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Here is what we know of Earth:

  • Mass: 5.972 sextillion metric tons
  • Diameter: 7,917.5 miles
  • Density: 5.51 grams per cubic centimeter
  • Rotation: 24 hours
  • Revolution: 365 days
  • Core: 760 miles wide, 1,355 miles thick, 84% iron, 6% nickel
  • Mantle: 3,958 miles deep, 45% silicon, 41% magnesium, 8% iron, 3% aluminum, 2% calcium
  • Crust: 3-30 miles thick, 58% silicon, 16% aluminum, 8% iron, 7% calcium, 4% magnesium, 3% sodium, 2% potassium
  • Water: 71%
  • Median surface temperature: 58.62 degrees Fahrenheit
  • Current axial tilt: 23.5 degrees

In an alternate universe, Earth is all of that...

enter image description here

...except that the bright light in this photo isn't the moon, but a gas giant the size of Jupiter (86,881 miles wide, 1.33 grams per cubic centimeter and 1,898,000,000,000 trillion metric tons in mass). Which means that this alternate Earth isn't a planet, but a moon. It orbits the gas giant so far away that its size on the sky is the same as our moon is back home. Is this far enough for Moon-Earth to retain the features listed above, or would it instead have to contend to tidal locking and intense radiation?

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  • $\begingroup$ I hope I can answer this once I have time, but doing quick calculations, the moon is ~31 arc min of angular diameter with 2100 mile actual diameter. For Jupiter to have the same apparent size, it would put it at 10 million miles, that is by my quick calculation. Not sure if there would be very much tidal locking going on for the first few million years. Jupiters radiation belt is ~600k to 2m miles toward the sun and 600m miles behind. Not sure how much effect it would have at 10m miles. $\endgroup$ – Sonvar Jul 18 at 3:32
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    $\begingroup$ Aside from the mixture of miles, centimeters, grams and Fahrenheit, what is a sextillion? Could you turn it into scientific notation? $\endgroup$ – L.Dutch - Reinstate Monica Jul 18 at 5:12
  • $\begingroup$ @L.Dutch No, because that would require special tricks on the keyboard. $\endgroup$ – JohnWDailey Jul 18 at 11:06
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    $\begingroup$ 5E6 doesn't require any special trick on the keyboard $\endgroup$ – L.Dutch - Reinstate Monica Jul 18 at 11:09
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In order to have the similar apparent size in the sky, the Jupiter-like body has to be at a distance that satisfies the following relationship

angular size comparison

$\phi_{moon}\over D_{Moon}$$=$$\phi_{Jupiter-like}\over D_{Jupiter-like}$

which gives us

$D_{Jupiter-like}=$$\phi_{Jupiter-like} \cdot D_{Moon} \over \phi_{Moon}$$=142984\cdot400000\over3476$$=16\cdot10^6 km$.

This is about $1/10$ of an AU.

Jupiter magnetopause is located somewhere between $3\cdot 10^6$ and $7\cdot 10^6$ km from Jupiter. This would make the Earth safe from the radiation. However, to keep its hold on the planet, Jupiter could not be at 1 AU from the Sun. As Mike Scott pointed out

If Jupiter was 1 AU from the sun, its Hill Sphere would only be about 10 million kilometers, so it couldn’t hold on to a moon at that distance.

This in turn would affect life on the Earth like planet.

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  • $\begingroup$ If Jupiter was 1 AU from the sun, its Hill Sphere would only be about 10 million kilometres, so it couldn’t hold on to a moon at that distance. $\endgroup$ – Mike Scott Jul 18 at 6:09
  • $\begingroup$ @MikeScott, thanks, fixed $\endgroup$ – L.Dutch - Reinstate Monica Jul 18 at 6:16
  • $\begingroup$ 1/10 of an AU is 15 million km, not much more than the Hill Sphere. Make the sun a bit brighter and move the planet/moon a bit further out, and it would work. Say the sun is 1.2 solar masses. This would make its luminosity twice as great as the Sun, which means that the planet/moon would have to be 1.4 AU away to receive the same amount of light and heat. That would make the diameter of the Hill Spere about 20 million km, making the Earth orbit well within it. $\endgroup$ – Klaus Æ. Mogensen Jul 18 at 7:34
  • $\begingroup$ @Klaus Æ. Mogensen it is probably simpler to make Jupiter heavier to expand its Hill sphere. $\endgroup$ – Alexander Jul 18 at 17:39
  • $\begingroup$ @Alexander: Since the radius of a Hill Sphere is proportional to the cube root of the mass, to double the radius you would need to octuble the mass of the "Jupiter". This might well increase the radiation coming from it beyond safe limits, even at the greater distance. $\endgroup$ – Klaus Æ. Mogensen Jul 19 at 7:40
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SHORT ANSWER:

It is impossible to have a situation exactly as you describe. So it may be necessary for you to modify the mass, density, and diameter of the body that your Earth like moon orbits. You may need to get someone to do the calculations for you.

