Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

M = mass of the Sun

m = mass of the Earth

r = distance between the Earth and the Sun

The sun is converting mass into energy by nuclear fusion.

$F = \frac{GMm}{r^2} = \frac{mv^2}{r} \rightarrow r = \frac{GM}{v^2}$

$\Delta E = \Delta M c^2 = (M_{t} - M_{t+\Delta t}) c^2 \rightarrow \Delta M = \Delta E / c^2$

$\rightarrow \frac{\Delta r}{\Delta t} = \frac{G}{v^2 c^2}.\frac{\Delta E}{\Delta t}$

Sun radiates $3.9 × 10^{26} W = \Delta E/\Delta t$

Velocity of the earth $v = 29.8k m/s$

There is nothing that is stopping the earth from moving with the same velocity so for centripetal force to balance gravitational force $r$ must change.

Is $r$ increasing? ($\Delta r/ \Delta t = 3.26070717 × 10^{-10} m/s $)

share|cite|improve this question
As written, the prediction is for Earth to get closer to the Sun because $\Delta E$ is negative, not positive. – Mark Eichenlaub Nov 4 '10 at 10:40
@Mark Eichenlaub I have written $M_{t} - M_{t+\Delta t}$ and not $M_{t+\Delta t} - M_{t}$. – Pratik Deoghare Nov 4 '10 at 11:11
Okay, but in that case the interpretation of $\Delta r/\Delta t$ is backwards, and a positive $\Delta r/\Delta t$ translates to a decreasing Earth-Sun distance. Just look at your proportionality relation. If $M$ is going down as time increases, so is $r$. – Mark Eichenlaub Nov 4 '10 at 11:46
Energy has gravity same as it's equivalent mass. It is only the radiation of the energy outside the shell of earth's orbit that is important. – ja72 May 14 '14 at 14:49
Also the sun causes tides also (smaller that the moon's) and when those collide with continents the energy is lost. – ja72 May 14 '14 at 14:50

12 Answers 12

up vote 11 down vote accepted

I think the reasoning has an error. It assumes $v$ is constant, but instead we ought to assume the angular momentum is constant.

By dimensional analysis that leads to

$r \propto \frac{L^2}{GM}$

so as $M$ decreases, $r$ increases (the original post had $r \propto M$, not $r \propto 1/M$.

On the other hand, assuming a circular orbit seems dubious.

As the other commenters said, this effect is minute. A significant effect on the orbit of the moon around the earth is tidal evolution, which does actually push the moon further away. See

share|cite|improve this answer
$r = \frac{k(mvr)^2}{GM} \rightarrow r = \frac{GM}{m^2v^2k}$ ? – Pratik Deoghare Nov 4 '10 at 14:02
@TheMachineCharmer: but $v$ is also a function of $r$, so you can't conclude that $r \propto M$. You could write $rv^2 = GM/m^2k$, and since the right side is independent of $r$, that tells you that $rv^2 \propto M$ - but $rv^2 \sim \frac{1}{r}$ because angular momentum is conserved. – David Z Nov 4 '10 at 17:52

Such a small magnitude makes this process negliable among other factors; indeed this result literally means that you don't need to care about this process until you build extreamly accurate theory of the full solar system dynamics. To the extent that is probably unreachable due to deterministic chaos.

share|cite|improve this answer

The Sun is also losing mass due to the solar wind. Again the fraction of mass lost is very small compared to the mass of the Sun, so the effect is very small.

There are these relevant papers that I think you will find interesting:

Orbital effects of Sun's mass loss and the Earth's fate

Astrometric Solar-System Anomalies

share|cite|improve this answer

That is basically correct, however that change is not very significant. The orbit of the planets in the solar system is chaotic over long periods of time (2 - 230 million years according to this wikipedia entry), and this effect is relatively minor. Other causes for change of orbit include gravitational pull from other planets, collisions with asteroids, solar wind and other variables.

share|cite|improve this answer

The distance is increasing due to friction from the tides. Besides the moon, the sun has a tidal component for the earths oceans, and when that crashes into continents the energy absorbed comes from the potential energy of the sun-earth system. I am not sure how this compares to the distance loss due to the mass and energy radiation of the sun you mentioned.

share|cite|improve this answer
This is a different mechanism then the one the OP proposes (stellar mass loss through fusion). Anyone care to do a BOTE calculation of the magnitude of the tidal angular momentum transfer? – dmckee Nov 18 '10 at 20:35
aghhhhhh ... (runs screaming away) – ja72 Nov 19 '10 at 7:00

This is an old question, but I thought it might be worth chewing on a bit. The loss of mass due to fusion in the sun is piffle. The Earth’s orbital radius will change more likely due to interactions with the other planets. The first order perturbations in the orbital elements of the Earth are its eccentricity and right ascension. The change in the orbital radius or the sem-major axis distance is higher order. However, that can occur and there is an over all orbital drift in planetary orbits which is chaotic in nature. The Earth is in a near 1/12 orbital resonance with Jupiter. The Earth may over the next billion years shift away from this and enter into a near 1/11 orbital resonance with Jupiter, where our orbital radius is about 1.06AU.

