Discussion
Started 18th Nov, 2018
• Bee Research Pty Ltd

# Newtonian Orbital Velocity Defies Common Sense

The derivation of orbital velocity is presumably well understood. One method is to set the centripetal force equal to the gravitational force and solve for v.
Mv^2/r = GMm/r^2
for which orbital velocity becomes v = sqrt(GM/r)
Now let's assume we have a spacecraft in stable orbit around a body at some distance r(1) and want to move the craft to a higher orbit r(2), to do this it must fire it's engines, i.e. accelerate the craft (a) for some time (t), and presumably increase its velocity as ∆v = at, however Newtonian theory tells us that the velocity has indeed decreased as r(2) is larger than r(1).
So I would like to know what kind of Hokus Pokus is normally applied to explain this problem.

## Most recent answer

8th Aug, 2021
Abdul Malek
Technologie DMI, Montréal , Canada
@Steven Sesselmann
Newton theory of extraterrestrial gravity is plain wrong! It only works on or near the surface of the earth. For your problem, please try the following gravitational potential:
E(p) = m(a/r^2 - GM/r - Cr^2,
Where G is Gravitational constant (determined on earth), C is yet to be determined cosmic constant (or a function), M and m are the masses of the sun and any planet and r is the distance. Please refer to the following link:
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## All replies (40)

19th Nov, 2018
Richard Epenoy
Centre National d’Etudes Spatiales
Hi, suppose that your spacecraft is on a first circular orbit with a radius r1. If you want to put it on a higher circular orbit of radius r2, you need at least two maneuvers because one maneuver will put it on an intermediate elliptic orbit. This classical transfer is called Hohmann tranfer (see attached Lecture).
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19th Nov, 2018
Steven Sesselmann
Bee Research Pty Ltd
Thank you for the reference. If I am reading it correctly on page 3, the space craft in LEO was travelling 7713 m/s and after firing its engines twice in the direction of travel it obtained a final velocity in GSO of 3072 m/s,
Speeding up to slow down seems to contradict common sense, so my question is, how do we explain this without magic?
1 Recommendation
19th Nov, 2018
Richard Epenoy
Centre National d’Etudes Spatiales
Hi, the first maneuver increases the velocity to put the spacecraft on an elliptical orbit with higher energy and the second one will decrease this velocity leading at the end to a decrease of the velocity from radius r1 to r2
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19th Nov, 2018
Steven Sesselmann
Bee Research Pty Ltd
I see, so do the thrusters fire in the opposite direction for the second stage?
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19th Nov, 2018
Richard Epenoy
Centre National d’Etudes Spatiales
Yes, to decrease the velocity the maneuver is applied in the opposite direction of the velocity. You can check this by implementing the formula of the Hohmann transfer and testing it on given data.
2 Recommendations
20th Nov, 2018
Steven Sesselmann
Bee Research Pty Ltd
Richard,
I quote from page 3 of Lecture 17, "the energy of the transfer orbit is greater than the inner orbit, and smaller than the energy of the outer orbit" , which of course makes perfect sense, and also means that both first and second thruster burns are in the same direction.
What I am focusing on here is the counter intuitive teachings, telling us that a craft accelerating to a higher energy level ends up actually moving slower. The only way to explain this is to offset some of the velocity for potential energy.
If I may suggest a better way to deal with this problem, it would be to redefine the direction of travel, let me explain;
Relative to an observer on the ground a rocket taking off is moving away from the observer and therefore ought to have negative velocity. Putting a negative sign in front of the velocity means that a spacecraft in orbit will have retrograde motion, and when it fires its thrusters the retrograde motion will increase, which now makes perfect sense.
Moving off topic here...
Early astronomers observed moving planets against the fixed background stars, and consequently came up with the 1/r^2 rule, but this is just an illusion, if there were no background stars and the sky was completely black, the planets would indeed appear to be moving backwards. Redefining orbital direction to be retrograde, makes it obvious that all planets are indeed moving backwards at faster and faster retrograde motion with increasing height.
In an expanding Universe we can think of orbiting bodies as nuts on a threaded rod where direction of rotation is not arbitrary, rotating a nut moves it one way or the other, but never does it stand still.
By making this small change we see that orbital velocity with respect to earth becomes closer and closer to escape velocity with increasing radius, the difference of course is only the factor √2.
Whats the point, why fix it if it's not broken?
