Is the Sun really burning and getting smaller? If so, because there is no air in space, what can be keeping it burning and what materials get burnt? If no air is required, what can speed up or can stop the burning process?


I won't bother answering how the Sun "burns" its fuel — that is already answered in this question. Your second question is whether the Sun is getting smaller; the answer depends on what you mean by "smaller". We'll analyze this in two ways:


If we define the radius of the Sun by distance between the core and photosphere, then the Sun is not getting smaller. In fact, it is growing over time, as demonstrated in this graph:

enter image description here

Image courtesy of Ribas (2009), colored by user RJHall on Wikipedia under the Creative Commons Attribution-Share Alike 3.0 Unported license

The height of the photosphere is increasing ever so slightly, even in the Sun's main sequence stage; specifically, Table 3 of Sackmann et al. (1993) states that at the Sun's zero-age main sequence, it was $0.9R_\odot$ in radius, and at its terminal age main sequence it will be $1.6R_\odot$. That means that, over the course of about 10.9 billion years, the Sun's radius will have increased by $0.7R_\odot$ — about 70% its current radius.

If we do the math $\left (0.7R_\odot \over 10.9 \ \text{Gyr} \right )$, we could find that the Sun's average increase in radius (from beginning to end of main sequence) is about 4.5 centimeters each year (1.77 inches). This calculation shouldn't be taken too seriously, since the growth rate is not linear, but it demonstrates how slowly the Sun has been growing.

The Sun will continue to grow at a very slow rate until it leaves its main sequence stage. Once it enters the red giant branch, it will grow many times its current size, to approximately 1 AU.


The Sun is losing mass for two reasons: the nuclear fusion in its core converts some mass into energy, and high energy protons and electrons are constantly ejected from the Sun's corona as solar winds. Wood et al. (2002) shows that the Sun's mass-loss rate $\dot{M}$ is (approximately) inversely proportional to the square of its age $t$:

$$\dot{M} \propto t^{-2.00 \pm 0.52}$$

As of now, $\dot{M} = 10^{-14}\frac{M_\odot}{\mathrm{yr}}$. After the Sun leaves the main sequence, it will lose its mass at a much faster rate. As stated in Sackermann et al., the rate at which it a post-main sequence star loses mass can be generalized with Reimers' law:

$$\dot{M} = -\eta (4 \times 10^{-13}) \frac{LR}{M}$$

where $\eta$ is a value depending on the star (derived to be ~0.6 for the Sun), $M$, $L$ and $R$ are in solar units, and $\dot{M}$ is in solar masses per year. This is simplified, but gives us an idea of how the Sun's phase affects its mass-loss rate.

By the time the Sun enters the red giant branch, it will have only 72.5% its current mass, and its luminosity and radius will reach $2300L_\odot$ and $170R_\odot$, respectively. Plugging those values in, we find that the rate of mass loss will be $1.3 \times 10^{-7} \frac{M_\odot}{\mathrm{yr}}$. It will lose mass even faster in the asymptotic giant branch, when its mass will become ~54.1% of what it is now, meaning $\dot{M}$ will be $2.5 \times 10^{-7} \frac{M_\odot}{\mathrm{yr}}$.

  • $\begingroup$ This chart doesn't look real because how come it's Now when they are all same!! $\endgroup$
    – Archi
    Feb 5 '17 at 14:32
  • 6
    $\begingroup$ @Archi Look at the left-hand axis. The graph describes the properties relative to the current Sun. Therefore, they should all be plotted as one at the present day. $\endgroup$
    – HDE 226868
    Feb 5 '17 at 14:46
  • $\begingroup$ Doesn't the equation you cite from Wood et al say that the mass-loss rate is (approximately) proportional to the inverse square of its age, rather than directly to its age? $\endgroup$
    – Brian Tung
    Feb 23 '17 at 16:40

The Sun is not burning, but undergoing nuclear fusion at the core of the star. There, under great pressure, and high temperature, Hydrogen is converted into Helium and energy (light).

If you recall Einstein's equation $E = mc^2$, energy is equivalent to mass, meaning a small amount of mass can be converted into a large amount of energy.

The actual process of converting the Hydrogen to Helium is complicated.

  • $\begingroup$ The Sun's mass is not strictly related to its radius. The density need not be constant. $\endgroup$ Feb 5 '17 at 6:53

The only other detail I can add to what has already been said above is about the actual fusion process inside he Sun - its engine.

4H -> He + neutrinos + gamma-ray photons

Mass of 4H = 6.693e-27 Kg Mass of He = 6.645e-27 Kg

Mass lost per single fusion reaction = 0.048e-27 Kg

This generates 4.3e-12 J

If 1Kg of H is converted into He, a mass of 0.007 kg (0.7% of total initial mass) is lost in the process, generated 6.3e14 J. This is equivalent to 20e6 kg of coal.

We know that the Sun's luminosity is 3.9e26 J/s which when divided by the energy/kg defined above gives us the mass of H converted per second: 600e12 Kg.

  • 1
    $\begingroup$ If you're going to throw around phrases like "the actual fusion process", you should probably link to proton-proton chain and CNO cycle. :-) The Sun is mostly powered by proton-proton chain fusion. $\endgroup$
    – Brian Tung
    Feb 23 '17 at 16:37
  • $\begingroup$ The mass is not "lost", of course. $\endgroup$ Mar 22 '19 at 12:44

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