### Sun constantly converts mass into energy, will this cause its gravity to decrease?

• If the sun is constantly converting the mass into energy, then will its gravitational field continue decreasing?

You might be interested in this question https://astronomy.stackexchange.com/questions/18539/does-the-loss-of-mass-create-an-observable-change-in-a-comets-orbit -- the answer links to a paper which describes how comet orbits change as the solar wind carries mass away from the Sun; it turns out that the change in orbits due to solar mass changing are pretty small compared to other effects on orbits.

A few points about relativity that don't seem to have been directly addressed by the existing answers: (1) Both photons and neutrinos are effectively massless (energy $\gg$ mass). (2) The source of gravity is not mass, nor is it mass+energy; it's the stress-energy tensor. (3) If the sun were only converting massive particles into massless particles such as photons, and *retaining* those massless particles, then its stress-energy tensor would not change. (The stress-energy is locally conserved.) So what matters is the rate at which mass-energy is *escaping* the sun.

• If the sun is constantly converting the mass into energy, then will its gravitational field go on decreasing?

It's a very interesting question and the answer is yes!

The solar constant indicates the mean solar radiation of electromagnetic waves (mostly in visible and near infrared light and I'll answer based on that.

While the conversion of mass matter to energy in the Sun's core now represents a loss of mass proper matter, it turns out that that energy (trapped in the Sun and slowly diffusing towards the surface) will have the same gravitational attraction as the matter it came from until it actually escapes the Sun!

There is some prompt mass and energy loss via neutrinos and it's significant, perhaps several hundred keV per neutrino I simply don't know the number yet. I'll ask a separate question about it. I'm guessing that losses due to the stellar wind are small, but I'll update here as soon as the following is answered:

update: the answer there is that loss via neutrinos is only about 2.3% of the radiative loss, and on average loss via solar wind and coronal mass ejections is about 4E+16 kg/year, or about another 30% relative to the radiative loss described below.

The value $$I$$ is about 1360 Watts per square meter at $$R$$ = 1 AU which is about 150 million kilometers or 150 billion meters. So the total energy lost per second $$P$$ is

$$P = 4 \pi R^2 I$$

Taking the time derivative of $$E = m c^2$$ we get

$$\frac{dE}{dt} = P = \frac{dm}{dt} c^2$$

so

$$\frac{dm}{dt} = \frac{1}{c^2} \ 4 \pi R^2 I$$

That means that the value of the mass that we use to calculate the Sun's gravitational attraction changes by about 4.3E+09 kilograms per second, or 1.3E+17 kilograms per year.

The Sun's current mass is about 2.00E+30 kilograms, so this effect changes by a very tiny fraction per year, about 6.7E-14. Over the age of the Earth of 4.5 billion years, that's 3E-04, or about 0.03% if the Sun's output were constant. It has probably changed over this time of course, so this is just a rough estimate.

Thanks to @S.Melted's answer for clarifying this.

added to my answer by @Tosic:

The Earth feels no torque from any force during this (the force from radiation is radial), which means its angular momentum is conserved. This means
$$R_1 v_1 = R v$$
$$R_1 \sqrt{\frac{GM_1}{R_1}} = R \sqrt{\frac{GM}{R}}$$
$$M_1 R_1 = M R$$
We can see the Earth's orbital radius would change by a factor of 0.03% as well (M1 and M are solar masses).

Should you not consider the "R" as radius of the SUN, since from that surface energy is radiated and then equivalent mass is reduced? How is Radius of 1 AU is relevant for the question

@MohanMone I think this is correct. The 1360 Watts/m^2 applies to 1 AU. If you read the link that I've included there you will see the explanation. At Earth's average distance from the Sun, above the atmosphere, the number is 1360, it's about 800 at the surface. It's a measure of the Sun's total electromagnetic output. The distance of 1 AU is defined precisely, the radius of the Sun is hard to define exactly.

@MohanMone thinking further, "Why does the solar constant use 1 AU as the reference?" might be an *excellent new question* to ask!

This answer is great imo, I just added the orbital radius part, hope you don't mind

@Tosic not at all! But until I can stop and read it and think about it I'll label it as an edit. *Thanks!*

@Tosic If I read your edit right, decreases in the mass of the Sun mean Earth's orbital radius will increase, right?

"effect changes by a very tiny fraction per year, about 6.7E-14" - is this a percent change? Or is the word "fraction" before it just throwing me?

@Bilkokuya if $m$ is the mass of the Sun at some time, and the mass is $m - \Delta m$ one year later, then the fraction $\Delta m / m$ is 6.7E-14. $\Delta m$ = 6.7E-14 $m$. It's the fraction of the total mass that's lost in one year.

This seems like a great answer that could be even better if it included other mass loss/gain mechanisms such as 1-The solar wind and 2- Infalling objects.

@James which is of course the reason I've asked exactly that question and have already indicated that I'll edit and update once that becomes available.

