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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?

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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. –  HDE 226868 Aug 6 '14 at 17:06
Okay, I relent. +1 for an interesting idea. –  HDE 226868 Aug 11 '14 at 18:15

4 Answers 4

up vote 2 down vote accepted

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:

$$\frac{L}{L(s)} = \left(\frac{M}{M(s)}\right)^a$$

Where $L$ and $M$ are the star's luminosity and mass, and $L(s)$ and $M(s)$ and the luminosity and mass of the Sun. For stars with $M < 0.43M(s)$, $a$ 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

$$m = m(s) - 2.5 log \left(\frac {L}{L(s)}\left(\frac {d(s)}{d}\right) ^ 2\right)$$

where $m$ is the apparent magnitude of the star, $m(s)$ is the apparent magnitude of the Sun, $d(s)$ 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.


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.

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It's an interesting answer! It sounds as thought there would be very little effect in regards to extra light or temperature. –  Maelish Aug 11 '14 at 17:55
Thanks. Just made an edit. –  HDE 226868 Aug 11 '14 at 17:56
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 Aug 11 '14 at 20:42
  1. Sun-Earth distance: 1AU
  2. Earth-Jupiter distance (at the conjunction): 4AU

So Lucifer will be four times farther than Sun when it is nearer (six times when it is farthest), and at the same time it is a thousand times smaller. This is approx 40 times more light than full moon concentrated in a tiny point on sky.

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I'm not sure this answers the question. And @Envite, how does your link prove anything? –  HDE 226868 Aug 6 '14 at 19:31
@HDE226868 The link is the reference for the mass relation between Sun and Jupiter –  Envite Aug 7 '14 at 7:42
Right, @Envite, but mass and size aren't necessarily correlated. And Jupiter still doesn't have anywhere near enough mass to begin fusion. –  HDE 226868 Aug 7 '14 at 13:50
Look, I feel that the whole exercise is futile. If Jupiter turned into a star - even a red dwarf - we would have a lot of problems with gravity. The solar system would become unstable, and there's a chance that some of the planets would be flung out of the solar system. We can't calculate the energy output because we can only guess at what type of star Jupiter would become, and we can't come up with any definite answer. There are dozens of possibilities; not a single one of them has any more merit than any of the others. Does the book specify? –  HDE 226868 Aug 7 '14 at 13:53
@HDE226868 Aboslutely False. We will not have any problems with gravity if Jupiter "magically" (as expressed by the OP) becomes a star with its own mass. –  Envite Aug 7 '14 at 14:34

Ignoring the impossibility of Jupiter going solar:

Assume that Jupiter turns into duplicate of the Sun in terms of energy output. Energy transmitted to the earth follows an inverse-square law. Since Jupiter is, at best, 4 times farther from the Earth than the Sun, Jupiter will supply the Earth with, at most, 1/16 the energy that the Sun supplies, for an increase of a bit more than 6%, at the most.

By comparison, between aphelion and perihelion, the Sun-Earth distance increases from around 147 million kilometers to around 152 million kilometers. This implies a seasonal energy input change of about 7%, that we experience now every year...

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And I'm pretty sure Lucifer's energy output was far less than the Sun's, so the increase would be even smaller. –  Keith Thompson Aug 7 '14 at 20:41

In reality, Jupiter doesn't have nearly enough mass to initiate stellar ignition or sustain it if we could somehow start it going.

Even the smallest star would require on the order of some 80 to 90 times the mass of Jupiter just to put out a faint red glow.

Even to become a brown dwarf proto-star, Jupiter would require a mass increase on the order of at least 10-fold or so.

Lucifer is simply not possible unless Jupiter collides with something to provide the extra mass it needs to go stellar and even then, it would be a red dwarf at best and quite faint, like a red-hot nail glowing in the dark.

But one can dream.


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Just a correction to make: a brown dwarf isn't a proto-star; it's a "failed star" - that is, it began as a proto-star but simply didn't have the mass necessary to enter the main sequence. I hate to be a nit-picker, though. +1 for a good, logical explanation. –  HDE 226868 Aug 10 '14 at 16:08

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