I have heard it said that charged planets could not orbit a massless (low mass) oppositely-charged star based on electromagnetic attraction the same way they can with gravitational attraction, because Maxwell's laws dictate that accelerating (orbiting) charges produce electromagnetic waves and therefore lose energy which would lead to the planets slowdown and eventually crash. But it occurred to me that something similar would seem likely with gravitational waves in real-life gravity-based orbits.

Is it true than that planets' orbits are decaying slowly and turning that energy into gravitational waves? If not, how can that be, given we know gravitational waves exist and surely expend energy in the same way the production of electromagnetic waves do?

  • $\begingroup$ Related: astronomy.stackexchange.com/questions/13716/… $\endgroup$
    – Pere
    Commented Mar 25, 2020 at 14:27
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    $\begingroup$ I've removed a number of comments that were attempting to answer the question and/or responses to them. Please keep in mind that comments should be used for suggesting improvements and requesting clarification on the question, not for answering. $\endgroup$
    – David Z
    Commented Mar 27, 2020 at 4:17

5 Answers 5


Yes, but undetectably. The Earth-Sun system radiates a continuous power average of about 200 watts as gravitational radiation. As Wikipedia explains, “At this rate, it would take the Earth approximately $1\times 10^{13}$ times more than the current age of the Universe to spiral onto the Sun.”

The Hulse-Taylor binary (two neutron stars, one a pulsar) was the first system in which the gravitational decay rate was measurable. It radiates $7.35\times 10^{24}$ watts as gravitational radiation, about 1.9% of the power radiated as light by the Sun.

  • $\begingroup$ Probably worth pointing out the differences between gravity and electromagnetism and how their fields work. (Tensor vs. scalar? I don't understand in enough detail to actually explain it). And how moving electromagnetic charges would indeed radiate more EM power than accelerating masses radiate grav-wave power, given enough charge to produce similar circular acceleration around a massless fixed charge. "Gravitational charge" (mass) is like electrical charge to some degree. $\endgroup$ Commented Mar 24, 2020 at 12:11
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    $\begingroup$ @PeterCordes tensor vs. vector, which require a changing quadrupole moment and dipole moment (or higher), respectively. Scalar waves require a changing monopole (imagine a spherically symmetric sound speaker emitting pressure waves spherically)...while a spherical symmetric charge distribution, even with radially moving charges, doesn't radiate. $\endgroup$
    – JEB
    Commented Mar 24, 2020 at 15:38
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    $\begingroup$ @mrmineheads Thank you for suggesting an edit. However, I rejected it because that number is part of a quote from Wikipedia. I think quoted material from sources should remain as in the source, even though I agree with you that the $1\times$ is questionable. It’s probably there to indicate that the quantity is accurate to one significant digit rather than to one power of ten. $\endgroup$
    – G. Smith
    Commented Mar 25, 2020 at 22:06

Yes, two bodies orbiting each other like this will indeed emit gravitational waves, regardless of whether or not they're compact objects like neutron stars or black holes. Obviously, most exoplanets will not emit strongly; a planet-star system generally involves large separations and non-relativistic speeds. Therefore, as G. Smith noted, while all such systems emit gravitational waves, the radiation is largely insignificant.

It's been proposed (Cunha et al. 2018) that some exoplanets with extremely small semi-major axes ($a\sim0.01$ AU) could be sources of gravitational waves that would detectable in the near future. As in most of these cases $a$ is large compared to the sources LIGO has observed so far (compact objects in the process of merging), these waves would be relatively low-frequency ($f\sim10^{-4}$ Hz) and would fall in the regime of long-baseline space-based interferometers like LISA, not ground-based interferometers like LIGO. Some exoplanets could reach peak strains of $h\sim10^{-22}$, which is indeed above LISA's sensitivity curve at those frequencies. (Compare this to the binary systems LIGO has observed so far, with $f\sim10^2\mathrm{-}10^3$ and $h\sim10^{-22}\mathrm{-}10^{-21}$ at peak.)

