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Back home, Earth's moon is 2159.2 miles wide and orbits 238,900 miles from its parent.

But let's pretend that the moon is 2500 miles wide and orbits 200,000 miles from Earth. Would the nightscape look any different? How much would tides and axial tilt be affected?

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  • $\begingroup$ The perigean tides will become significantly higher than usual, increase in brightness wise is difficult to judge due to lack of reference however it will definitely appears bigger in the night sky and to affect Earth's axial tilt I reckon billions or perhaps trillions times the force of differential gravitational effect such as impact from colossal planetoid would be needed. I don't know Newton law of universal gravitation at all so no working. 🌜🌍 $\endgroup$ – user6760 Sep 15 '15 at 2:51
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To simplify, lets calculate some of the basic ratios up front

Avg. orbital distance ratio (center to center) = 238,900 / 200,000 = 1.1945
Distance squared ratio = 1.1945^2 = 1.43683
Distance cubed ratio = 1.704349

Diameter ratio = 2500 / 2159.2 = 1.157836
Area ratio = diameter ratio squared = 1.340585
Volume ratio = diameter ration cubed = 1.552178

Assume mass ratio equals volume ratio, i.e. moon density is unchanged - this would not be true given identical materials as the matter would be more compressed with the additional pressure. So the mass ratio would be larger than the volume ratio. We don't know enough about the moon to calculate this accurately, not that I could model this change accurately without a lot of work.

Moonlight is brighter because it has 134% of its current surface area to reflect sunlight (and Earth-shine). But it is also closer and thus 143% of its intensity. Total moonlight received would be 1.912787 times the current moonlight.

Eyesight response to light is non-linear, so it won't appear 91% brighter. Due to variations in orbits, the full moon varies from about mean value of -12.74 magnitude to -12.9 at it brightest. Under this scenario mean brightness would be around -13.5 magnitude. Reading by moonlight would be easier, etc. More stars would be obscured during a full moon.

Tides are larger - roughly speaking, tides are proportional to mass / distance^3 so the lunar tides are 2.64552 times as large as current tides. (Solar tides are about 45% those of the moon currently). Actual local tides vary quite a bit, but in general, they would be over twice as strong. This would affect ocean life as well as seaports, etc.

Axial tilt is complicated - I have seen frequently quoted that without the moon the axial tilt would vary up to 85 degrees, but I've also read that the axial tilt would vary by no more than 10 degrees based on newer calculations that include the effects of the other planets. My guess is that the axial tilt would be a bit more tightly regulated than today -- lots of non-linear factors. For example, if Jupiter were a lot closer, the moon would actually be a destabilizing influence on axial tilt.

Earth's rotational slowdown would also be proportional to the tidal forces, i.e., the earth's current daily rotation slows down by around 20 millionths of a second per year. With this scenario, that change to about 50 millionths of a second per day per year -- more frequent leap seconds.

Moon orbital speed increases. For circular orbits, orbital velocity is proportional to 1/radius, so the moon will speed up from 0.635 miles/sec to 0.759 miles per second. Since the orbital path is also 19.45 percent shorter, the orbital time is 42.68% shorter -- i.e., sidereal month (360-degree revolution) changes from 27.3 days to 19.1 days. A synodic month (new moon to new moon) changes from 29.5 days to 20.15 days

Solar eclipses are more frequent and longer in duration. The Moon orbits the Earth more frequently - thus more opportunities for passing between us and the Sun. Many moon passes do not have the moon shadow crossing the earth because the moon's orbit is inclined to the earth orbit around the Sun so the shadow passes under or over the earth. With the smaller orbit, there will be fewer misses one a percentage basis. The Sun is about 400 times the distance of the moon and about 400 times the diameter of the moon, so eclipses are almost perfectly matched so we sometimes see total eclipses and sometimes see annular eclipses -- not anymore, there will be no annular eclipses. Currently, under ideal conditions, the maximum full umbra of the moon is about 166 miles wide on the earth allowing for a little over 7 minutes of maximum full totality. Full eclipses over 7 minutes are very rare, everything has to be lined up in a nearly optimal fashion.

Shadow width on earth is now larger because the moon is larger and closer. These combine to make the shadow significantly larger - about 230 miles across. The increased orbital speed of the moon decreases transit time (offsetting the increased shadow size due to distance). Without complicated math and lots of new assumptions, exact figures are not possible, but I believe that assuming the eclipses would be about 15.7% longer on average because of the increased lunar diameter is a good first approximation. The larger umbra projected on the earth also means that around 30% more people will see each solar eclipse.


Lunar surface gravity is now 15.7% greater than before and the gravity well is 55% deeper so lunar missions just got harder. On the plus side, the trip won't take quite as long.

The earth-moon barycenter has been moved from about 4700 km to 6100 km from the earth's center -- still within the earth though. This will result is a slightly more pronounced wobble (and more frequent) in the earth's orbit.

The barycenter change may also cause slightly more seismic activity, especially when combined with the increased tidal stress. Perhaps slightly more weather variation too - tidal flows are thought to affect weakly some aspects of our weather.

Many animals cycles will be affected due to brightness and frequency of the lunar cycles. Some people will also have their sleeping habits changed. Currently, there is little correlation of crime to the lunar cycle, but the increased brightness could deter crime.

Everything based on lunar calendars will either change or ignore the changed lunar mechanics. Jewish, Christian, Islamic and Chinese calendars would be included.