Yes at first glance our current methods work just fine we can put spacecraft into orbit and dock with the space station and all is fine..., but lurking in the distance is a huge problem called "Dark Matter"
The dark matter problem came about because of a problem with the Keplerian rotation curves of galaxies, and dark matter was invented to solve the problem.
The problem is there is no problem, the galaxies are just fine and there is no need for dark matter, all we need to do is reverse the direction.
Rotation rates of galaxies are measured by doppler shift of light spectra on opposite sides of the galactic disc, and are therefore relative to earth, not relative to the fixed background stars.
By changing the sign a flat rotation curve is obtained and the need for dark matter evaporates.
1 Recommendation
22nd Nov, 2018
Guido Colasurdo
Sapienza University of Rome
Dear Steven,
your first two sentences are correct, and my friend Richard, who perfectly knows the problem, was too fast in his answer.
The following part of your last comment requires time and I will try to understand your point of view.
G.
2 Recommendations
23rd Nov, 2018
Nainan Varghese
It is so due to its apparent nature. In reality, it is physically impossible for a free macro body to revolve around another moving body in any type of geometrically closed path. Central bodies are moving, planetary bodies are free macro bodies and revolving around requires geometrically closed paths. Results of above equation is apparent with respect to static central bodies. These results may be used only to predict cyclical (yearly) phenomena due to relative positions of central and planetary bodies. See:
23rd Nov, 2018
Steven Sesselmann
Bee Research Pty Ltd
Thanks for replies. I have been studying this problem for a while now and it seems like astronomers have not yet discovered handedness, something which engineers have understood for a long time. One can not simply assume that one body is orbiting another is doing so in the positive direction, as there are two possibilities, left handed helix and right handed helix.
Take the case of the moon orbiting the Earth, we have always assumed it's velocity vector to be positive, but in fact we know that the moon is moving away from us by about an inch or two every year, which means it is tracing out a left handed helix to the observer, this in turn means it's velocity vector should be written with a negative sign.
As such, we understand space is expanding everywhere and so when we peer out into the Universe and observe stars rotating in galactic discs we should likewise assign the negative vector to this.
I did some quick modelling and from what I can see, there really is no anomaly in the observed galaxy rotation curve when you plot it in the negative quadrant, which might turn out to be rather disappointing to the dark matter theorists.
Astronomers need to understand how observing an orbit from the bottom up is not the same as observing the same orbit from the top down, because the vector obviously changes sign.
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24th Nov, 2018
Guido Colasurdo
Sapienza University of Rome
Dear Steven,
I owe you an answer: your point of view (considering the relative motion) is satisfactory for you in the short time, but complicates the problem and may be misleading.
For instance, by studying Hohmann transfer in a rotating frame, two additional inexistent (that is, only apparent) forces must be added: centrifugal and Coriolis' force; be aware that they are not the same if you change the angular velocity of the reference frame.
As an example, let me discuss your third sentence. the first DV increases centrifugal force and the spacecraft moves outwards. When the maximum distance is achieved, the centrifugal force is too low in order to obtain a circular motion and one has to increase again the velocity. (Note that I am considering a reference frame rotating, at variable speed, with the spacecraft: r-axis from earth to S/C). Is it a simple approach? Not at all.
Concerning the planet motion; the maior complication is that their apparent motion is due to both Earth rotation and revolution. If one observer lives on a "moon-like" Earth, which points always the same face/continent towards the sun (alternatively, the observer is on an aircraft that flies keeping itself always at local midnight time), he would see only exterior planets moving very slowly (neither backward nor foreward before one defines a privileged direction). Neverthless there is no reason to adfirm that his point of view is the best to study the Universe.
Believe me, Astrodynamics prefers non rotating frames (and spacecraft energy split in kinetic and potential energy).
Guido
24th Nov, 2018
Steven Sesselmann
Bee Research Pty Ltd
Guido, I agree it can get confusing with too many moving bodies, but we can simplify it by looking at the observers potential vs. the orbiting bodies potential. The one constant we are relatively confident about is expanding space, so when the observer is at lower potential than an orbiting planet the vector must be negative, which is almost always the case unless we are using the fixed background stars as a reference point as Kepler and Newton presumably did.
1 Recommendation
27th Nov, 2018
Bishal Banjara
Ronin Institute
Hello Steven,
I request you to see my preprint for your answers so far I believe.
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27th Nov, 2018
Steven Sesselmann
Bee Research Pty Ltd
Bishal,
Thanks for suggesting your preprint. I read it briefly but it's not clear to me how it solves my problem of negative vs positive orbital velocity, maybe you can explain.
