4
$\begingroup$

When I studied physics some time ago my teacher explained that if we consider the gravitational atraction not instantaneous, such as the General Relativity says, the planets would be attracted towards the position were other planets were before and not the current one, getting spiral orbits instead of ellipses.

What was the solution to this problem? Because we do observe elliptic orbits. I guess when we use speed "c" we can't continue using newton laws and we need the realtivity.

Many articles, even in the Wikipedia, say that the speed of the gravitons must be much higher than "c".
Is it a complet nonsense or some academics believe it?

regards

$\endgroup$
1
  • 3
    $\begingroup$ Which Wikipedia article? $\endgroup$ – Qmechanic Sep 28 '13 at 13:33
4
$\begingroup$

Stable closed orbits are not possible according to general relativity. The orbits which we observe are not exactly elliptical according to GR but only approximately so. This approximation holds quite well for weak gravitational fields and lower velocities ($v<<c$). as per the predictions of GR the orbiting object loses energy in the form of gravitational waves and this loss of energy causes it to (slowly) spiral inwards. Though gravitational waves have not yet been directly detected till date, the effect on binary pulsars has been observed and is in agreement with the predictions of GR.

$\endgroup$
4
  • $\begingroup$ +1 Wasn't the precession of the orbit of Mercury an example of an unstable orbit due to GR, or is that something different? $\endgroup$ – Mike Dunlavey Sep 28 '13 at 14:52
  • $\begingroup$ @MikeDunlavey: There are two effects here: $\endgroup$ – guru Sep 28 '13 at 21:05
  • 1
    $\begingroup$ The precession of mercury is not an example of an unstable orbit, rather it is an example of an orbit which is not closed. Such calculations typically assume that the orbiting object can be treated as a test mass - meaning that its own mass can be neglected or ignored. In the test mass approximation, the effect of gravitational wave emission is ignored and the orbit is stable. According to Bertrand's theorem in classical mechanics, the only two forces which can result in closed non-circular orbits are the inverse square law and Hooke's law. So any departure from the inverse square law is $\endgroup$ – guru Sep 28 '13 at 21:55
  • 1
    $\begingroup$ going to result in orbits which are not closed. As per GR, all orbiting objects emit gravitational waves and would gradually spiral inwards. But this effect is indeed negligible for solar system objects and the time taken for mercury to fall into the sun as a result of this would probably work out to be larger than the age of the universe. However, this effect is of considerable importance for more massive dense objects such as neutron stars or black holes. The wikipedia article may interest you en.wikipedia.org/wiki/… . $\endgroup$ – guru Sep 28 '13 at 22:18
-2
$\begingroup$

Celestial orbits are sure spiraled.The proof starts by refusing the area law of Kepler. In fact when the work equation is considered with its vectorial components, we write, with vectors, $$W=W_{radial}+W_{perpendicular}.$$ Then, $L_p$ being the distance, $$W_p=L_p\cdot F_p,$$ then the differential of $W_p$ is $$dW_p=dL_p\cdot F_p+L_p\cdot dF_p$$ where $F_p\cdot dt=m\,dV_p$ (Newton's law). When $F_p$ is replaced, we have $$dW_p=dL_p\cdot(m\, dV_p/dt)+L_p\cdot d(m\,dV_p/dt).$$

We know from physics $dW_p=0$,and for this we have to write $(dV_p/dt)=0$, which means when integrating $V_p=Constant$. So, Kepler area law $(\frac12r\,V_p=Ct)$ is wrong.

On other hand, when the energy conservation equation is written we get, from the solution of a differential form, a new celestial orbit movement equation, $$r=-4t^2+4tT-\frac{2}{3}T^2$$ ($t$=real time, $T$=life-time of the body) This equation do not indicate an ellipse,but a parabola on Cartesian or a spiraled finite orbit on Polar. Newton has discovered this spirals, but commented wrongly, saying the orbits could not be spirals as it is going on "ad infinitum". He was induced by his period law.But period law do not exist in astronomy. And the orbits are on "ad finitum". See Newton's PRINCIPIA (page 296 by Andre Motte). In astronomy a new time law is valid for whole sun system: $r*V_p^2=Constant$.

You must control this law with the known data of the celestial bodies. For Earth ($r=149597890$ km;$V_p=29,78607371$ km/sec) and the Constant=$1,32725E+11$.

Same constant for Mars,or Halley,or Pluto ,or for comet ISON. Try the evaluation, you will believe that orbits are not elliptical, no area law, no aphelion, no perihelion, no period. All celestial bodies are born from the inside of the sun(when $t=0$, $r<0$). Celestial bodies have a birth date,a living time and a death time. For Earth the actual time is approx 4600000000 years and the life time is 9263192008 years. By years we mean cycles around the sun, years are not 365 days for each cycling.Then what do you think about lightyear distance? Is that a correct definition?

$\endgroup$
2
  • $\begingroup$ In English, a comma (,) has a space after it; similarly for the period (.). Please note that for future use. $\endgroup$ – Kyle Kanos Jan 3 '14 at 15:13
  • $\begingroup$ Except Vp=const is consistent with Kepler's law. The above analysis is not only wrong, it's contrary to observation. $\endgroup$ – Brent Meeker Jun 26 '20 at 3:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.