitational parameter µ of the planet. But the radius of an orbit in which any body (a satellite, a spacecraft etc) rotates around the planet is related with its orbital speed accordingly to the formula √(µ/Ro) (if we neglect the ellipticity of the orbit Ro). That is, the orbital speed of a body related to the maximum radius of the sphere of influence of the planet must be minimal. And I thought: could this minimum speed also exist in the micro-world of the planet, that is, in atoms, nuclei and electrons?

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    but no reader had even showed interest in continuing this idea. If readers do not want to understand this unusual idea, then it may indeed be better to end this thread. I understand your desire for me to state everything absolutely precisely and clearly, but how can this be done if my idea is absolutely new and unusual. I am only at the beginning of the journey. It may take years for its development, and so I would like to public intermediate results that, I am sure, are indirect evidence of the connection between the microcosm and the macrocosm by means of fundamental physical constants, in order other researchers could try to apply them.

    (LaTeX for some reason does not work in my browser.)

    Quote Originally Posted by Shaula View Post
    If you don't want to talk about this aspect of your claims then I suggest you just say "It is out of scope" and stop discussing it.
    Ok, I propose to abandon the secondary topic of the radius "2.3584∙109 m" and return to the main topic, that is, to the distance between the Earth and the Sun.

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    So, what is the coefficient "C= 1 m/s"? I got it by developing my model of electron according to which, as a first approximation, an electron is a ball, the spherical shell of which is formed by the turns of an electron's sub-neutrino ("the pramatter") moving with the speed of light and passing the poles of the electron, like on the figure below:

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    Accordingly to my model, the total length of all the turns is exactly equal to the Compton wavelength of electron multiplied by 2, the sub-neutrino's mass equal to the electron rest mass, and the number of the turns is equal to or a multiple of the integer "299792458" i.e. to the numerical value of the speed of light. This is a key feature! That is I suppose that the radius of an electron, in a first approximation, can be got using the equations:

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    where me is the electron rest mass, c is the speed of light, h is the Planck constant, λe is the Compton wavelength of electron, Re is the electron's radius, n is some integer, C = 1 m/s is a matching coefficient, α ≈ 1/137.04 is the fine-structure constant, Rc is the Rydberg constant (in Hz). It is here that the coefficient C is introduced. It is needed just to simplify the calculation formula - in order to get rid of the number "299792458".
    Now about the number n. It can only be an integer since the number of turns of the electron neutrino can only be an integer. But how to determine this number? In general, this is a difficult question. I tried to take into account various factors and came to the conclusion that the optimal choice would be n = 16. It is with such a number that the radius of the electron "1.6101∙10-22 m" is obtained close to the experimental value obtained by Hans Dehmelt ("Experiments with an isolated subatomic particle at rest"). Thus, if it turns out that the radius of the electron is really equal to "1.6101∙10-22 m", this will mean that the decision to set the speed of light with an integer was absolutely correct.

    Oh, I think I entered a lot of discussion information today. If you are still interested and you are all not tired of my English, I am ready to continue later.

    Quote Originally Posted by Van Rijn View Post
    Anyway, you're welcome to argue any hypothesis you would like, but if you're arguing limitations to gravity's range, there's abundant evidence already available against it.
    I do not know such evidence.

    Quote Originally Posted by Hornblower View Post
    If the spheres of influence were hard limits, the planets would not perturb each other.
    It all depends on the radii of the spheres of influence of the planets or luminaries. If we assume that the formula

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    is correct, then it follows that, for example, the Sun gravitationally influences Mercury, but Mercury does not gravitationally influence the Sun because the Mercury's sphere of influence (accordingly to the given formula) is approximately 7 times smaller than the average distance between Mercury and the Sun.

  27. #27
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    Quote Originally Posted by Nikolay Sukhorukov View Post
    but no reader had even showed interest in continuing this idea.
    That is correct, Nikolay Sukhorukov. No one here and probably no one else wants to continue your idea because it is obviously wrong. That is been explained to you.
    A theory about the electron is not a theory of solar system formation that might give the distance of the Earth from the Sun.
    Your formula gets the wrong distance of the Earth from the Sun.
    Altering your formula gets the wrong value for the sphere of influence of the Earth.
    Your sources are images of a textbook in Russian which you may not understand correctly, e.g. the first one looks like a Hill sphere equation.

    Errors in this post:
    The idea of an electron having an extent ("an electron is a ball") was abandoned in the 1920's when we found that the measured spin of an electron gives a surface of an extended electron moving many times faster than the speed of light. That was before we had measurements of the upper limit to the extent of an electron which makes that speed even larger.
    There is no such thing as a "sub-neutrino". Actual neutrinos do not interact with electrons and cannot be inside a hypothetical electron ball. Actual neutrinos have a tiny mass much less tan an electron mass.