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27th Nov, 2018
Bishal Banjara
Ronin Institute
Hello Steven,
" negative vs positive orbital velocity" I am not getting your point how to analyze the negative velocity and for what reference it is indeed needed?
1 Recommendation
27th Nov, 2018
Guido Colasurdo
Sapienza University of Rome
Kepler and Newton were great because dei discovered 99,99% of the phenomena "using the fixed background stars as a reference point". And this is enough in order to travel in space.
1 Recommendation
28th Nov, 2018
Steven Sesselmann
Bee Research Pty Ltd
Guido,
Your statement is indeed correct, but you are missing my point.
We have two options when it comes to measuring velocity, one is with respect to the fixed background stars and the other is with respect to the observer.
Either method is an acceptable way to compare velocities, but using both at the same time does not work. This is exactly what astronomers have been doing and what I believe has given rise to the anomalous rotation curves of galaxies.
Astronomers measure galactic disc velocity using spectroscopic doppler shift, this method measures velocity with respect to earth, then they use Kepler/Newton to predict and compare the outcome...... FAIL - FAIL - FAIL
When measuring the galactic disc velocity from earth, we are measuring negative velocity and there is supposed to be a negative sign in front.
We must then use an inverse version of Kepler which is the normal orbital term less escape velocity.
(orbital velociity) = √GM/r - √2GM/r
This inverts the rotation curve and puts it in the right quadrant, and produces a flat rotation curve in line with expectations.
So Dark Matter scientists can turn off the lights and go home :)
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28th Nov, 2018
Steven Sesselmann
Bee Research Pty Ltd
Here is a clear example showing a comparison between the two methods and it's pretty obvious that my rotation curve is a far better fit than Sir Isaac Newton.
There is NO anomaly in the Galaxy rotation curves, it's just a plain old stupid mistake and no one has picked it up yet..
Can we get some astronomers to try my method please?
1 Recommendation
29th Nov, 2018
James Garry
Red Core Consulting ltd.
The first equation shows the relationship between v and r for a circular orbit.
In your example, a ship fires its engine briefly, and indeed speeds up.
It now coasts, with its radius vector to the 'parent' growing longer.
And as it coasts, it slows (relative to the fixed stars, and the original body around which it travelled).
After half an orbit, it will be at apoapsis, and may be moving quite slowly.
It now lights its engines again - and speeds up - so as to circularize its orbit.
In that circular orbit it will be moving more slowly than in its original circular orbit.
Work has been done by the engine to lift the craft out of the potential well.
There is no hocus-pocus.
<mumble: vis-viva equation>
2 Recommendations
30th Nov, 2018
Steven Sesselmann
Bee Research Pty Ltd
@James Garry
Thanks for reply, I guess the point I am trying to make here is that there is a difference in the way we measure velocities.
If you consider a rocket taking off from the Cape Canaveral as heading for the stars, then yes, it has positive velocity with respect to the stars, however if you are standing on the launch pad, then the rocket is accelerating away from you with negative velocity, just basic relative motion, nothing new or exciting here.
The root of the problem, I believe, is that astronomers are using spectral doppler shift to measure rotation of the galactic disk, and then making a false assumption that this velocity is positive.
If you follow my logic, the doppler shifted spectra method, is equivalent to standing on the launch pad watching a rocket take off, it has its reference point on earth, ergo a negative sign needs to be added.
Unfortunately there isn't any simple way for astronomers to measure galaxy rotation with reference to background stars, it would take too long to see the stars move, but using my method above we can just flip the sign and use the inverse Kepler to plot the rotation curve and compare.
As you can see my simple plot above produces a far better prediction than the normal Kepler/Newton method.
Personally I would rather believe in a logical explanation, than some mystical DM which we can neither see nor detect?
1 Recommendation
30th Nov, 2018
James Garry
Red Core Consulting ltd.
" The root of the problem, I believe, is that astronomers are using spectral doppler shift to measure rotation of the galactic disk, and then making a false assumption that this velocity is positive. "
I do not think that there is such an assumption - the velocity field can be measured across the plane of the viewed galaxy (if it is somewhat inclined to the line of sight).
Viz:
figure reproduced in:
Note figure 1 in that second link.
Half the plots (roughly) show negative-going plateaux, half positive.
As one might expect from viewing randomly arranged rotating bodies.
2 Recommendations
30th Nov, 2018
Steven Sesselmann
Bee Research Pty Ltd
James, thanks for the references looks like some good information I can use. Planning to write a paper on this subject over the Christmas break.
