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

Question

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?

Answer 1

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.

Answer 2

I think it's a fun question, if impossible. The only way to turn Jupiter into a star that's even remotely practical is to add to it's mass. Ignoring brown dwarfs that are very limited in energy output, to get a red dwarf going, you'd need to add at least 75-80 or so Jupiter masses (a bit more than 24,000 earth masses). You'd want to add a fair percentage of hydrogen, but some rocky debris wouldn't hurt the mix.

Anyway, assuming the impossible is done, there are several things to consider. The greater gravity (75-80 times) would significantly alter all the planets' orbits. Predicting exactly how is hard, but that much more mass, and the planets' orbits - certainly all the inner ones - would wobble a lot more, and some might get pulled completely out of their orbit, likely thrown out of the solar system.

You might think that the planets nearer to Jupiter would be the most affected, but it really has more to do with tidal synch than anything else. Any of the 4 inner planets could get tugged into a new orbit. You'd also likely see the earth's orbit elongate in resonance with Jupiter, perhaps increasing the ice age/ice melt cycle. Precise answers are hard, and none of these things would happen over 1 orbit, but over time, certainly. Orbital changes to all the inner planets and perhaps Saturn as well would be inevitable if Jupiter becomes a red dwarf. Imagine if Saturn was pulled closer to the earth, into an orbit between Mars and Jupiter, or Mercury was pulled out past the earth. Odds are it wouldn't hit us, but we might want to keep an eye on it.

http://en.wikipedia.org/wiki/Stability_of_the_Solar_System#Mercury.E2.80.93Jupiter_1:1_resonance

The second thing to consider is magnetism and solar flares. Young stars tend to spin very fast due to conservation of angular momentum when the stars form, and this creates enormous magnetic fields and huge solar flares, much bigger than we get from the sun. It's strange to think that a tiny red dwarf, 4 times as far from our sun as the sun would create solar flares to worry about, but it's possible. Whether it would need high angular momentum for this to happen, I'm not sure, but we could see larger solar flares from star-Jupiter than from the sun.

http://en.wikipedia.org/wiki/Flare_star

Brightness, heat, and visibility were covered above, but I'll touch on that again. Brightness of -6.25 would be 5-6 times brighter than Venus and you'd see it at night, Venus isn't seen in peak darkness, so it would be significantly brighter than any other star/planet in the sky, but significantly less bright than the moon, like, you couldn't make your way with just that star's light the way you can see things around you in moonlight. But when I run the numbers, I think it would be quite a bit brighter than that.

Mass to luminosity is to the power of 3.5 - quick estimate, so, let's say the red dwarf has a mass of 80 Jupiters. That's 0.076 Suns. 0.076^3.5 = about 1/8,000, so 4.2 times as far away at closest point (square of that), 1/8000th as bright, we're looking at 1/140,000 times the light we get from the sun - not very much and likely less than that in its early stages and because the smaller stars tend to fall off, so let's estimate 1/200,000 - 1/300,000 the apparent brightness of the sun as a ballpark estimate. That's not enough to heat the earth at all, but that's still brighter, (a little bit) than the full moon which is about 1/400,000 the brightness of the sun. It would be enough light to see your way around, but I wouldn't want to try to read by it. It would also be distinctly reddish light. Not the white light we're used to getting from the day or night sky.

Finally size - a red-dwarf star of 80 Jupiter masses would actually be slightly smaller than Jupiter due to the gravitation, so it would appear like a planet - not quite a point in the sky, but almost a point, but a bit brighter than the full moon and red. That's likely bright enough to see during the day too. I don't think it would be hard to look at or hurt your eyes, but it would shine like a tiny bright red flashlight in the distance.

http://www.space.com/21420-smallest-star-size-red-dwarf.html

I don't think I like star-Jupiter.Let's not plan on doing this. :-)

Answer 3

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...

Answer 4

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.

:)

Answer 5

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.