### What would the effects be on Earth if Jupiter was turned into a star?

• In Clarke's book 2010, the monolith and its brethren turned Jupiter into the small star nicknamed Lucifer. Ignoring the reality that we won't have any magical monoliths appearing in our future, what would the effects be on Earth if Jupiter was turned into a star?

At it's closest and furthest:

How bright would the "back-side" of the earth be with light from Lucifer?

How much heat would the small star generate on earth?

How many days or months would we actually have night when we circled away behind the sun?

How much brighter would the sun-side of earth be when Lucifer and the sun both shine on the same side of the planet?

While this is an interesting question, I don't know if there's a proper way to answer it. Jupiter's mass is far less that that of the smallest brown dwarfs, also dubbed "failed stars". Brown dwarfs don't have enough mass to sustain hydrogen fusion, and don't emit a whole lot of light. I don't think that there's any way that you could realistically do the calculations for a Jupiter-star scenario, because of the impossibility of it beginning hydrogen fusion. Still, it's an interesting idea.

Okay, I relent. +1 for an interesting idea.

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Jupiter can burn as brightly as you want it to depending on how much mass you add to it. If you somehow put a very massive core at the center of Jupiter, the total mass of the system would determine how much fusion can take place. It can probably range from a supernova if you put in a neutron star just below the Chandrasekhar limit inside to a very weak red dwarf if you just add enough mass to make fusion start.

How do you know we won't have any "magical" monoliths appearing in the future? It's as good of a scenario of first contact as any.

• HDE 226868 Correct answer

8 years ago

Before I start, I'll admit that I've criticized the question based on its improbability; however, I've been persuaded otherwise. I'm going to try to do the calculations based on completely different formulas than I think have been used; I hope you'll stay with me as I work it out.

Let's imagine that Lucifer becomes a main-sequence star - in fact, let's call it a low-mass red dwarf. Main-sequence stars follow the mass-luminosity relation:

$$\frac{L}{L_\odot} = \left(\frac{M}{M_\odot}\right)^a$$

Where $L$ and $M$ are the star's luminosity and mass, and $L_\odot$ and $M_\odot$ and the luminosity and mass of the Sun. For stars with $M < 0.43M_\odot$, $a$ takes the value of 2.3. Now we can plug in Jupiter's mass ($1.8986 \times 10 ^{27}$ kg) into the formula, as well as the Sun's mass ($1.98855 \times 10 ^ {30}$ kg) and luminosity ($3.846 \times 10 ^ {26}$ watts), and we get

$$\frac{L}{3.846 \times 10 ^ {26}} = \left(\frac{1.8986 \times 10 ^ {27}}{1.98855 \times 10 ^ {30}}\right)^{2.3}$$

This becomes $$L = \left(\frac{1.8986 \times 10 ^ {27}}{1.98855 \times 10 ^ {30}}\right)^{2.3} \times 3.846 \times 10 ^ {26}$$

which then becomes

$$L = 4.35 \times 10 ^ {19}$$ watts.

Now we can work out the apparent brightness of Lucifer, as seen from Earth. For that, we need the formula

$$m = m_\odot - 2.5 \log \left(\frac {L}{L_\odot}\left(\frac {d_\odot}{d}\right) ^ 2\right)$$

where $m$ is the apparent magnitude of the star, $m_\odot$ is the apparent magnitude of the Sun, $d_\odot$ is the distance to the Sun, and $d$ is the distance to the star. Now, $m = -26.73$ and $d(s)$ is 1 (in astronomical units). $d$ varies. Jupiter is about 5.2 AU from the Sun, so at its closest distance to Earth, it would be ~4.2 AU away. We plug these numbers into the formula, and find

$$m = -6.25$$

which is a lot less brighter than the Sun. Now, when Jupiter is farthest away from the Sun, it is ~6.2 AU away. We plug that into the formula, and find

$$m = -5.40$$

which is dimmer still - although, of course, Jupiter would be completely blocked by the Sun. Still, for finding the apparent magnitude of Jupiter at some distance from Earth, we can change the above formula to

$$m = -26.73 - 2.5 \log \left(\frac {4.35 \times 10 ^ {19}}{3.846 \times 10 6 {26}}\left(\frac {1}{d}\right) ^ 2\right)$$

By comparison, the Moon can have an average apparent magnitude of -12.74 at full moon - much brighter than Lucifer. The apparent magnitude of both bodies can, of course, change - Jupiter by transits of its moon, for example - but these are the optimal values.

While the above calculations really don't answer most parts of your question, I hope it helps a bit. And please, correct me if I made a mistake somewhere. LaTeX is by no means my native language, and I could have gotten something wrong.

I hope this helps.

Edit

The combined brightness of Lucifer and the Sun would depend on the angle of the Sun's rays and Lucifer's rays. Remember how we have different seasons because of the tilt of the Earth's axis? Well, the added heat would have to do with the tilt of Earth's and Lucifer's axes relative to one another. I can't give you a numerical result, but I can add that I hope it wouldn't be too much hotter than it is now, as I'm writing this!

Second Edit

Like I said in a comment somewhere on this page, the mass-luminosity relation really only works for main-sequence stars. If Lucifer was not on the main sequence. . . Well, then none of my calculations would be right.

It's an interesting answer! It sounds as thought there would be very little effect in regards to extra light or temperature.

In answer to the edit you made to your comment: Yep. Not a big difference. At least, not on Earth. An interesting follow-up would be to see if it could indeed cause conditions on Europa to change in favor of life.

@HDE 226868 Just for fun did you think anymore about what it would take to make Europa habitable for the aliens (I know, it depends on the alien). Jupiter couldn't get "too hot" obviously. I love A. C. Clarke, but he did need to ignore science for the sake of the story sometimes (ie. humans wouldn't survive in Jupiter's orbit due to the magnetic field).