Which will be brighter when it supernovas - Eta Carinae or Betelgeuse?

  • Context: Undergrad 
  • Thread starter Thread starter swampwiz
  • Start date Start date
Click For Summary
SUMMARY

Betelgeuse will be significantly brighter than Eta Carinae when both stars undergo supernova events. Betelgeuse is over 10 times closer to Earth than Eta Carinae, which means it will appear more than 100 times brighter due to the inverse square law. Although Eta Carinae is potentially 10 times more massive, its intrinsic brightness cannot compensate for the distance disadvantage. The discussion highlights the complexities of supernova types, including the potential for Betelgeuse to produce a superluminous supernova (SLSN) or hypernova due to its mass and circumstellar material.

PREREQUISITES
  • Understanding of the inverse square law in astronomy
  • Familiarity with stellar evolution and supernova classifications
  • Knowledge of Wolf–Rayet stars and their characteristics
  • Basic concepts of circumstellar material and its impact on supernova brightness
NEXT STEPS
  • Research the properties and lifecycle of Wolf–Rayet stars
  • Explore the mechanisms behind superluminous supernovae (SLSN)
  • Study the differences between type Ib and Ic supernovae
  • Investigate the role of metallicity in stellar evolution and supernova outcomes
USEFUL FOR

Astronomers, astrophysics students, and anyone interested in stellar phenomena and supernova mechanics will benefit from this discussion.

swampwiz
Messages
567
Reaction score
83
It seems that Betelgeuse is a lot closer, but even with its gargantuan size, it's still a lot smaller than Eta Carinae.
 
Astronomy news on Phys.org
swampwiz said:
It seems that Betelgeuse is a lot closer, but even with its gargantuan size, it's still a lot smaller than Eta Carinae.
I'm pretty sure Betelgeuse will be a lot brighter. It is more than 10 times closer than Eta Carina. Because of the inverse square law, if they were the same intrinsic brightness, Betelgeuse would be more than 100 times brighter. As you say, Eta Carina is perhaps as much as 10 times more massive, but even if it is intrinsically 10 times brighter, it would still be 1/10 as bright as Betelgeuse.
 
Here is what Wikipedia says:

As a single star, a star originally around 150 times as massive as the Sun would typically reach core collapse as a Wolf–Rayet star within 3 million years. At low metallicity, many massive stars will collapse directly to a black hole with no visible explosion or a sub-luminous supernova, and a small fraction will produce a pair-instability supernova, but at solar metallicity and above there is expected to be sufficient mass loss before collapse to allow a visible supernova of type Ib or Ic. If there is still a large amount of expelled material close to the star, the shock formed by the supernova explosion impacting the circumstellar material can efficiently convert kinetic energy to radiation, resulting in a superluminous supernova (SLSN) or hypernova, several times more luminous than a typical core collapse supernova and much longer-lasting. Highly massive progenitors may also eject sufficient nickel to cause a SLSN simply from the radioactive decay. The resulting remnant would be a black hole since it is highly unlikely such a massive star could ever lose sufficient mass for its core not to exceed the limit for a neutron star.

So we don't know.

If you read Wikipedia first, why do you think our answer would be different?
If you didn't read Wikipedia first, why not?
 
  • Like
Likes   Reactions: Dragrath

Similar threads

  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 35 ·
2
Replies
35
Views
34K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 8 ·
Replies
8
Views
16K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 109 ·
4
Replies
109
Views
7K
  • · Replies 45 ·
2
Replies
45
Views
10K
  • · Replies 2 ·
Replies
2
Views
2K