BTW, I do understand how they observe and measure the doppler measurement, and how half the disc approaches us and the other half moves away from us, but the final averaged velocity is incorrectly assumed to be positive, which it can't possibly be.
Recall, how space is expanding, so any doppler measurement out into space will be red shifted, which of course means negative velocity.
Any kind of orbital rotation therefore traces out a helix of growing length with respect to the observer on the ground.
I'm not suggesting to change any laws of physics, just applying in a way that I believe is correct, try seeing it my way.
1 Recommendation
30th Nov, 2018
James Garry
Red Core Consulting ltd.
Steven Sesselmann " but the final averaged velocity is incorrectly assumed to be positive, which it can't possibly be "
I am puzzled.
Sources that recede from us (let's drop the notion of positive and negative velocities for now) are seen to be reddened.
The ensemble light from a galaxy is seen to be red-shifted. So it recedes at some speed from us. Superimposed on that speed is a modulation to and fro - depending on which bit of the galaxy one looks at.
Do we agree on that?
<I think that it's not unreasonable that if everything is receding from us, to call all such motions positive velocities. One wastes a lot of ink by always writing -k for the rate of change of position, rather than 'k'>
30th Nov, 2018
Steven Sesselmann
Bee Research Pty Ltd
@James Garry
JG: The ensemble light from a galaxy is seen to be red-shifted. So it recedes at some speed from us. Superimposed on that speed is a modulation to and fro - depending on which bit of the galaxy one looks at. Do we agree on that?
Yes that sounds reasonable.
Although it may be unwise dropping negative signs to save ink, especially if it produces so much dark matter that we are literally drowning in it ;)
Do you agree that Kepler's equations are for motions measured against the fixed background stars?
Do you also agree that Galaxy disc velocities calculated from doppler shifts are referenced by earth (or the sun, as astronomers presumably compensate for earth rotation) and not by the fixed background stars?
And no doubt you also agree on the velocity of a ball, when I throw a ball to you, I see it receding and you see it approaching.
So if I want to use the doppler shift method to measure how fast the ball is approaching you, I have to flip the sign to get your point of view, do you agree
1 Recommendation
30th Nov, 2018
James Garry
Red Core Consulting ltd.
" Do you agree that Kepler's equations are for motions measured against the fixed background stars? "
It depends on the frame of reference. The 'V' in your first equation is pertinent to the rest frame of the Earth.
" Do you also agree that Galaxy disc ..."
I doubt that the rotation of the Earth (a mere 0.5 km/s at most) is accounted for - whereas the orbital motion of the Earth is noted (IIRC) and depending on the viewed object, may or may not be significant.
" And no doubt you also agree ..."
Yes.
" So if I want to use the doppler shift method ..."
Yes. If you describe 'the speed at which the ball approaches me' as having the same value as the 'speed at which the ball recedes from you'.
V_approach = -V_recede (being frame dependent)
speed is identical in both cases (being a scalar)
Naturally.
?:)
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30th Nov, 2018
Steven Sesselmann
Bee Research Pty Ltd
@James Garry
Great, all agree...
The universe is not static, The Hubble redshift tells us that space is stretching. The general understanding is that everything started at one point (BB theory), but my personal theory is that earth or more specifically, ground potential is falling (see my Ground potential theory), but why isn't relevant to this argument, just that it is.
Basically anything staying up there in orbit ought to have negative velocity with respect to us unless it's a meteorite on a trajectory to ground.
The nice part about negative velocity is that it doesn't fall off at 1/r^2, instead it increases with altitude and flattens out which seem more sensible.
The inverse Kepler equation for orbital velocity as measured from the ground then becomes;
V(tangential) = √(GM/r) - √(2GM/r)
CONCLUSION
Measuring orbital tangential velocity with respect to the fixed stars produces a positive vector and is inconsistent with measuring tangential velocity with respect to earth which produces a negative vector.
My understanding is that astronomers use spectral doppler shift to measure tangential velocity in galaxies, then arbitrarily convert the negative velocity to speed by dropping the negative sign, which means they are plotting the rotation curve in the wrong quadrant.
Astronomers then use Kepler's law of orbital motion against the fixed background stars to model the orbital velocity and find that it does not match observations.
To fix the problem theorists invent Dark Matter which still doesn't quite fix the problem, so they have to further invent the Dark Halo and as we all know it has produced one big messy problem.
Let's bring some sanity back into physics : )
1 Recommendation
3rd Dec, 2018
James Garry
Red Core Consulting ltd.
