I'm just going to go ahead and steal this question question directly of Reddit since I have more trust in the answers I get on this site. So, if the moon was twice as big but also twice as far way, would there be any difference in the effects that the moon now has on earth, like tides and such? Also, would there be any difference in the moons orbital motion apart from the fact that the time it takes for the moon to rotate around earth would be longer? What will increase is the mass and diameter(by a factor of 2) of the moon. The density of the moon will stay the same.

  • $\begingroup$ What does "twice as big as it was far away" mean? Twice as big AND twice as far away? Something else? $\endgroup$ Jan 6 '14 at 3:54
  • $\begingroup$ @BrandonEnright 5 O'clock in the morning might not be the best time to ask questions :), but I fixed it. $\endgroup$
    – Reds
    Jan 6 '14 at 4:05
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    $\begingroup$ Presumably "big" means physical extent (diameter). What about mass? Should we assume the same density, or what. It matters a lot to the tidal effects and a little to the orbital period. $\endgroup$ Jan 6 '14 at 4:10
  • $\begingroup$ For the sake of argument, lets assume that moon were to be scaled up while the density stay the same, So it's the mass and diameter that will increase. The diameter will increase by a factor of two. $\endgroup$
    – Reds
    Jan 6 '14 at 4:15
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    $\begingroup$ Could you edit these clarifications into the question? In particular, if "twice as big" means the diameter increases by a factor of two, that is extremely important to mention. $\endgroup$
    – David Z
    Jan 6 '14 at 4:30

Increasing the diameter and distance of the Moon by a factor 2 would lead to a number of very subtle differences. I will list the ones I came up with:

Apparent size

If $R_m$ is the radius of the Moon and $D_m$ its geocentric distance (that is, the distance between the centre of the Moon and the centre of the Earth), then its geocentric angular diameter is $$ \delta = 2\arcsin\left(\frac{R_m}{D_m}\right). $$ Obviously, changing both $R_m$ and $D_m$ by a factor 2 wouldn't change the geocentric angular size. However, we're not observing the Moon from the centre of the Earth, but from a particular location on the surface on the Earth: when the Moon is above the horizon, we're slightly closer to it than the geocentric distance. In particular, if the Moon is directly overhead, it extends an angle $$ \delta' = 2\arcsin\left(\frac{R_m}{D_m-R_e}\right), $$ where $R_e$ is the radius of the Earth. So if we change $R_m$ and $D_m$ by a factor 2, we'd measure an angle $$ \delta'' = 2\arcsin\left(\frac{2R_m}{2D_m-R_e}\right) < \delta', $$ in other words, the Moon would actually appear slightly smaller from a particular location.

Diurnal parallax

The diurnal parallax is the apparent change in position of a foreground celestial object with respect to distant stars, as viewed from two different locations on Earth (or the same location at different moments). A special case is the lunar equatorial parallax: $$ P_m = \arcsin\left(\frac{R_e}{D_m}\right). $$ If the distance of the Moon increases by a factor 2, its parallax decreases accordingly.

Orbital motion

Of course, the most noticeable effect will be the change in the lunar orbital period. From Kepler's Third Law, $$ T^2 = \frac{4\pi^2 a^3}{G(M_e +M_m)}, $$ with $a=384\,748\;\text{km}$ the semi-major axis. Thus $$ T'^2 = \frac{4\pi^2 (2a)^3}{G(M_e + 8M_m)}, $$ which gives a new sidereal period $T' = 74.2$ days. This is the orbital period with respect to the stars, but if we want to know the period between two Full Moons, we need to know the lunar period with respect to the Sun, the so-called synodic period $S'$. From $$ \frac{2\pi}{S'} = \frac{2\pi}{T'} - \frac{2\pi}{J}, $$ with $J=365.256$ days the sidereal year, we find $S'=93.1$ days.

Tidal effects

Tides due to the Moon are caused by a difference in gravitational acceleration between the Earth's centre and the surface: $$ \Delta a = -\frac{GM_m}{D_m^2} + \frac{GM_m}{(D_m-R_e)^2}\approx 2R_e\frac{GM_m}{D_m^3} - \ldots $$ So in first order, the average tidal forces remain the same. But the Sun also causes a tidal force, and the amplitude of the total tide depends on the relative position of the Sun and Moon, from spring tides (at Full Moon and New Moon) to neap tides (at First and Last Quarter).

In other words, the strength of the tides depends on the synodic lunar period $S'$, and also on the position of the Moon above or below the ecliptic (the tides are strongest during an eclipse, i.e. when Sun, Moon, and Earth are aligned).