    You seem to think that the sphere of influence is a limit to gravitational influence which is wrong. This is sphere of influence of a body
    A sphere of influence (SOI) in astrodynamics and astronomy is the oblate-spheroid-shaped region around a celestial body where the primary gravitational influence on an orbiting object is that body. This is usually used to describe the areas in the Solar System where planets dominate the orbits of surrounding objects such as moons, despite the presence of the much more massive but distant Sun. In the patched conic approximation, used in estimating the trajectories of bodies moving between the neighbourhoods of different masses using a two body approximation, ellipses and hyperbolae, the SOI is taken as the boundary where the trajectory switches which mass field it is influenced by.
    I highlighted the important words primary and dominate.

    Using your example: the Sun gravitationally influences Mercury and Mercury gravitationally influences the Sun but Mercury's gravitation dominates inside Mercury's sphere of influence and the Sun's gravitation dominates outside Mercury's sphere of influence. If we want to plan a trip from Mercury to Earth, we can use an approximation. We neglect the Sun inside Mercury's sphere of influence and ignore Mercury neglect Mercury's sphere of influence.
    Last edited by Reality Check; 2018-Dec-05 at 10:17 PM.

  28. #28
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    Quote Originally Posted by Nikolay Sukhorukov View Post
    Ok, I propose to abandon the secondary topic of the radius "2.3584∙109 m" and return to the main topic, that is, to the distance between the Earth and the Sun.
    The equation you presented only gave an approximate result for the distance of the Earth from the sun, and didn't work at all for other planets. On the other hand, Newtonian physics will work well for almost anything in the solar system (General Relativity is important close to the sun, or where extreme accuracy is desired). General applicability make Newtonian physics and GR useful. How is your equation useful?

    Quote Originally Posted by Nikolay Sukhorukov View Post
    Quote Originally Posted by Van Rijn View Post
    Anyway, you're welcome to argue any hypothesis you would like, but if you're arguing limitations to gravity's range, there's abundant evidence already available against it.
    I do not know such evidence.
    It's everywhere, from the perturbations found in orbits and spacecraft trajectories, to gravitational interaction of galaxies, to detection of gravitational waves from six billion light years away:

    https://phys.org/news/2018-12-scient...collision.html

    If you're going to make arguments involving gravity, you really need to do some research on it.

    It all depends on the radii of the spheres of influence of the planets or luminaries. If we assume that the formula

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    is correct,
    Why would we assume that?

    then it follows that, for example, the Sun gravitationally influences Mercury, but Mercury does not gravitationally influence the Sun because the Mercury's sphere of influence (accordingly to the given formula) is approximately 7 times smaller than the average distance between Mercury and the Sun.
    Mercury (and other planets) do gravitationally influence the sun. In fact, one of the methods for detecting planets around other stars is by looking for the wobble of the star caused by the planets on their sun. So again, based on the evidence, this argument is wrong.

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  29. #29
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    Hans Dehmelt did not meaure the electron radius. He measured the g-factor very accurately in 1991. He then used (as he said himself) a simple model to estimate limits of substructure (derived in 1980) to estimate an upper limit of the electron's size. Which came out at ~10-22m. So it wasn't measured and was an order of magnitude estimate of a limit. Making your claim that your result is close to a measured value false. You could just have easily used the 10-18m result from first order uncertainty theory or the 10-15m classical radius which has the benefit of actually being a radius, not an upper limit.

    References:
    Drehmelt lecture
    Brodsky and Drell paper

    I won't comment on the original topic because, honestly, I don't see anything to really comment on. You've basically noticed that one number is about 10000 times smaller than another. Meaningless. Unless you can show otherwise, which you have not done. As an example - did you know that:



    Where alpha is the fine structure constant, is the mass of my car and is the height of Carrie-Ann Moss? Q is a constant to make the units work, rather like your C. This is true to about the same level of accuracy as your orbital calculations. So apparently my car and one of the stars of the Matrix films have a deep, fundamental relationship to the universe and QED particularly. I'm scared now that if I buy a different car the Sun may stop shining as alpha changes.

    So what is it you want to discuss? A numerological coincidence (which is actually not very coincidental, you've contrived it)? An unpresented model of the electron? An unpresented model for how the solar system is related to the fine structure constant?

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    Quote Originally Posted by Nikolay Sukhorukov View Post
    Hello, All.

    During developing my model of electron, I have accidentally derived a math formula with help of which the average distance between the Earth and the Sun can be calculated:

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    where µ ≈ 3.986∙1014 m^3/s^2 is the Earth's gravitational parameter, α ≈ 1/137.04 is the fine-structure constant, C = 1 m/s is a matching coefficient.

    So I have a question for space flight specialists. If we replace "8" with "1" in the obtained formula, we can get a value close to the radius of the sphere of influence of the Earth (not the Hill sphere!):

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    where a is the semi-major axis of the Earth's orbit, mE is the Earth's mass, MS is the Sun's mass.


    If one clicks on your link "Source: Calculation of the distance from the Earth to the Sun" there is NO derivation. There is no "calculation".

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