" Basically anything staying up there in orbit ought to have negative velocity with respect to us unless it's a meteorite on a trajectory to ground. "
If I see the ISS rising over the horizon, it has a substantial speed towards me. And then, if it passes over my zenith, it recedes from me.
I do not understand why you think that objects in orbit are always receding from an observer.
" Measuring orbital tangential velocity ..."
I honestly don't know what you mean with this paragraph.
The 'positive' or 'negative' nature of the velocity vector of an object is entirely dependent on how you orient your frame of reference (which is utterly arbitrary).
Can you rephrase your notion in terms of 'approaching' and receding'?
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3rd Dec, 2018
Steven Sesselmann
Bee Research Pty Ltd
James
[JG: Can you rephrase your notion in terms of 'approaching' and receding'?]
Easy, a free falling body with positive velocity will eventually hit you in the head, a free falling body with no velocity will go nowhere, and a free falling body with negative velocity will spiral out to infinity.
Stars in galaxies and planets around stars generally have negative velocity when observed from below, i.e. they are slowly receeding (like our moon).
I believe this recessional velocity is caused by a cooling (shrinking) earth, but those in the other camp have more imagination than me and believe dark energy is blowing the universe apart.
My complete lack of vivid imagination tells me, an earth with it's hot molten core, suspended in cold space (2-3 K˚) ought to radiate energy out into space, which in turn means that the energy per unit charge E/Q or J/C ought to be falling. It's a long story, but essentially it is this which I believe is responsible for the Hubble red shift.
1 Recommendation
4th Dec, 2018
James Garry
Red Core Consulting ltd.
How unusual!
Were we not discussing circular motion?
In the first case, the body is approaching 'a'. A short time later, it is receding. In the first case the distance 'r' is shrinking, a little while later, increasing.
No?
We agree that distant galaxies, as a whole are receding from us. At a given speed. The distance to them grows larger with time.
If you choose to orient your reference frame such that +x points to that galaxy, then its distance grows with time: v = r_dot. And is positive.
If you choose to orient your frame such that -x points to that galaxy, then its distance still increases with time (distance being the modulus of the radius vector).
If the Earth is contracting, it does so at a very very slow rate. And yet we infer speeds of many hundreds of km/s for some galaxies (distant) and far smaller speeds for closer galaxies. How can a contracting Earth (mm/millenium?) lead to two simultaneously different relative speeds?
Furthermore, we can observe redshift in galaxies from orbiting telescopes. Even if, by some unknown physics, the shrinkage of the Earth 'causes' redshift, these orbiting telescopes would be immune.
Is it not simpler to say that the redshift is caused by recession? This *can* be demonstrated to be true - both in the lab, and with planetary craft*.
* Cassini (famously) had its transceiver blocks built without being aware of the Doppler shift - and a mad scramble was had 6 months before Saturn when this was discovered.
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4th Dec, 2018
Steven Sesselmann
Bee Research Pty Ltd
@ James Garry
[JG: Were we not discussing circular motion?]
The simple case would be to put the observer at the centre of a circular orbit, in this case a slight tilt of the velocity vector changes the velocity from positive to negative.
One might at the same time question weather Euclidean geometry applies to free fall. Would it not also be valid to call the trajectory of a cannon ball, fired with sufficient velocity to reach orbit, for a straight line?
I believe the proper term is Geodesic.
We could say that the moons trajectory around earth is a geodesic and that it's velocity along this trajectory is negative (away from the observer).
Now on the subject of redshift.
How earth shrinking a few mm can result in a redshift.
To understand that we start by understanding how potential is energy per charge E/q, for which the SI unit is Volts (J/C), the Earth has a fixed number of charged particles, but due to radioactive decay of Th, U, K etc. heat is produced and lost to space. Earth's temperature is 1000's of degrees inside and space is freezing cold, so the flow has to be outwards.
So if energy is falling while charge stays the same, ground potential must be falling. As it happened I was able to calculate the exact value for ground potential today, it is about 930 Million Volts, and has fallen around 8 Million volts since the beginning of time.
Assuming the Universe is around 13 something billion years old, we can estimate the rate of change to be 0.0005 Volts per year. This doesn't seem much, but if we go back 120 million years to the age of Dinosaurs, ground potential would have been 60 kV higher, earth would have had a larger radius as well.
It turns out that the proper definition of time itself is ∆V or change in potential, which makes a lot of sense when you think about it.
Redshift is therefore caused by our potential falling.
A 3.0 eV photon emitted from a star 1000 light years away travels along a geodesic and reaches us after our potential has fallen 0.5 volts, and we see it as 2.5 Volts, giving the appearance of expanding space.