The authors note that in these systems, orbital decay is indeed occurring, but at lower rates than, say, famous orbiting compact objects like the Hulse-Taylor binary pulsar. Over long timescales, this decay should be detectable. In a few systems, the period decay are comparable to the Hulse-Taylor binary, within a factor of a few, although the gravitational wave luminosities remain lower by a couple orders of magnitude or more.

  • $\begingroup$ Looks like such a planet would be close to its Roche limit, around ~2.5 times the star's radius: Radius of the Sun is 7E8 m, Roche limit then 1.75E9 m. One AU is 1.5E11 m, so that 0.01 AU = 1.5E9. (It's possible I got some number wrong, of course, but it looks ok to me ...). This is somewhat interesting because it limits the properties of objects in orbits small enough to produce detectable GW: Very dense. "Normal" stars would disintegrate too early. $\endgroup$ Commented Mar 25, 2020 at 7:19
  • $\begingroup$ I realize that the 2.5 factor for the Roche limit is dependent on the satellite's density, so for a dense planet like Earth it's much less (<1). $\endgroup$ Commented Mar 25, 2020 at 7:29
  • $\begingroup$ SO does every piece of matter emit a wave? Like a human being? $\endgroup$
    – Cloud
    Commented Mar 25, 2020 at 12:18
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    $\begingroup$ @Cloud Matter with a time-varying quadrupole moment will radiate gravitational waves. $\endgroup$
    – HDE 226868
    Commented Mar 25, 2020 at 13:34
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    $\begingroup$ @Cloud A human being on an otherwise perfectly spherical, rotating planet would emit gravitational waves much like an asymmetrical black hole, just less. Over time, that would slow the planet down. Don't you feel the braking force already? Anyway, get a coffee while you are waiting. $\endgroup$ Commented Mar 26, 2020 at 13:21

G.Smith and HDE 226868 gave good answers.

I would add that, in the Solar system case, the gravitational waves are clearly not the dominant factor in changing the (Keplerian parameters of) orbits. Momentum exchange between planets, solar radiation pressure, solar wind effects, tidal effects - every one of these (and probably more that I cannot recall right now) are orders of magnitude stronger than the orbit decay because of gravitational waves radiation.


As noted in @G.Smith's answer, Wikipedia gives a figure of $\sim 200 \, \mathrm{W}$ for Earth/Sol orbital radiation.

Wikipedia didn't clearly cite the source, but this PDF is cited not long thereafter and may be it. That PDF claims that the radiated energy for a non-relativistic binary system is about $$ \frac{\mathrm{d}E}{\mathrm{d}t} ~=~ - \frac{32 \, G^4}{ 5 \, c^5 \, r^5} {\left(m_1 \, m_2\right)}^{2} \, {\left(m_1 + m_2\right)} \tag{24} \,, $$ where the masses $m_1$ and $m_2$ are separated by a radius $r .$ The numbers seem to sync up, so I'm guessing that it may be the source.

