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Here's a video showing what the Moon would look like if it were as close to Earth as the International Space Station is.
The io9 article quickly points out that if the Moon were that close, it would break apart and form a ring (like Saturn's) because of the Roche limit. Here's another io9 article with pictures of what that would look like.
But here, let's assume the Moon doesn't break apart. Here's a bird'seye view looking down at the north pole at the moment when both the ISS and the Moon are about to rise over the horizon.
Show Instructions
Show Instructions
Patrick Truchon, 19 October 2013, Created with GeoGebra 
Note that:
Now let's try to answer a few interesting questions…
Why is the ISS going faster if the Moon is at the same distance?
Even though the surface of the Moon is 420 km away from the surface of the Earth (like the ISS), its centre of mass isn't, which means it's in a different orbit. It's not taking more time to go around because it's bigger and more sluggish. If that were true, the astronauts in the ISS would want to orbit faster than the station itself. That would be weird…
The only thing that affects orbital speed is the distance from the planet. The further away an object is, the longer it takes to go around it. For example, Neptune takes about 165 human years to go around the sun because it's so much further away from it than we are.
So because the Moon's centre of mass is further away, it revolves around the Earth more slowly:
Period of Revolution  

ISS  1 hr 33 min 
Moon  2 hr 11 min 
Show Calculations
Show Calculations
To do these calculations, we'll need some (but not all) of the following numbers. Can you think of which one(s) we don't need?
Radius  Mass  

Earth  6370 km = 6.37 x 10^{6} m  5.97 x 10^{24} kg  
Moon  1740 km = 1.74 x 10^{6} m  7.35 x 10^{22} kg  
ISS  72.8 m x 108.5 m x 20 m  4.5 x 10^{5} kg  
Distance to Surface  420 km = 4.2 x 10^{5} m 
The orbital period is the time (in seconds) it takes to go around the planet. To find it, we start by setting the gravitational force equal to the centripetal force:
Here's what all those variables mean:
can calculated by taking the total distance around the Earth (the circumference of a circle) divided by the period of revolution, , which is what we're really after:
Thus, the previous equation becomes:
Simplifying and solving for , we get:
Again, notice that the only mass that matters is that of the Earth, not the Moon or the ISS.
So for the ISS, the orbital period is:
And for the Moon, it's:
The reason I've underlined some digits is because they are the last significant ones. I kept a few digits afterwards so I can reuse these results in subsequent calculations without accumulating rounding errors.
How long would we see them in the sky?
Since the Moon would take about twice as long to go around the Earth as the ISS, it's reasonable to think that it would be visible for about twice as long. But that close to Earth, the Moon's size is very big, and if we calculate the time from when we “start” seeing it to the time when we completely lose it below the horizon, we would see it four times longer than we'd see the ISS!
At best (if they rise straight over us, and ignoring atmospheric and optical issues):
Visible for at most  

ISS  10 min and 30 sec 
Moon  41 min and 30 sec 
Show Calculations
Show Calculations
We know how long it takes for the ISS and the Moon to go around Earth (360°). Now, we need to find the fraction of their orbit when they are above the horizon.
If we draw a right triangle from the centre of the Earth to the ISS and to us, we see that
The Moon
For the Moon, it's a bit more complicated since we can't treat it as a point (like the ISS). Instead, we have to find out where the Moon's centre is when its tip breaks above the horizon.
I've seen the video, but how do you explain how it would “look” like?
During its flyby, the following three aspects change:
At the horizon  Overhead  

Distance from us  1425 km  420 km  (3.4x closer) 
Angular size  66.7°  107.3°  (1.6x more) 
Visible surface area  22.5%  9.7%  (2.3x less) 
Because the Moon is about 3.4 times closer to us when it's overhead compared to when it's at the horizon, we'd see craters getting bigger and bigger as the Moon rises.
As soon as the Moon is completely over the horizon, what we see would cover about 67° of our field of vision. To put this number into perspective, looking at the horizon and then looking straight up covers 90°. If the Moon was rising right behind the Eiffel tower (which is 300 m high), you'd only have to be 129 m away from the tower to see the Moon at the same eight as the tower.
As the Moon rises, the disk that we see in the sky would grow in angular size to 107° (1.6 times more). This is crazy! It means that looking at the horizon, we would only be able to see 36° of sky before we see the Moon. In every direction! It would be like being in a room where the ceiling is the Moon and the walls are open.
As soon as the Moon is completely over the horizon, we would see about 23% of its surface area (instead of about 50% for our real Moon).
However, as it rises straight overhead, what we see would drop to 9.7% (2.3 times less). That's because as it gets closer to us, more of the Moon's surface gets hidden behind its curvature.
Show Calculations
Show Calculations
Before calculating specific cases, we need to derive a general formula for calculating the percentage of visible surface area as a function of the angle.


The first step is to find all sides and angles of this right triangle.
Next we find the angles:
The same calculations are much simpler in this case.
That's it for this weekend. Here are a few more ideas I have sketched on scrap paper that I want to add (maybe next week).
Here, I'll find out if how much lighter the moon fly by would make us feel.
I'll also look at it from the perspective of an astronaut on the moon. That part will be fun since the moon is revolving so fast!
Spoiler Alert
Spoiler Alert
On Earth, gravity would increase by 0.2% when the Moon is on the other side of the Earth, and it would decrease by 0.7% when the Moon is straight overhead. I can't even start to imagine how that would affect the tides though… Maybe it would revolve too fast?
On the Moon however it's a very different story! Assuming that the Moon is still tidally locked, we'll see that an astronaut on the “far side” of the Moon would barely be able to stay on the surface, but an astronaut on the “near side” would fall towards Earth. Although, I don't know if they'd fall *on* Earth or into a lower orbit…
I want to see if I can add sun rays to the Geogebra animation so that we can see which part of the moon would be lighted and which part of Earth would be in Moon's umbra depending on the time of day.