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For questions sake assume that this moon is a perfect recreation of Earth on January 1st 1660 and that the planet looks identical to our moon on the same day.

The Planet would have a mass of 81.3 earths and would have identical orbital charteristics(orbital period, inclination, eccentricity, etc) to Earth while orbiting a Sun identical to our own. This Planets rotational period and axial tilt would be equal to our own moon. The size of this planet would be adjusted so that it looks the same size from the moon as the moon does from the surface of the Earth, density of the planet would be adjusted accordingly to preserve mass.

The moon would have a mass of 1 Earth and would have identical orbital charcteristics to our moon with the exception of orbital distance which would be adjusted to make this moon orbit with the same orbital period as our own moon. The rotational period and axial tilt of this moon would be the same as earth.

How would someone armed with all the resources of the time and records of the past manage to deduce that something had changed. Would they be able to do so? If not, what date would a theoretical earth be able to deduce it? Assume that tidal forces are not a factor.

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    $\begingroup$ If you point a laser at real-Moon from real-Earth, it takes 2.6 seconds for light to make the round-trip. From orbiting-Earth to big-planet it would take over 11 seconds. But we didn't have lasers in 1660, so I don't think they could have noticed the increased distance that way. However, see Who first measured the distance to the moon and how was it done? and also Want to measure the distance to the Moon yourself? Now you can! $\endgroup$
    – Stef
    Mar 28, 2023 at 9:40
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    $\begingroup$ Waaaaaait wait wait wait wait. Big-planet's mass = 81.3 times Earth's mass. And Earth's mass = 81.3 times Moon's mass. And you want the apparent diameter of big-planet from Earth to be equal to the apparent diameter of real-Moon from real-Earth. For that you need the distance between big-planet and orbiting-Earth to be cubicroot(81.3 squared) times the distance between real-Earth and real-Moon. This is inconsistent with the fact that the orbiting period of orbiting-Earth should be the same as that of real-Moon. So you need to sacrifice either the apparent diameter or the orbiting period. $\endgroup$
    – Stef
    Mar 28, 2023 at 13:04
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    $\begingroup$ The usual Moon shows always the same face, to the naked eye. What happens to that? $\endgroup$
    – Pablo H
    Mar 28, 2023 at 13:56
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    $\begingroup$ 81.3 is highly specific. Why that number? $\endgroup$
    – RonJohn
    Mar 28, 2023 at 18:38
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    $\begingroup$ @RonJohn it's the ratio of the Earth's mass to the moon's. $\endgroup$
    – phoog
    Mar 28, 2023 at 21:32

3 Answers 3

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There is no need for any kind of technology; because . . .

The path of the sun in the sky would be noticeably different

The primary's mass is about 80 times as large as Earth's mass. Since the satellite must have the same orbital period as our Moon, the radius of the orbit must be about 10 times as large as the distance between Earth and our Moon, or about 1/40 a.u.

The orbit of the moon is inclined by about 5° with respect to the plane of the ecliptic, which means that the sun would wobble up and down by about half a degree over the course of a month.

In 1659 the sun did not do that. In 1660 is started doing it. By February 1660 everybody would have noticed that the sun is doing a very strange dance, and all their sundials stopped working.

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  • $\begingroup$ Not to mention our days will start to be 28 days long. $\endgroup$
    – workerjoe
    Mar 28, 2023 at 14:07
  • $\begingroup$ @workerjoe: "The rotational period and axial tilt of this moon would be the same as earth." $\endgroup$
    – AlexP
    Mar 28, 2023 at 14:13
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    $\begingroup$ may be worth noting that half a degree is similar to the angular width of the sun, moon, and this planet for anyone reader who, like me, doesn't have much intuition on how big angular widths actually are $\endgroup$
    – Tristan
    Mar 28, 2023 at 15:40
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There is no need for any kind of technology, because...

The big planet will appear 4.33 times bigger than the real Moon

The apparent diameter of the planet from the orbiting fake Earth will be much larger than the apparent diameter of the Moon from the real Earth.

The apparent diameter can be calculated if we know the diameter of the big planet and the distance between the big planet and the fake Earth.

The diameter of the big planet is easy to figure out. You specified that the big planet was 81.3 times more massive than the Earth; and the Earth is 81.3 times more massive than the Moon; hence the big planet is $81.3^2$ times more massive than the Moon; hence its diameter is $\sqrt[3]{81.3^2}$ times the diameter of the Moon.

