In 4 billion years, when our Sun sheds all of its outer gas layers and turns into a white dwarf, how much mass will the white dwarf have compared to what the sun has today?

Will the planets still orbit in the same way, or will the reduced mass cause the planets' trajectories to change, so that they eventually leave the solar system?

  • 1
    $\begingroup$ Well it's not really 4 billion years. It's closer to 7-8 billion. $\endgroup$ Apr 17, 2022 at 14:58

2 Answers 2


Short answer:

The Sun will lose about half of its mass on the way to becoming a white dwarf. Most of this mass loss will occur in the last few million years of its life, during the Asymptotic Giant Branch (AGB) phase. At the same time the orbital radius of the Earth around the Sun will grow by a factor of two (as will the outer planets). Unfortunately for the Earth, the radius of the Sun will also reach to about 2 au, so it will be toasted.

There is the possibility that the decreased binding energy and increased eccentricity of the Earth and the outer planets will lead to dynamical instabilities that could lead to planetary ejection. This is highly dependent on the exact time dependence of the late, heavy mass loss and the alignment or otherwise of the planets at the time.

Long answer:

Stars with mass less than about 8 solar masses will end their lives as white dwarfs on a timescale which increases as their main sequence initial mass decreases. The white dwarfs that are formed are of lower mass than their progenitor main sequence stars, because much of the initial mass of a star is lost through stellar winds (particularly during the thermally pulsating asymptotic giant branch phase) and final ejection of a planetary nebula. Thus, the current distribution of white dwarf masses, that peaks between $0.6$ and $0.7 M_{\odot}$ and with a dispersion of $\sim 0.2 M_{\odot}$, reflects the final states of all main sequence stars with $0.9 <M/M_{\odot}<8 M_{\odot}$, that have had time to evolve and die during our Galaxy's lifetime.

The most reliable information we have about the relationship between the initial main sequence mass and final white dwarf mass (the initial-final mass relation or IFMR) comes from measuring the properties of white dwarfs in star clusters of known age. Spectroscopy leads to a mass estimate for the white dwarf. The initial mass is estimated by calculating a main sequence plus giant branch lifetime from the difference between the age of the star cluster and the cooling age of the white dwarf. Stellar models then tell us the relationship between main sequence plus giant lifetime and the initial main sequence mass, hence leading to an IFMR.

A recent compilation from Kalirai (2013) is shown below. This shows that a star like the Sun, born with an initial mass of $1M_{\odot}$ (or maybe a per cent or two more, since the Sun has already lost some mass), ends its life as a white dwarf with $M = 0.53 \pm 0.03\ M_{\odot}$. i.e. The Sun should lose approximately 50% of its initial mass in stellar winds and (possibly) planetary nebula ejection.

IFMR from Kalirai (2013)

A comprehensive treatment of what happens to solar systems when the central star loses mass in a time-dependent way is given in Adams et al. (2013). The simplest cases are initially circular orbits where the mass loss takes place on much longer timescales than the orbital period. As mass loss proceeds, the gravitational potential energy increases (becomes less negative) and thus the total orbital energy increases and the orbit gets wider. Roughly speaking, $aM$ is a constant, where $a$ is the orbital radius, which is a simple consequence of conservation of angular momentum: so the Earth would end up in a 2 au orbit.

However, in the presence of a non-zero eccentricity in the initial orbit, or in the case of rapid mass loss, such as that which occurs towards the end of the AGB phase, then things become altogether more unpredictable, with the eccentricity also growing as mass loss proceeds. This has a knock-on effect when considering the dynamical stability of the whole (evolved) solar system and may result in planetary ejection. The faster the mass loss, the more unpredictable things get.

The radius of an AGB star can be calculated using $ L = 4\pi R^2 \sigma T_{eff}^{4}$. Stars at the tip of the AGB branch have luminosities of $\sim 10^{4} L_{\odot}$ and $T_{eff} \simeq 2500\ K$, leading to likely radii of $\sim 2$ au. So it is quite likely that unless the Earth is ejected or has its orbit significantly modified by some dynamical instability that, like the inner planets, it will end up engulfed in the outer enevelope of the AGB star and spiral inwards...

Even should it narrowly escape this immediate fate, it is then quite likely that tidal dissipation will rapidly extract energy out of the orbit and the Earth will spiral in towards the envelope of the giant Sun... with the same result.

  • $\begingroup$ To add a point of physical interest to that excellent and complete answer, note that the circular orbital radius being inversely proportional to the stellar mass is a consequence of maintaining a fixed orbital angular momentum as the central star loses mass. $\endgroup$
    – Ken G
    Apr 30, 2018 at 13:25

Well, put simply, the Sun will certainly lose at least a fourth of its original mass. This is because most of the mass of the Sun is centered in its core. And since a white dwarf is just the remnant core of a star . Oh, and before the Sun becomes a white dwarf, it goes through the "red giant" phase, where it grows to almost about the size of the orbit of Mars. All the planets will either get sucked into the expanding Sun, leave the solar system or get vaporized when the Sun's nova occurs.

  • $\begingroup$ Planets don't simply stop orbiting. And the Sun is far too small to die in a supernova. As Rob said, it will eventually become a white dwarf, which will continue to glow for trillions of years. $\endgroup$
    – PM 2Ring
    May 2, 2020 at 14:17

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