The oscillations in tidal strength have effects on the Earth's rotation: small fluctuations in the daily rotation and periodic changes in the axial tilt (nutations). So the periods of these effects would change. But non-periodic long-term effects, like lunisolar precession or tidal acceleration, would remain the same.


enter image description here

When a solar eclipse occurs, a total eclipse can be seen in locations inside the umbra. If the size and distance of the Moon increase by a factor 2, the umbra remains (almost) the same (actually, it decreases slightly, because I said earlier that the angular size of the Moon as seen from a location on Earth does decrease slightly, but this effect is very small).

But the penumbra would increase by about a factor 2. This means that the fraction on the Earth's surface where you can see a partial eclipse increases. It also means that the ratio of partial vs total solar eclipses increases. And since the orbital velocity of the Moon is also slower, the average duration of an eclipse is also longer.

However, the chances of solar eclipses would become much rarer, because an eclipse can only occur when both Sun and the Moon are sufficiently close to a lunar node. The Sun, Earth and nodes are aligned twice a year, so twice a year there is a window of opportunity (about two months) for an eclipse, if the Moon is also near a node. But since the orbital period of the Moon is now longer than two months, it could miss this window of opportunity entirely.

Lunar eclipses would become even rarer, not just for the same reason above, but also because a lunar eclipse occurs when the Moon passes inside the Earth's umbra.

enter image description here

But if the Moon's distance increases by a factor 2, the apparent size of the umbra decreases at that distance. So the chance that the Moon passes inside the umbra also decreases.

Earth-Moon barycentre

If $D$ is the distance between the Earth's centre and the Earth-Moon barycentre, then $$ D = \frac{D_m M_m}{M_e+M_m}, $$ so the new distance becomes $$ D' = \frac{16D_m M_m}{M_e+ 8M_m}, $$ which is about 69000 km, well outside the Earth's radius. We wouldn't directly notice the motion of the Earth around this barycentre, but I think that in principle it would be measurable as a small periodic Doppler shift in the radio signals from spacecraft in the solar system.

Lagrangian Points

Changes in the lunar mass and distance would affect the positions of the Lagrangian points.

  • $\begingroup$ I guess you covered all of it, except I have one question. What in your opinion will happen to the tidal locking of the moon? Will it be disrupted, so that we may be able to see the dark side of the moon from Earth. I understand that this one takes a lot of parameters to ascertain, so a general opinion will be helpful enough. $\endgroup$
    – Cheeku
    Jan 7 '14 at 2:21
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    $\begingroup$ @Cheeku I suspect that a bigger moon would still be tidally locked, but the process would've taken longer. So the transfer of energy and angular momentum between the Earth and the Moon in the past would've been different. $\endgroup$
    – Pulsar
    Jan 7 '14 at 9:15

If the diameter would be increased by a factor of 2, then if the density stays the same, the mass would increase by a factor of $2^3 = 8$, since volume is proportional to the cube of the diameter.

The main effects of doing this are:

  • the Moon's pull on the Earth would be double of what it is now, since the gravitational force is proportional to the inverse of the square of the distance $(F_1\sim M/r^2~~,~~F_2\sim8M/(2r)^2 \rightarrow F_2=2F_1$
  • The tidal forces would however be roughly the same, since the tidal force is proportional to the cube of the distance (see the wiki for details)
  • the Moon's orbital period would be roughly $\sqrt{8}\approx2.8$ longer, since that is proportional to the distance raised to the power 3/2 (see here; note that I'm ignoring smaller effects due to the mass increase here for simplicity. Doing the more complete calculation gives a factor of 2.715).

So, the Moon would look exactly the same, total solar eclipses would still occur, there would not be any significant effects on the tides, etc. However, the cycles of Lunar phases would be slowed down by a factor of $\sim\sqrt{8}$.

The only difference when choosing the density such that also the mass would just double (thus not increase by a factor of 8, but by a factor of 2), is that the tides would be roughly four times as weak as they are now. Obviously, this would have major effects on the biosphere. Another important effect would be that the amount of tidal heating caused by the Moon would also be four times less, thus decreasing the Earth's internal temperature and decreasing the amount of volcanism and tectonic activity. I can't find any numbers on that, but I suspect it could have severe long-term consequences.

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    $\begingroup$ Tides would not be twice as strong. Tidal forces are proportional to the cube of the distance (in first order). $\endgroup$
    – Pulsar
    Jan 6 '14 at 9:46
  • $\begingroup$ @Pulsar: you are very right, I forgot the details. Editing... $\endgroup$ Jan 6 '14 at 10:21

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