Issue 033 - Astronomy - Explosion energetics
How much kinetic energy was carried away by the exploding star?
Astronomers observed the death of a massive Wolf-Rayet star from its initial shock breakout through almost three months of follow-up. Reuters reported ejecta moving at a bit more than 10% of the speed of light, and the collapse probably left behind a black hole.
The problem
Estimate the kinetic energy carried by the material expelled in the supernova.
Use the result to compare the explosion with one or two familiar energy scales, such as the magnitude 7.1 Japan earthquake from an earlier problem, humanity's annual energy use, or the Sun's energy output.
As a secondary estimate, determine how far the fastest ejecta would travel during the roughly three months astronomers followed the explosion.
Because Fermi problems target an order of magnitude, I normally use no more than two significant digits and write most calculations in scientific notation; the Fermi reference explains both conventions.
Before checking sources
Matt's first pass
I assumed 70% of the star's mass was ejected during the supernova, with the other 30% becoming the black hole. The star was about 30 times the mass of the Sun, and I remembered the Sun's mass as about 2 x 10^30 kg.
star mass ~= 30 x 2 x 10^30 kg
~= 6 x 10^31 kg
ejected mass ~= 0.7 x 6 x 10^31 kg
~= 4.2 x 10^31 kgThe ejecta reached about 10% the speed of light. Since light speed is about 3 x 10^8 m/s, I used 3 x 10^7 m/s.
KE ~= 1/2 x m x v^2
~= 1/2 x 4.2 x 10^31 x (3 x 10^7)^2
~= 1.9 x 10^46 JThat is an enormous number. Comparing it with the energy released by the Sun, I remembered that about 4 x 10^9 kg of matter are converted to energy in the Sun every second:
solar energy/second ~= m c^2
~= 4 x 10^9 x (3 x 10^8)^2
~= 3.6 x 10^26 J/s
solar time ~= 1.9 x 10^46 / 3.6 x 10^26
~= 5 x 10^19 seconds
~= 1.7 x 10^12 yearsI also remembered a statistic that humanity uses about 5.8 x 10^20 J per year worldwide. If that is correct, the supernova ejecta kinetic energy would equal about:
humanity-years ~= 1.9 x 10^46 / 5.8 x 10^20
~= 3.3 x 10^25 yearsIf scientists tracked the progression for three months, that is about 90 days and 86,000 seconds per day, or about 7.8 x 10^6 seconds. Moving at about 3 x 10^7 m/s, the ejecta would cover:
distance ~= 3 x 10^7 m/s x 7.8 x 10^6 s
~= 2.3 x 10^14 m
~= 2.3 x 10^11 kmCalibration Score
Matt's Calibration Score: 50 / 100
Higher is better: earn points for accurate pegs, sound models, correct math, and a result close to the sourced answer. The image shows percent full of it: 100 minus the Calibration Score.
Pegs: 0/30. Ejecta mass and average-speed assumptions were too high.
Model: 30/30. Kinetic energy was the right model, and relativistic correction at 0.1c is small for Fermi work.
Math: 10/10. The arithmetic was clean.
Result: 10/30. The result was near the edge of order-of-magnitude correctness for the presumed answer.
Grounding facts
The simple kinetic-energy equation explains most of the scale. One solar mass moving at 0.1c already carries about:
KE ~= 1/2 x 2 x 10^30 x (3 x 10^7)^2
~= 9 x 10^44 JThat single line is the whole shock of the story: stellar masses plus even a tenth of light speed produce energies no Earth-scale event can approach.
After checking sources
Check and recalibrate
The biggest correction is that the reported "a bit more than 10% of light speed" describes very fast ejecta, not necessarily the average velocity of all expelled stellar mass. The original star may have been around 30 solar masses, but a Wolf-Rayet star has already lost its hydrogen and helium layers, and much of the remaining core can collapse into the remnant.
Published modeling for SN 2026gzf gives ejecta masses and kinetic energies in the broad-lined Type Ic supernova range. One useful Fermi setup is a few solar masses moving at a characteristic speed below the fastest 0.1c material:
ejecta mass ~= 2 solar masses
~= 4 x 10^30 kg
characteristic speed ~= 2 x 10^7 m/s
KE ~= 1/2 x 4 x 10^30 x (2 x 10^7)^2
~= 8 x 10^44 JIf you instead use 2 solar masses at the fastest 0.1c speed:
KE ~= 1/2 x 4 x 10^30 x (3 x 10^7)^2
~= 1.8 x 10^45 JSo a good corrected range is roughly 1 x 10^45 J, give or take a factor of a few. The published values are often reported in ergs; 10^52 erg = 10^45 J.
At 0.1c, relativistic corrections do not change the Fermi answer very much. The Lorentz factor is only about 1.005, so ordinary kinetic energy is within about 1% of the relativistic value.
Now compare with familiar scales:
Japan M7.1 earthquake energy ~= 3 x 10^15 J
supernova / earthquake ~= 1 x 10^45 / 3 x 10^15
~= 3 x 10^29
world energy use/year ~= 6 x 10^20 J/year
supernova / humanity ~= 1 x 10^45 / 6 x 10^20
~= 2 x 10^24 years
Sun luminosity ~= 4 x 10^26 J/s
Sun time ~= 1 x 10^45 / 4 x 10^26
~= 2.5 x 10^18 s
~= 8 x 10^10 yearsEven after reducing Matt's kinetic-energy estimate by about an order of magnitude, the comparisons are still absurdly large: hundreds of billions of years of present solar output, and far more than all human energy use over any meaningful civilization timescale.
For the fastest ejecta distance, Matt's calculation was right:
three months ~= 90 days
~= 7.8 x 10^6 s
distance ~= 0.1c x time
~= 3 x 10^7 x 7.8 x 10^6
~= 2.3 x 10^14 mThat is about 2.3 x 10^11 km, about 1,500 AU, or roughly 9 light-days. The star is 500 million light-years away, so this is still a tiny angular expansion from Earth's perspective.
Post-check reflection
Matt's reflection
Looks like I assumed way too much of the star's mass would become ejecta, and that the speed given was the average ejecta speed rather than a maximal one. All subsequent calculations were correct if the assumptions were correct, and my answers were close to being within about an order of magnitude from the presumed correct answer.
My thoughts on the topic are not considerably changed. When we are talking about observable astronomical phenomena, the energies, distances, speeds, and timescales involved are so enormous that they are difficult to conceptualize.
Recommended memory peg
For supernova energy estimates, remember 1 solar mass ~= 2 x 10^30 kg, 0.1c ~= 3 x 10^7 m/s, 1 solar mass at 0.1c ~= 9 x 10^44 J, and 10^52 erg = 10^45 J.
Reader results
Bars show how submitted estimates sort into the answer choices from the gut-check prompt.