This early Earth may have been at .83AU relative to today’s orbit very early on. This is an orbital resonance of about 16 with Jupiter. The sun had a power output of 70% of current power. If you factor these together you get a solar irradiance on the Earth comparable to today. If the Earth had the same orbital radius as today, even factoring in a $CO_2$ atmosphere temperatures would be $30C$ cooler than today. Curiously if Earth does drift outwards this delays the solar death of the Earth. If Earth remains at the current radius temperatures will become intolerant in 500 million years for complex life.

Some numerical analyses of this I have run. The interaction with Jupiter results in a periodic oscillation, and a computation over a longer period of time result in a drift which pushes the Earth outwards on average by about $4.2km/sec$.

Perburbation of Earth orbit by Jupiter Long term drift of Earth orbit


This is in part due to alpha Centuri’s commets. One big uncertainty is with understanding the early Earth. I did some homework on this and at 1AU about the warmest the Earth could have been is about -25C with various estimates. Of course this is my interpretation of geo-modelling.

The orbital dynamics is based on computer modeling. This is a general plot of 45,000 years. I should have posted this image. This illustrates the “signal” in these long runs, where the low frequency stuff has the largest amplitude. This is the main signal for an outwards drift.

enter image description here

This does extend the future for life on Earth. If this planet stays at 1AU the prognosis becomes grim about 500 million years from now. The planet will start to reach temperatures 30C higher than today and complex life will begin to die out, and further in a billion years oceans will start to boil. That will really foul things up. However, with the outwards drift these time frames are almost doubled. The luminosity increase in the sun will accelerate faster in time and over take this. The outwards range on this is 2.5 billion years before the oceans start boiling. Once the oceans start boiling this planet will transform into a 400C version of Venus. So I figure complex life on this planet, life which emerged with the Cambrian revolution 550 million years ago, might have a good 750 to maybe 1000 million years ahead of it.

When I first read about the future time frame of life on Earth my mind instantly questioned what happened going back in time. It implies a very cold early Earth; one where it seems the development of life would have been far more difficult.

share|cite|improve this answer
Thats very interesting. How solid is the oribital mechanics for a significantly smaller early earth solar distance? The faint young sun has been a problem for planetary atmospheres/climate, as you say adding CO2, and methane isn't enough (although changing the planets albedo could also have an effect). Currently the planet reflects roughly 30% of sunlight, but under early earth, with a radically different atmosphere, and very little land we don't habe a good handle on the albedo. – Omega Centauri Feb 5 '11 at 15:53
in a "very cold early Earth" life is impossible (I use the Artic versus tropical distribution of life). Oldest bacteria live at thermal vents ( at temperatures ranging from 60 to as high as 464 °C ). – Helder Velez May 15 '14 at 13:09

I think your reasoning is correct, but the values involved are very small. In one year r will increase by 1mm, so in 1 billion years it will have increased by 1000 km or by about0.01%

share|cite|improve this answer
Agreed... Now just change "you're" to "your" please, it's really bugging me. ;) – Noldorin Nov 4 '10 at 14:43
Downvoted because the reasoning actually is not correct. – Mark Eichenlaub Dec 4 '10 at 20:28
@Noldorin: Yeah, I'm a "grammar grump" too. – Mike Dunlavey Oct 21 '11 at 14:13

Right now? No, it's decreasing, until we get to perihelion (91.4M miles, near the January 3rd), and then it'll start increasing again 'til aphelion (94.5M miles, near July 4th).

(As Mark Eichenlaub pointed out -- Earth's orbit is not circular)

share|cite|improve this answer

Definitely Yes.
The Earth is moving away from the Sun at a rate ($0.57H_{0}$) that can not be explained by any of the current official models.

