How much of the earth would remain intact if the sun expanded into a red giant encompassing its orbit?

Question

It seems to me that part of the earth might survive being absorbed by a red giant. How long would it take the heat, plasma, magnetic field, etc. of the sun to eat away at the earth? How long could it orbit while inside the outer edge of the sun? Do we have any idea what would happen to the earth and how long it would take?

Would it be a year in orbit? 100 years? A million?

Answer 1

About a century, once the process really start going. Before that there will be a few million years of spiralling inward towards the surface of the sun.

It takes time $\tau_{sweep}\sim M_{earth}/\pi R^2_{earth} \rho v$ to sweep out one Earth-mass of gas, which would significantly slow down the planet. Red giant envelopes have density around 0.1 kg/m$^3$ and if we use $v=30$ km/s I get $\tau_{sweep}\approx 0.5$ years. As the planet loses momentum it sinks into denser regions, speeding up the process. So the whole inspiral once the planet is inside the envelope takes a few years. A bit more careful analysis in (Goldstein 1987) gives a timescale about 200 years.

Since there is a massive solar wind there will also be gas drag before moving inside the sun. This very relevant paper estimates an inspiral time of about 14 million years from the gas drag, speeded up by tidal effects. Once the planet gets inside it plunges very quickly.

While the envelope density is low, it is also heated by the planet moving through it supersonically. That means it is going to erode the planet rapidly by vaporization. It should be noted that brown dwarfs and gas giants under some circumstances can survive the episode, becoming terrestrials. But terrestrials erode quickly.

Answer 2

Let's calculate a few figures:

The average temperature of the solar plasma is about $10^7 K$ . If the solar radius expands from its present value to the earth's orbit, the new radius would have increased by a factor 215. This means the gravitational potential energy, which for a gas sphere in hydrostatic equilibriumm is given by

$$U=-\frac{GM^2}{R}$$

would decrease by a factor $1/215$, hence also (according to the virial theorem) the average kinetic energy, i.e. the temperature would now be about $4.7\cdot 10^4 K$. Even though the density of the expanded sun would now have decreased by a factor $1/215^3$ (which would be less than 1% of the density of the earth's atmosphere at sea level), this temperature would be more than 10 times that required to vaporize any material.

So it is safe to say that the earth would be completely vaporized more or less instantly.

---

EDIT: below a numerical estimate of the time required to vaporize the Earth:

latent heat for vaporization of silicon = $13$ Mega Joule/ kg (see https://en.wikipedia.org/wiki/Latent_heat )

mass of Earth = $6 \cdot 10^{24}$ kg

=> energy required to vaporize Earth = $10^{32}$ Joule

energy of hydrogen atom at $10^5$ K = $1.4\cdot10^{-11}$ Joule

particle density of hydrogen gas at $10^{-7}$ air density = $3\cdot 10^{18}/m^3$

speed of hydrogen atom at $10^5$ K = $4\cdot 10^4$ m/sec

=> number of hydrogen atoms hitting $ 1 m^2 = 3\cdot 10^{18}\cdot 4\cdot 10^4 = 1.2 \cdot 10^{23} / sec$

surface area of Earth = $5\cdot 10^{14} m^2$

=> energy transfer to Earth = $1.4\cdot10^{-11}\cdot 1.2 \cdot 10^{23} 5\cdot 10^{14}$ = $10^{27} Joule/ sec$

=> Earth vaporized in $10^5$ sec = $28$ hours

Of course, the Earth will get smaller whilst the outer layers are getting vaporized (and thus the energy transferred will get less), but on the other hand the core of the Earth is already basically at or above the boiling point (only kept together by the high pressure), so not much energy is needed to vaporize the latter.