LONG ANSWER:

There have been a lot of questions on this site about hypothetical habitable Earth-sized exomoons orbiting gas giant planets that orbit in the habitable zones of their stars.

And when I answer those questions I often cite this article:

"Exomoon Habitability Constrained by Illumination and Tidal heating" by Rene Heller and Roy Barnes Astrobiology, January 2013.

https://www.ncbi.nlm.nih.gov/pmc/articles/PMC3549631/1

You may be pleased to know that they considered it to be possible that even a slowly rotating exomoon could have a magnetic field strong enough to deflect charged particles trapped in the giant planet's magnetic field, and so it is possible that a habitable exomoon would not need to be outside of the giant planet's radiation zone.

You didn't specify the length of your planet/moon's "month", which would be the length of time it took to orbit its planet. But there is apparently an upper limit to the length of an exomoon's orbit around its planet compared to the length of their planet's orbit around their star. IN secton 2 Habitability of Exomoons, Heller and Barns say:

The longest possible length of a satellite's day compatible with Hill stability has been shown to be about Pp/9, Pp being the planet's orbital period about the star (Kipping, 2009a)

Kipping D.M. Transit timing effects due to an exomoon. Mon Not R Astron Soc. 2009a;392:181–189. [Google Scholar]

So the longest possible orbital period of a moon orbiting a planet that has a year of 365.25 days would be about 0.1111 Earth years or about 40.5833 Earth days.

Around Jupiter, an orbit of 40.5833 Earth days would be beyond that of Callisto, 1,882,706 kilometers from Jupiter, at 16.689 days, and inside that of Themisto, 7,393,216 kilometers from Jupiter, at 129.87 days.

https://en.wikipedia.org/wiki/Moons_of_Jupiter2

Even if an orbit around Jupiter with a period of 40.5833 days was as large as that of Themisto, Jupiter, with an equatorial diameter of 142,984 kilometers, would appear to have an angular diameter of about 1.108 degrees, much larger than that of the moon.

Around Saturn, an orbit of 40.5833 Earth days would be beyond that of Hyperion, 1,481,010 kilometers from Saturn, at 21.27661 days, and inside that of Iapetus, 3,560,820 kilometers from Saturn, at 79.3215 days.

https://en.wikipedia.org/wiki/Moons_of_Saturn3

Even if an orbit around Saturn with a period of 40.5833 days was as large as that of Iapetus, Saturn, with an equatorial diameter of 120,536 kilometers, would appear to have an angular diameter of about 1.939 degrees, much larger than that of the moon. And that's not counting the rings.

Around Uranus, an orbit of 40.5833 Earth days would be beyond that of Oberon, 583,520 kilometers from Uranus, at 13.463 days, and inside that of Francisco, 4,276,000 kilometers from Uranus, at 256.66 days.

https://en.wikipedia.org/wiki/Moons_of_Uranus4

Even if an orbit around Uranus with a period of 40.5833 days was as large as that of Francisco, Uranus, with an equatorial diameter of 51,118 kilometers, would appear to have an angular diameter of about 0.686 degrees, a little larger than that of the moon.

Around Neptune, an orbit of 40.5833 Earth days would be beyond that of Triton, 354,759 kilometers from Neptune, at 5.877 days, and inside that of Nereid, 5,513,818 kilometers from Neptune, at 360.13 days.

https://en.wikipedia.org/wiki/Moons_of_Neptune5

Even if an orbit around Neptune with a period of 40.5833 days was as large as that of Nereid, Neptune, with an equatorial diameter of 49,528 kilometers, would appear to have an angular diameter of about 0.5156 degrees, about the same as the Moon, which has an angular diameter of 0.52 degrees as seen from Earth.

So the angular diameter of the giant planets as seen from the outermost stable orbit around them seems to decrease with increasing density of the planet.

Therefore it might be a good idea for you to replace the Jupiter sized planet with a sub Neptune or super Earth type planet, less massive and more dense than Neptune.

So you may need to calculate if it is possible to have an orbital period of 40.5833 days or less around a sub Neptune or super Earth type planet where the primary will appear to have the same angular diameter as the Moon as seen from Earth.

On the other hand, you may need to replace the Jupiter sized planet with an even more massive planet than Jupiter. It is believed that Jupiter has almost the largest diameter possible for a giant planet. If more matter is added to a Jupiter sized planet it will grow in size for a little while and then if even more matter is added it will start to shrink as the gravity of that matter compresses it more and more densely.

So you may need to calculate if it is possible to have an orbital period of 40.5833 days or less around a super Jupiter or a brown dwarf where the primary will appear to have the same angular diameter as the Moon as seen from Earth.

And if it is possible for your Earth like moon to orbit around either a sub Neptune/super Earth or a super Jupiter/brown dwarf with an orbital period of 40.5833 days or less and the primary appearing to have the same angular diameter as the Moon as seen from Earth, you will have to make other calculations. You will want to calculate whether your moon will be tidally locked to a primary of that mass at that distance, and calculate whether it will be in a stable orbit for billions of years at that distance from its primary and from its star.

Or more likely you may want to ask someone to do those calculations.

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