Supernova Expansion Estimator

Estimate how large a supernova remnant has grown and whether a telescope could separate it from a point source.

Worked example — Supernova Expansion Estimator

Convert speed	3.00 Mm/s
Convert age	31.6 Gs
Free-expansion radius	10.0069 ly
Convert distance	6.15×10¹⁹ m
Project onto sky	635.1 arcsec
Estimated remnant radius	10.0069 ly
Estimated remnant diameter	20.0138 ly
Estimated angular diameter	635.1 arcsec

These are the values the calculator shows when the page opens. Change any input above to recompute them.

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R = vt describes a remnant for only a few centuries 🖖

Distance equals speed times time is exact while the ejecta are still coasting through near-vacuum. That free-expansion phase is short — a few hundred years at most, until the shell has swept up more interstellar gas than it started with. After that the remnant enters the Sedov–Taylor phase, where the radius grows as t to the power 2/5 instead of t. The consequence is easy to feel: doubling the age from there multiplies the radius by only 1.32, not by 2. Later still, once the gas cools enough to radiate away its energy, expansion slows further before the remnant merges into the interstellar medium entirely. So a radius extrapolated linearly from today’s velocity is a snapshot pushed backwards through physics that has already changed.

Why distance shrinks a giant remnant 🖖

At heart this tool uses the plainest formula in physics: distance equals speed times time, so the remnant's radius is R = vt. The same shell looks tiny or huge depending on how far away it sits. Because angular size is roughly θ ≈ 2R/d, moving a remnant twice as far away halves the angle your telescope must resolve.

The Crab expands faster than 1054 allows 🖖

Run the Crab Nebula's motion backward at today's speeds and the filaments converge around the year 1140 — roughly 90 years after the supernova China recorded in 1054. The remnant is therefore accelerating, not coasting: the central pulsar's wind keeps pumping energy into the shell. So a constant-speed R = vt would date the explosion too late, a rare case where reality outruns the simple model.

Problem solved in full

The shell's radius at 3000 km/s and 6500 light years away 6 steps
3000 km/s, 1000 years, 6500 light years away. Find the shell's radius and how wide it looks — then check whether "free expansion" is still allowed.
1. A light year is defined as the distance light covers in a year, so dividing a speed by c and a time by a year cancels every unit at once. No metres, no seconds.
  
  $R = v t; 1 ly = c \times 1 yr ⟹ \frac{R}{ly} = \frac{v}{c} \times \frac{t}{yr}$
2. Evaluate. 3000 km/s is almost exactly 1% of light speed, which makes the radius almost exactly 1% of the age in light years — a coincidence of the chosen numbers, not of physics.
  
  $\frac{v}{c} = \frac{3000}{299 792.458} = 0.0100069 = 1.0007% ⟹ R = 0.0100069 \times 1000 = 10.0069 ly$
3. Diameter is trivial; the age in seconds is not needed for anything above, and is shown only because the panel reports it.
  
  $D = 2 R = 20.0138 ly; t = 1000 yr = 3.156 \times 1 0^{10} s = 31.6 Gs$
4. Angular size is the diameter over the distance. The small-angle approximation is safe here because the ratio is 0.003, and the panel says so explicitly.
  
  $θ = 2 arctan \frac{R}{d} \approx \frac{2 R}{d} = \frac{20.0138}{6500} = 3.0791 \times 1 0^{- 3} rad$
5. Convert to arcseconds by multiplying radians by 206,265 — the number of arcseconds in a radian, and the same constant that defines the parsec.
  
  $3.0791 \times 1 0^{- 3} \times 206265 = 635.1 arcsec = 10.6 arcmin$
6. Now test the model's own assumption. Sweep out the volume of the shell, fill it with one hydrogen atom per cubic centimetre, and weigh what the blast has had to push.
  
  $M_{swept} = \frac{4}{3} π R^{3} ρ_{ISM}, n = 1 c m^{- 3} : \frac{4}{3} π (9.467 \times 1 0^{16})^{3} \times 1.67 \times 1 0^{- 21} = 5.9 \times 1 0^{30} kg = 3.0 M_{⊙}$
Answer

10.0069 ly across a radius, 635 arcsec on the sky — and no, free expansion ended long ago. At that radius the shell has swept up about 3 solar masses of interstellar gas, which is more than a Type Ia ejects in the first place. Once the swept mass exceeds the ejecta, the shell is being braked by what it collects and the constant-speed model is finished: the remnant enters the Sedov–Taylor phase, where the radius grows as t^(2/5) rather than t. The tool says at the top that it assumes constant speed, and it is right to; what it does not say is that the assumption expires after a few hundred years, which for a 1000-year-old remnant is most of its life. The pretty part of the arithmetic is step 1, where a light year does all the unit conversion for free.

$\frac{R}{ly} = \frac{v}{c} \cdot \frac{t}{yr} = 10.0069, θ = \frac{2 R}{d} = 635. 1^{''}$

References (1)

The Crab filaments converging on 1140 rather than 1054: V. Trimble, "Motions and Structure of the Filamentary Envelope of the Crab Nebula." The Astronomical Journal 73, 535, 1968 — proper motions extrapolate back to AD 1140 ± 10, about 90 years after the recorded supernova.

By the 42Math team · Updated 2026-05-31

Reviewed against published sources. Corrections welcome. Editorial policy · Corrections

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Example problems

Crab-like remnant - 1500 km/s for 971 years gives a radius of 4.86 light-years, spanning 5.14 arcminutes at 6500 ly. The age is set to what the 1054 record implies.
Tycho-like remnant - Faster and older than the Crab, with a radius of 7.56 light-years, and further away at 2500 pc. It still appears larger: 6.37 arcminutes against the Crab's 5.14.
SN 1987A-like - A 0.39 light-year radius spanning under one arcsecond, because 50 kpc is a long way. Twelve times smaller than the Crab in size and three hundred times smaller in the sky.
Nearby tiny toy case - A deliberately absurd case: 1 km/s for a tenth of a year is a shell about ten light-seconds across, a hundredth of a light-year away. It still subtends 13.8 arcseconds, fourteen times SN 1987A.