Once again I point to a MODEL in which the data is consistent with the theory. This model is public since 2002 (and my personal knowledge since 1982) in anticipation of the factual finding that the Earth was moving away from the Sun.
It is very problematic to place the origin of life in a cold world. With this model the life has started on an Earth full of energy.
I think that someday this model will be studied with due attention. I am not the author of the model. I'm just a messenger (preaching in the desert?).

by G. A. Krasinsky and V. A. Brumberg, 2004
Secular Increase of astronomical unit from analysis of the major planet motions, and Its Interpretation

$\frac{dAU}{dt}=15\pm4\: m/cy$

at present there is no satisfactory explanation of the detected secular increase of AU

WeiJia Zhang, ZhengBin Li and Yang Lei, 2010
Experimental measurement of growth patterns on fossil corals: Secular variation in short distances ancient Earth-Sun

both the modern and ancient leaving rates could be measured with high precision, and it was found that the Earth has been leaving the Sun over the past 0.53 billion years. The Earth’s semi-major axis was 146 million kilometers at the beginning of the Phanerozoic Eon, equating to 97.6% of its current value. Measured modern leaving rates are 5–14 m/cy, whereas the ancient rates were much higher. Experimental results indicate a special expansion with an average expansion coefficient of $0.57H_{0}$

share|cite|improve this answer
The Krasminsky value has been superceded by more modern measurements, which show there is no change within the limits of measurement. – DavePhD May 13 '14 at 18:10
@DavePhd. Recently I've read the 2011 Pitjeva paper, but I was not convinced about the methods/conclusions. (I assume $G$ constant) – Helder Velez May 15 '14 at 13:04
@DavePhd. The same measure is not the same 'as invariant': "So, we can say that the atomic length unit is a fixed multiple of the Bohr radius; if the latter varies, so will bodies' length and the unit of length, holding invariant the measures of bodies' length. from SE – Helder Velez May 15 '14 at 13:20
Neither Krasinsky nor Pitjeva was referring to the actual Earth-Sun distance, which of course is constantly changing. Instead, they were measuring the pre-2012 "astronomical unit", which has since been redefinied to be exactly 149,597,870,700 meters (see The pre-2012 astronomical unit was defined without respect to the Earth as explained in – DavePhD May 15 '14 at 14:14
the pre-2012 astronmical unit definition was "the radius of an unperturbed circular Newtonian orbit about the Sun of a particle having infinitesimal mass, moving with a mean motion of 0.017 202 098 95 radians per day", independent of Earth, this is what Krasinsky and Pitjeva measured. – DavePhD May 15 '14 at 14:24

I think even currently mass loss due to the solar wind is orders of magnitude greater than mass loss due to $E=mc^2$. But even so it isn't much. Ideally you would have enough mass loss, so that a planets radius would increase at just that rate that kept the stars luminosity divided by r squared constant, i.e. if that were the case the planet would stay within the habitable zone, even as the star brightened due to stellar evolution. Alas the solar wind is much too weak to accomplish the task. But we should be moving further out in any case.

share|cite|improve this answer
Could someone edit this post ! It is illegible (one very long monospaced line) ? – Frédéric Grosshans Nov 17 '10 at 17:29

The answer depends heavily on the time frame you are taking, and also on the definition of the distance between the Sun and the Earth.

The Sun will expand after a while (a long while, but still...). And its surface will get way closer to the actual position of Earth. Will Earth still be near this position remains to be seen, but it is womehow likely, but not guaranteed. So the Sun and Earth will be closer together, but they will probably start by very slightly get further apart first...

To be honest, on the long run, there is no good and "correct answer", for the very fundamental reason that we are talking about a very chaotic system: the Solar System is a complexified version of the famous three body problem. Since we don't have enough information on the state of the System, we cannot predict its evolution. It may very well be that Earth, in a billion or 2 years, will be way way further or closer...

No equation, no clear answer... Sorry, but this question has, as of today, no clear answer.

share|cite|improve this answer

Certainly the sun loses mass by radiation, and also by solar wind, but this is only one side of the story. Mass falling into the sun must also be considered. Comets routinely fall into the Sun, like one below: enter image description here

Subsequent to this question being asked, E. V. Pitjeva published VALUES OF SOME ASTRONOMICAL PARAMETERS (AU, GM⊙, M⊙), THEIR POSSIBLE VARIATIONS FROM MODERN OBSERVATIONS.., where $GM_{sun}$ was determined to be changing by less than 1 part in $10^{13}$ per year.

share|cite|improve this answer
Is there net increase or decrease in mass of the Sun? – Pratik Deoghare May 13 '14 at 17:53
the reference says $\delta GM_{sun}/GM_{sun}$ is $-5 \pm 4 \times 10^{-14}$ per year so I would say it is unknown – DavePhD May 13 '14 at 18:04

protected by Qmechanic May 13 '14 at 11:09

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

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