Back to home

Meteors - A Voice from the Dark

Have you ever wondered why meteorites were created? These chunks of rock sail across the vast expanse of space and their trajectory occasionally crosses earth’s orbit.

Image of meteorite on ice

The photo above, taken from Katherine Joy’s blog, shows a meteorite at the Antarctic.

When a meteorite is within the solar system, its temperature can be predicted just like any planet.

Estimating the temperature of an object based on sunlight

The temperature of the sun can be computed based on its color. Blue stars are hotter than red ones. The power radiated out by the sun (Ps) can be calculated using the Stefan–Boltzmann law:

Ps = 4 π R2 σ T4
= 3.84 * 1026 W

(where R is the radius of the sun, 6.957 * 108m; σ is the Stefan–Boltzmann constant, 5.67 * 10-8Wm-2K-4; and T is the temperature of the sun, 5778K.)

A blackbody in thermal equilibrium is defined as an object that absorbs all radiation that is incident upon it and emits it out as heat.
We can estimate the surface temperature of Pluto by assuming it to be a blackbody in thermal equilibrium. The sunlight incident upon it (Pi) is

Pi = Ps * π rp2 / (4π D2)
=3.87 * 1012 W

(where rp is the radius of Pluto, 1.19 * 106m; and D is its distance from the sun, 5.91*1012m.)
By using the Stefan–Boltzmann law, its temperature needs to be 44K for thermal equilibrium. Actual measurements indicate that its surface temperature is about 36K (Stern, et al., 1993).
Pluto reflects a significant portion of the sunlight it receives. By neglecting its albedo, the temperature is over-estimated.
Earth’s core is much hotter than its surface, and occasionally there are volcanoes. In the same way, Pluto too may have a much hotter core, and by neglecting the effect from its core, the temperature is under-estimated.

Estimating the temperature of an object based on starlight

What is the temperature of objects that are not near a star, and only have faint starlight falling on them? Most of space is not close to any star.

An AU (Astronomical Unit) is the distance between the earth and sun. Pluto is ~40 AUs from the sun. In contrast, Alpha Centauri, our closest star is 270000 AUs away. To put this in perspective, if one pixel on a large screen TV represents the distance between Pluto and the sun, the corners diagonally opposite represent the distance of the closest star, Alpha Centauri to the sun.

In other words, if dark matter is uniformly distributed, most of it is not close to any star, and its temperature depends on the cumulative starlight it receives!

Even though the received power falls off as the square of distance, the number of stars increases proportional to the distance cubed. If we know the average number of stars per unit volume and power emitted by them, we can estimate the temperature of an object from the stars in a given volume.

Sphere of radius r

Consider a spherical shell of radius ‘r’ and thickness ‘dr’. If stars are uniformly distributed and the number of stars per unit volume is ρ, the number of stars in that shell is 4πρr2 dr. If Pa is the average power emitted by a star, the power received at a distance r per unit area from an average star is Pa/4πr2. The power received from the shell (dPr) of thickness dr is then

dPr = 4π ρ Pa r2 dr / (4π r2)
= ρ Pa dr

Integrating, the power received per unit area at the center from all the stars within radius ‘r’ is

Pr = ρ Pa r

The HYG database (derived from the Hipparcos, Yale and Gliese stellar catalogs) is available from David Nash. It contains the distance of stars computed by parallax and their apparent magnitude. The apparent magnitude of a star is defined on a negative logarithmic scale where 5 units less corresponds to a brightness ratio of 100 times more. If there are 2 identical stars, one 5 light years away, and the other 50 light years away; their brightness ratio would be 100 since received power falls off with the square of distance. The apparent magnitude of the nearer star would then be 5 units less.

The apparent magnitude of the sun is -26.74

We can calculate the power received from each star, and keep adding the power received from other stars. For example, Sirius, the brightest star has an apparent magnitude is -1.44. The power received from it (Psirius), in terms of solar irradiance, i.e. the solar power received on earth per unit area (Lsolar) can be calculated by converting the logarithmic scale to a linear one:

Psirius = Lsolar * 10 (1.44-26.74)/2.5
= 7.59 * 10-11 * Lsolar

The table below shows the total power received from nearby stars upto Sirius. The rows are arranged according to the distance from earth. The last column is the total power received from all other stars within that distance (excluding the sun). In the case of Sirius, it is 1.02*10-10 (by adding the cumulative power from the previous row i.e. 2.62*10-11 + 7.59*10-11) solar irradiances.

Table of cumulative power received from stars

The plot of the total starlight received from stars within a distance for all stars closer than 50 parsecs from the HYG database is quite linear with a slope of 4.24*10-12 solar irradiances per light year which is the stellar density (ρ) multiplied by the average star power (Pa).

Cumulative power received from stars vs. Distance

The temperature of dark matter depends on primarily on cumulative starlight. If the universe was created ~6000 years ago, starlight from within a distance of 6000 light years would have reached us. The sun's apparent magnitude corresponds to an irradiance of 1361 W/m2. The graph above can be extrapolated to determine the cumulative starlight received per unit area from within 6000 light years (P6000.)

P6000 = 4.24 * 10-12 * 6000 * 1361 W/m2
= 3.46 * 10-5 W/m2

But dark matter doesn’t absorb all light falling on it. The albedo of meteorites has been measured varying from 0.2 to 0.6 (Piironen, et al., 1998). If we calculate the temperature of dark matter from the Stefan–Boltzmann law by assuming that their average albedo is 0.5, we get ~2.5K. The Cosmic Microwave Background (CMB) has a thermal black body spectrum of 2.73K. (Fixsen, 2009)

Conclusion

This analysis doesn't take into consideration bolometric errors, a correction that must be done to determine the total power emitted by a star. The essense of the argument is to show that by extrapolating what we observe nearby to the rest of the universe, we can estimate the age of the universe. Within the solar system we see planets and meteorites having certain albedos. The statistical average can be assumed for the rest of dark matter in interstellar space. Whatever reasonable corrections applied still show that the temperature of dark objects lit by starlight will be much more than 2.7K if the universe is older than 100 000 years.

If you have any alternate explanation (without defying the known laws of physics) for the observed 2.73K, please let me know.

References

  1. Fixsen, D. J., 2009. The Temperature of the Cosmic Microwave Background. The Astrophysical Journal, December, 707(2), pp. 916-920.
  2. Piironen, J. et al., 1998. Albedo measurements on meteorite particles. Planetary and Space Science, August, 46(8), pp. 937-943.
  3. Stern, S. A., Weintraub, D. A. & Festou, M. C., 1993. Evidence for a low surface temperature on pluto from millimeter-wave thermal emission measurements. Science, 24 September, 261(5129), pp. 1713-1716.

© Selva Harris
2019

Share