For the solar system, WolframAlpha calculates:
$ {\def\Calc{~~{{\color{darkblue}{\Large{🖩}}} \!\!} }} {\def\RowHeaderPrefix{ \textbf{Mercury} }} {\def\RowHeader{ {\phantom{\RowHeaderPrefix{\Calc}\textbf{:}~~}} }} {\def\EnergyColumn{ \phantom{0 {,}\, 000 {.}\, 000 {,}\, 000 {,}\, 000 {,}\, 000 {,}\, 00} }} {\def\PlanetEntry#1#2{ \rlap{ {\RowHeader} {\llap{\textbf{#1} \phantom{\Calc} \textbf{:}~}} {\rlap{~~#2}} }} {\def\CalculationLink{ \rlap{ \phantom{\RowHeaderPrefix} {\Calc} }}}} {\def\Placeholders#1{{ \color{lightgrey}{#1} }}} $$ {\rlap{\begin{array}{c} {\smash{\RowHeader}} \\[-25px] {\underline{\textbf{Planet}}}\phantom{:} \end{array}}} {\rlap{\RowHeader\begin{array}{c} {\smash{\EnergyColumn}} \\[-25px] {\underline{\textbf{Radiation}~\left(\mathrm{W}\right)}} \end{array}}} $
$\PlanetEntry{Mercury}{\phantom{0{,}\,0} 69 {.}\,}$$\CalculationLink$
$\PlanetEntry{Venus}{\phantom{0{,}\,} 658 {.}\,}$$\CalculationLink$
$\PlanetEntry{Earth}{\phantom{0{,}\,} 196 {.}\,}$$\CalculationLink$
$\PlanetEntry{Mars}{\phantom{0{,}\,00} {\Placeholders{0 {.}\,}} 276}$$\CalculationLink$
$\PlanetEntry{Jupiter}{\phantom{} 5{,}\,200 {.}\,}$$\CalculationLink$
$\PlanetEntry{Saturn}{\phantom{0{,}\,0} 22 {.}\, 54}$$\CalculationLink$
$\PlanetEntry{Uranus}{\phantom{0{,}\,00} {\Placeholders{0 {.}\, 0}}15 {,}\, 93}$$\CalculationLink$
$\PlanetEntry{Neptune}{\phantom{0{,}\,00} {\Placeholders{0 {.}\, 00}}2 {,}\, 349}$$\CalculationLink$
$\PlanetEntry{Pluto}{\phantom{0 {,}\, 00} {\Placeholders{0 {.}\, 000 {,}\, 000 {,}\, 000 {,}\, 00}}9 {,}\, 83}$$\CalculationLink$

To note it, these figures are theoretical; it remains to be seen if current theories work in contexts like this.

  • 1
    $\begingroup$ Good job posting all the numbers for the solar system. One idea for improvement though: It's relatively hard to tell the dot in $0.276 W$ apart from the comma in $5,200W $. I would suggest giving the thousands separator a good kick and using proper unit prefixes instead: Mars: $276 mW$, Jupiter: $5.2 kW$, etc. $\endgroup$ Commented Mar 25, 2020 at 10:58
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    $\begingroup$ "these figures are theoretical" only in the sense they are too small to be detected with current measurement abilities. The theory is tested for much more extreme cases and generally agrees with the calculations. $\endgroup$
    – fraxinus
    Commented Mar 25, 2020 at 13:26
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    $\begingroup$ You should take the eccentricities of the planetary orbits into account since some of them are significant. The formula for average power radiated gravitationally by an eccentric binary is here. I believe it was originally derived by Peters and Matthews in 1963. Today it is a common homework problem for a General Relativity course. $\endgroup$
    – G. Smith
    Commented Mar 25, 2020 at 18:06
  • $\begingroup$ @cmaster-reinstatemonica: Took a go at reformatting it; hopefully it'll be a lil more intuitive now. $\endgroup$
    – Nat
    Commented Mar 31, 2020 at 11:45
  • $\begingroup$ To note it, I was considering making the placeholder-$0\text{'s}$ light-grey. I wonder if that'd improve the presentation? $\endgroup$
    – Nat
    Commented Mar 31, 2020 at 11:51

If you had an interferometer that was accurate enough, you would be constantly in an ocean of gravitational waves. The frequencies of the planet's waves would be very low frequencies, about 1 period per year! Jupiter going to aphelion would vary in amplitude every 12 years. For the moment, 20Hz is the record for low frequency gravitational wave detection.

  • $\begingroup$ Shouldn't that be "about 2 periods per year"? I mean, gravitational waves don't care which emitting body is on which side of the common center of gravity. This is especially obvious if both bodies have the same mass, but I believe that gravity wave frequency should always be twice the orbital frequency. $\endgroup$ Commented Mar 25, 2020 at 10:54
  • $\begingroup$ The amplitude of the wave varies from maximum to minimum as the distant gravity mass is near or far on the course of it's orbit. The real timings are extremely complex (async orbits) month, year and decade periods. the gravitational force exerted by distant star systems is not completely unrelated to their luminosity, i.e. the gravity that the planets of alpha centauri exert upon us compared to our own, is not unrelated to the difference in apparent magnitude between the alpha centauri and sol. So building an interferometer to detect distant planets is like trying to sunbathe from starlight. $\endgroup$ Commented Mar 25, 2020 at 11:28

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