Actual night sky moon would be a little larger than calculated above as this was based on center to center distance as would be accurate when the moon is on the horizon. When the moon is directly overhead you are somewhat closer because of the earth's radius (whether the moon is 200,000 or 238,900 miles away). For the moon remaining a fixed size, its apparent diameter would appear 19.84% larger when directly overhead instead of 19.45% when at the horizon, so the effect is small.

A similar effect also occurs in that the surface of the moon is closer too -- light reflecting off the center of the moon will be just a little closer and correspondingly more likely to impact the earth and thus brighter. Again, the difference is minor.

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    $\begingroup$ On the other hand, solar eclipses would be far less impressive as the exact fit we currently have lets you see much coronal detail. The larger moon will swallow that up more. $\endgroup$ – Oldcat Sep 15 '15 at 23:53
  • $\begingroup$ @GaryWalker I'm curious as to what you think about Vincent's calculations which are 38% larger vs. 19% larger... $\endgroup$ – user3082 Sep 23 '15 at 1:38
  • $\begingroup$ The 19% larger figure I used is prefaced by "for the moon remaining a fixed size" -- I was only illustrating the minor error due to ignoring the difference in center to center vs. surface to center. The overall apparent diameter ratio is the product of the inverse distance ratio and the actual diameter ratio -- 1.1945 * 1.157836 = 1.383 -- Vincent and I agree. $\endgroup$ – Gary Walker Sep 23 '15 at 11:38
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The Moon would just appear a bit larger in the sky. How many larger?

Since the size correlates with the square of the distance and the area of the circle correlates to the square of the size (it is a sphere, but that is projected as a circle in our retinas), this means that:

$$ \text{actual size} = \frac{(2,159.2\text{ miles})^2}{(238,900\text{ miles})^2} = \frac{4,662,144.64\text{ miles}^2}{57,073,210,000\text{ miles}^2} = 0.0000816...$$

$$ \text{proposed size} = \frac{(2,500\text{ miles})^2}{(200,000\text{ miles})^2} = \frac{6,250,000\text{ miles}^2}{40,000,000,000\text{ miles}^2} = 0.00015625$$

$$ \text{elargement factor} = \frac{\text{proposed size}}{\text{actual size}} = \frac{0.00015625}{0,0000816...} = 1.9148...$$

I.E, The Moon would be seen on the sky almost with the double of the size in area (191.48% to be precise), which is roughly 38% larger in diameter (the square root of 1.9148 is 1.38377).

Let's suppose that our larger Moon has the same density as our common Moon. By which factor the Moon mass get larger?

Mass would be measured by $m = v \times d$, where $m$ is mass, $v$ is volume and $d$ is density. The volume of a sphere is $v = \frac{4}{3} \pi r^3$. So, we have that $m = \frac{4}{3} d \pi r^3$:

$$\text{Actual Moon mass} = \frac{4}{3} \pi (\frac{2159.2}{2})^3 d = \frac{(2159.2)^3 \pi d}{6} = 10,066,502,706.688 \frac{\pi d}{6}$$ $$\text{Proposed Moon mass} = \frac{4}{3} \pi (\frac{2500}{2})^3 d = \frac{(2500)^3 \pi d}{6} = 15,625,000,000 \frac{\pi d}{6}$$

$$\text{Mass elergement factor} = \frac{\text{proposed mass}}{\text{actual mass}} = \frac{15,625,000,000 \frac{\pi d}{6}}{10,066,502,706.688 \frac{\pi d}{6}} = 1,5521...$$

This means that the proposed Moon is 55% more massive than our actual one.

Now lets measure gravity attraction as $\text{gravity} = \frac{\text{mass}}{\text{distance}^2}$:

$$\text{Actual Moon gravity} = \frac{10,066,502,706.688 \frac{\pi d}{6}}{(238,900 \text{ miles})^2} =$$

$$= \frac{10,066,502,706.688 \frac{\pi d}{6}}{57,073,210,000\text{ miles}^2} = 0.17637... \frac{\pi d}{6} \text{miles}^{-2}$$

$$\text{}$$

$$\text{Proposed Moon gravity} = \frac{15,625,000,000 \frac{\pi d}{6}}{(200,000 \text{ miles})^2} =$$ $$= \frac{15,625,000,000 \frac{\pi d}{6}}{40,000,000,000\text{ miles}^2} = 0.390625 \frac{\pi d}{6} \text{miles}^{-2}$$

$$\text{}$$

$$\text{Gravity elergement factor} = \frac{\text{proposed gravity}}{\text{actual gravity}} = \frac{0.390625 \frac{\pi d}{6} \text{miles}^{-2}}{0.17637... \frac{\pi d}{6} \text{miles}^{-2}} = 2.21469...$$

This would also means that the gravity that the Moon would exerce onto Earth would have 221.46% of the strength, which means tides with a bit more than the double of the strength. Assuming that the proposed Moon has the same density than our actual Moon.

About the Earth's axial tilt. By searching in wikipedia:

Lunisolar precession is caused by the gravitational forces of the Moon and Sun on Earth's equatorial bulge, causing Earth's axis to move with respect to inertial space. Planetary precession (an advance) is due to the small angle between the gravitational force of the other planets on Earth and its orbital plane (the ecliptic), causing the plane of the ecliptic to shift slightly relative to inertial space. Lunisolar precession is about 500 times greater than planetary precession.

This would translate in a precession of equinoces where the lunisolar component would be significantly larger.

Otherwise, the changes in Earth are not very significative. Things would be probably much as they are like today.

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