The distance between the big planet and the orbiting Earth can be deduced from the orbiting period. You gave the mass of all bodies concerned, and you specified that the orbiting period of fake Earth around the planet should be equal to the orbiting period of real Moon around real Earth. From this information, we can deduce the exact distance between orbiting fake Earth and the planet.

Indeed, when a satellite orbits around a planet, the following equation must be satisfied by the mass $m$ of the planet, the distance $d$ between the two bodies, and the orbiting period $T$: $$d^3 = \frac{G m T^2}{4 \pi ^2}$$

Since you imposed $m_{\text{big planet}} = 81.3 \times m_{\text{Earth}}$, we can conclude: $$d_{(\text{big planet},\text{ Earth})} = \sqrt[3]{81.3} \times d_{(\text{real Earth},\text{ real Moon})}$$

So, the diameter of the planet will be $\sqrt[3]{81.3^2}$ times bigger than the diameter of the Moon; but the distance of the planet from the orbiting Earth will be only $\sqrt[3]{81.3}$ times bigger than the distance of the real Moon from the real Earth.

The ratio of these two factors, $\sqrt[3]{81.3} = 4.33$, will be exactly the ratio between the apparent diameter of the big planet seen from the orbiting Earth, and the apparent diameter of the real Moon seen from the real Earth.

If you want formal calculations to support that claim, here they are:

Now the apparent diameter $\alpha_{\text{Moon}}$ of the real Moon seen from the real Earth satisfies: $$\tan ({\frac{1}{2}\alpha_{\text{real Moon}}}) = \frac{1}{2}\frac{\text{diam}_{\text{real Moon}}}{d_{(\text{real Earth},\text{ real Moon})}}$$ And the apparent diameter $\alpha_{\text{big planet}}$ of the big planet seen from the orbiting fake Earth satisfies: \begin{align*} \tan ({\frac{1}{2}\alpha_{\text{big planet}}}) & = \frac{1}{2}\frac{\text{diam}_{\text{big planet}}}{d_{(\text{big planet},\text{ Earth})}} \\ & = \frac{1}{2} \frac{\sqrt[3]{81.3^2}}{\sqrt[3]{81.3}} \frac{\text{diam}_{\text{real Moon}}}{d_{(\text{real Earth},\text{ real Moon})}} \\ \tan ({\frac{1}{2}\alpha_{\text{big planet}}}) & = \sqrt[3]{81.3} \times \tan ({\frac{1}{2}\alpha_{\text{real Moon}}}) \end{align*}

We can take the arctangent to find $\alpha_{\text{big planet}}$. But actually, the $\alpha$ are so small here that we can use the approximation $\tan x \approx x$ and deduce directly: \begin{align*}\alpha_{\text{big planet}} & = \sqrt[3]{81.3} \times \alpha_{\text{real Moon}} \\ & = 4.33 \times \alpha_{\text{real Moon}} \end{align*}

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    $\begingroup$ this assumes the planet's density is the same as the moon's, when the question states the exact opposite: "The size of this planet would be adjusted so that it looks the same size from the moon as the moon does from the surface of the Earth, density of the planet would be adjusted accordingly to preserve mass." $\endgroup$
    – Tristan
    Mar 28, 2023 at 15:42
  • $\begingroup$ @Tristan Oh. So the planet is basically a big ball of air, then. $\endgroup$
    – Stef
    Mar 28, 2023 at 15:45
  • $\begingroup$ if the planet appears larger than the moon when it has the same density as the moon, if it's compressed to appear the same size as the moon it must be denser than the moon, not less dense $\endgroup$
    – Tristan
    Mar 28, 2023 at 15:46
  • $\begingroup$ @Tristan Sorry, I don't follow your last comment. $\endgroup$
    – Stef
    Mar 28, 2023 at 15:49
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    $\begingroup$ your calculation assumes the planet is as dense as the moon, and says that it will appear larger than the real moon. We want the planet to appear the same size as the real moon, but at the distance and mass you used in your calculation. To appear smaller at the same distance it must be smaller. If it is smaller and the same mass, it must be more dense, not less, so the planet is not a big a ball of air, but a big ball of lead (or rather something several times denser than moon rock) $\endgroup$
    – Tristan
    Mar 28, 2023 at 15:51
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Someone will notice on the first night.

Unless special care is taken in the transportation to replicate the night sky of Earth (or in destroying the original Earth and placing the new system in its place) , the expected visible planets will be identified as being in the wrong positions, or missing entirely, or replaced with new ones.

If the replica Earth was set up in another solar system, the constellations will be different.

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