@Bilkokuya it's probably okay to leave your comments and mine, I think they may be helpful to some future readers.

@uhoh: Thanks, I totally missed that part of your answer. Also, it would be interesting to compare the mass loss mechanisms to the mass gained from infalling matter.

@James I don't know much about that, it's an interesting point. This is slightly related and you might find some of it interesting: What is the origin of the dust near the sun?

@Nzall That's right. On a circular orbit approximation, Earth's specific angular momentum $h=\sqrt{GMR}$, where $M$ is the Sun's mass and $R$ is Earth's orbital radius, so $R\propto 1/M$.

Can we use just percentages? Assuming the sun's gravity right now is 100%, what was it 4.5BYA, and what will it be ~5BY from now? Will both be close enough to 100% not to bother? - "Earth's orbital radius would change by a factor of 0.03%" - every year? Every 4.5BY? Or, in ~5BY (the last time it will ever matter)? unclear. What's the trite answer? : Yes, but it doesn't mater, and here's by how many zeros behind the decimal point that it doesn't mater; time after which nothing maters.

@Mazura I've just read this excellent answer to my parallel question (linked above) which means I need to update my answer. I'll add a tl;dr or similar and see if I can do exactly that. Thanks!

FWIW, neutrinos move at almost the speed of light, so it only takes them a couple of seconds to escape the Sun. In contrast, it takes a *very* long time for the electromagnetic energy to diffuse through the Sun; various sites give figures ranging from 10,000 to 100,000 years or more.

@PM2Ring Thanks! thus the "(trapped in the Sun and slowly diffusing towards the surface)" bit.

Sure! I mostly just wanted to mention that the neutrino energy isn't trapped. That becomes very important in larger stars, when they start fusing oxygen, because the neutrinos quickly carry away a lot of energy, as mentioned in https://en.wikipedia.org/wiki/Oxygen-burning_process#Neutrino_losses

"energy ... will have the same gravitational attraction as the matter it came from until it actually escapes the Sun": doesn't it continue to have the same gravitational attraction even after it escapes the sun? Its effect on the earth only changes when it gets to be farther away from the sun's center of mass than the earth is (assuming radial symmetry).

@phoog the question asks about the "gravity of the Sun"; once light and other things leave the Sun they are no longer part of the Sun. Earth isn't mentioned in the question, but that said yes Newton's shell theorem would apply to any radially symmetric distribution and so on top of the roughly 100,000 years the energy took to leave the Sun it will have some minutes more in the form of free sunshine to affect the inner planets.

• I wanted to expand a bit on a point @uhoh made. (I would have made this a comment, but lack the reputation). The Sun is not converting mass to energy. The Sun is converting matter (not mass) into other forms of energy, such as light. As you noted, the energy still has gravitational attraction (i.e. mass).

If you use nuclear fuel and afterwards collect all the matter, you'll find that you have less. Some has been converted to other forms of energy and radiated away. However, if you had an impenetrable box that contained all forms of energy and use the fuel, you would find that mass is the same.

To expand, if you warm a cup of coffee, that same cup of coffee now weighs more. It has gained the mass of the energy of the heat it now contains.

Edit: I wanted to add a link to an article that was my "light bulb" moment on this years ago when I had some misconceptions about this:

https://www.fnal.gov/pub/today/archive/archive_2012/today12-01-13_NutshellMassReadMore.html

Thank you very much for this! We'll see what we can do about getting your reputation above 50 ASAP ;-) I'll point back to this from my answer.

Is the Sun not converting mass to energy? I thought that energy was not matter, but motion of matter (e.g. heat energy is matter vibrating), and that Einstein's Special Theory of Relativity(E=mc^2) held an equivalence between mass and energy, which presumably implies for transference from one to the other. Light is being produced, and is massless (or so I learned in college), so presumably hydrogen et. al that is consumed in fission can be reguarded as lost mass? I assume I am missing something...

I don't know why all the hair-splitting over mass vs. matter. Any matter ("frozen energy", if you will) has mass, which is convertible to and from energy. In Einstein's famous equation $E=mc^2$, my understanding is that "m" is _mass_, measured in something like kilograms.

@PhilPerry It's rest mass, which isn't something you could measure directly unless the object was completely still. The full equation is (from memory) $E^2 = m^2 c^4 + c^2 p^2$, which takes into account momentum too.

@sharur This one page article from Fermilab is written to lay people and explains it clearly. It has some really good explanations and analogies. But the bottom line is E = mc^2 isn't like trading paper money for gold, it's like having gold coinage. The gold is the coin and vice versa. But I defer to the experts: https://www.fnal.gov/pub/today/archive/archive_2012/today12-01-13_NutshellMassReadMore.html

Remember: if you have an "impenetrable" (read: perfectly isolating) box you don't now anything from the inside except its gravitation. It simply *cannot* lose weight/mass by any means whatsoever even if you annihilate equal amounts of matter and antimatter inside.

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