# 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 astronomy.stackexchange.com/questions/18539/… -- 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. – Eric Lippert Feb 25 '20 at 18:22
• 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. – Ben Crowell Feb 27 '20 at 23:40

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.

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).

• This answer is great imo, I just added the orbital radius part, hope you don't mind – Tosic Feb 25 '20 at 9:35
• @Tosic not at all! But until I can stop and read it and think about it I'll label it as an edit. Thanks! – uhoh Feb 25 '20 at 9:37
• @Tosic If I read your edit right, decreases in the mass of the Sun mean Earth's orbital radius will increase, right? – Nzall Feb 25 '20 at 11:49
• @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. – uhoh Feb 25 '20 at 13:22
• @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. – uhoh Feb 25 '20 at 13:26

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.

• 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. – Phil Perry Feb 27 '20 at 1:09
• @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. – wizzwizz4 Feb 27 '20 at 6:40