Hope that makes a bit of sense.
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4th Dec, 2018
Steven Sesselmann
Bee Research Pty Ltd
[side tracking slightly here]
As it happens the surface temperature of a body is a function of the outwards flow of energy which I describe as ∂V/∂t. The Sun is obviously radiating energy into space at a much higher rate than Earth and it's potential is falling a lot faster.
Global Warming, or change in surface temperature on Earth, is a direct function of ∂V/∂t , I doubt it has much to do with Co2. The correlation is probably due to increased overall energy use.
1 Recommendation
4th Dec, 2018
James Garry
Red Core Consulting ltd.
There is much in your reply that starts ' The simple case would be ...' which is wrong and I recommend Feynman's lectures as a good starting point.
a) The electrical potential of an object in a system is not the same as the gravitational potential in a system.
b) ' Redshift is therefore caused by our potential falling. '
No. That is empirically not true. If I charge a lamp, its spectrum does not change one iota.
I can generate quite large Doppler shifts in a laboratory from a *neutral* gas by simply warming it.
I think that I'll step away from this conversation. Thanks for the replies.
<Feynman, or any similar introductory physics text: Kip for Electrostatics served me well, with Flowers and Mendoza for basic properties of matter>
2 Recommendations
4th Dec, 2018
Steven Sesselmann
Bee Research Pty Ltd
@JG
In research we can only make progress if we explore and discuss new ideas.
When I hear things that don't make sense, like invisible dark matter that doesn't interact with any instruments etc. I find it challenging. Maybe you have heard the term Dark Matter so many times that you feel it is now okay and accepted, but for me it's "wrong"
Below is a link to my paper, where I describe the logical path that led me to the audacious claims I make.
PS: If you charge your lamp with enough energy it will fly off the surface of the earth like a rocket.
1 Recommendation
9th Mar, 2019
Ali Kamran
Institute of Space Technology
for any orbit the velocity to maintain it is a unique requirement and hence no orbit without desired velocity , hence the equation compared must fulfill the necessary condition first.
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21st Jun, 2019
Nainan Varghese
An alternative view: Currently, all parameters of planetary orbital paths are based on its apparent elliptical (circular) orbits around the sun (with respect to assumed static state of sun in space). But, by simple dynamics, it is physically impossible for a free macro body to revolve around another moving body in any type of geometrically closed path. Sun and planets in solar system are moving bodies. Elliptical orbits are geometrically closed paths. Therefore, planets (satellites) do not orbit in elliptical paths around the sun (planets). All macro bodies in a planetary system move together (each in its own wavy path) around galactic centre. Hence, real orbital parameters of planets are quite different from currently assumed values. See: http://vixra.org/abs/1008.0010, ‘MATTER (Re-examined)’ www.matterdoc.info .
1 Recommendation
28th Dec, 2019
Javad Fardaei
One Gravity of Sir. Newton can't be correct, if earth had gravity, earth would not and could not have rain. The perception of Einstein of gravity also is incorrect.
Gravity is inside of an atom (Mass) not outside of the mass
1 Recommendation
17th Jun, 2021
Jianan Wang
Shenzhen University
The Modification of Newton's Gravitational Law and its Application in the Study of Dark Matter and Black Hole: https://www.researchsquare.com/article/rs-373969/v1
The Physical Cause of Planetary Perihelion: Precession:https://www.researchsquare.com/article/rs-536456/v1
1 Recommendation
17th Jun, 2021
Javad Fardaei
All the formulas are 1D, or applies to two object and one distance, while universe is 3D with multiple distance. I.e. If sun is following GTR, it can not control over 100s of plants and moons in different D and different distance, not to mention Newtonian principle is not even working on our earth.
regards
17th Jun, 2021
Steven Sesselmann
Bee Research Pty Ltd
You disagreed with my hypothesis previously, but now there is a potential for my theory to be proven within a few years.
Patience is a virtue.
Steven
1 Recommendation
8th Aug, 2021
Abdul Malek
Technologie DMI, Montréal , Canada
@Steven Sesselmann
Newton theory of extraterrestrial gravity is plain wrong! It only works on or near the surface of the earth. For your problem, please try the following gravitational potential:
E(p) = m(a/r^2 - GM/r - Cr^2,
Where G is Gravitational constant (determined on earth), C is yet to be determined cosmic constant (or a function), M and m are the masses of the sun and any planet and r is the distance. Please refer to the following link:
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