So I read through this post and saw that he was doing things wrong! Because Duty Calls, I had to correct them, and slept no more than the recommended amount of <4 hours in the process _. The first day, that is (I may or may not have spent an entire week on this post, only to correct some random person on the Internet). It may also have something to do with that OP got 16k upvotes for not doing math, so i may be able to get... 7? upvotes for doing things properly.
Most calculations can be found in this spreadsheet.
So anyway, what do we know?
Measurements
My FOV is 103°, which results in an actual FOV of 90° because Overwatch is weird. My screen is 1920px wide, which makes the pixel radius 960px. I took screenshots of the Moon from Dorado and of the Earth from Horizon Lunar Colony as you can see here. The Moon height over horizon from Earth was such that the moon just touched the top of my screen, so the radius is 1080/2=540px.
Object 
Pixel radius 
Moon 
70.5px 
Earth 
323px 
Screen 
960px 
Moon height over horizon from Earth 
540px 
Earth height over horizon from Moon 
259px 
We also know the radius and mass of the OW earth from these screens on Horizon.
Pixels and angles
We are interested in the angular radii of all our objects, but we only have the sizes in pixels. So how do we convert between the two?
We can visualize the situation with this figure where x_o
is the pixel radius of the object, ɑ_o
is the angular radius of the object, x_s
is the screen radius and ɑ_s
is the angular radius of the screen, which is half the FOV, so 45°.
If we remember our trigonometric functions, we notice that
tan(ɑ_s) = x_s / z z tan(45°) = x_s z = x_s (Because tan(45°)=1)
We can also use tan
to calculate ɑ_o:
tan(ɑ_o) = x_o/z ɑ_o = atan(x_o/z)
Using this formula on the known pixel widths, we get this table:
Thing 
Pixel radius 
Angular radius 
Moon radius 
70.5px 
4.20° 
Earth radius 
323px 
18.60° 
Screen horizontal radius 
960px 
45° 
Moon height over horizon from Earth 
540px 
29.4° 
Earth height over horizon from Moon 
259px 
15.1° 
Because I will use it later, also note that cos(x + 90°) = sin(x)
and that sin(x + 90°) = cos(x)
.
Where things get complicated
First of all, the symbols I’ll be using are:
Thing 
Symbol 
Unit 
Moon radius 
x 
km 
Earth radius 
r 
km 
Moon angular radius 
α_m 
° 
Earth angular radius 
α_e 
° 
Moon height over horizon from Earth 
h_m 
° 
Earth height over horizon from Moon 
h_e 
° 
Earth gravity 
g_e 
m/s^{2}

Moon gravity 
g_m 
m/s^{2}

Earth jump duration 
t_e 
s 
Moon jump duration 
t_m 
s 
Latitude of Dorado 
φ 
° 
Our situation is this figure where E and M are the center of the Earth and Moon respectively and that D and H are Dorado and Horizon Lunar Colony ^{remember this, I will use it a lot}. The lengths of f
and g
are derived from that
sin(ɑ_m) = x / f f = x / sin(ɑ_m)
Using the law of cosines, we can write d^{2} in two ways: using the triangles △MED and △MEH. Moving things around, we get a quadratic equation. Solving this equation yields the value of x
.
Now knowing x
, d
is now solvable using the law of cosines on either △MED or △MEH. (I used △MED in my spreadsheet.)
All this yields the following table:
Thing 
OW 
Real 
Ratio (Real/OW) 
Moon radius 
1203 km 
1737 km 
1.444 
MoonEarth distance 
16428 km 
384400 km 
23.4 
Acceleration, mass and gravity
The position of an object under constant acceleration is s = 1/2 a t^2 + v_0 t
. Plugging in the numbers at the end of the jump on Earth:
v_0 t_e + 1/2 g_e t_e^2 = 0 v_0 + 1/2 g_e t_e = 0 (dividing by t_e) v_0 = 1/2 g_e t_e
Plugging in the numbers at the end of the jump on the Moon yields:
1/2 g_m t_m^2 + v_0 t_m = 0 1/2 g_m t_m  1/2 g_e t_e = 0 (substituting in v_0) g_m t_m = g_e t_e (multiplying by 2 and adding by g_e t_e) g_m = g_e t_e / t_m
Using Newton’s law of universal gravitation and Newton’s second law:
F_g = gm = GmM/r^2 g = GM/r^2 M = gr^2/G
Plugging in the numbers yields:
Thing 
OW 
Real 
Ratio (OW / Real) 
Moon gravity 
2.43 m/s^{2}

1.62 m/s^{2}

1.5 
Moon mass 
5.270e16 kg 
7.342e22 kg 
1 / 1 400 000 
The rest of the maths
Using the law of sines, we can calculate φ, the latitude of Dorado, remembering our figure:
sin(φ) / f = sin(h_m + 90°) / d φ = asin( cos(h_m) f/d )
which turns out to be ~44.78756°, which is around the latitude of Minneapolis, US or Bordeaux, FR. Not exactly Mexico...
Anyway, the thing OP thought was the big deal. Tidal acceleration is calculated using the formula 2 G m r / d^3
, so calculating the effect of the moon on the earth and vice versa yields this:
Tidal acceleration on 
OW 
Real 
Ratio of OW/Real 
Earth 
5.338×10^{9}

1.10e6 
1/206 
Moon 
0.1142 
2.44e5 
4680 ^{this isn't good}

The effects
As you might imagine, the Earth will be just fine with only having smaller tides: basically the exact opposite of OP:s conclusion. The Moon is in bigger danger, but it will probably just result in tidal heating (albeit a lot of it) instead of something cataclysmic, but I really do not know^{Scott Manley plz}. I don’t know what else to say really, I can’t make an anticlimactic conclusion more interesting. But it was totally worth it just to say that OP was incorrect.
TLDR:
The Earth is definitely not in danger, but Mexico is in Minnesota.
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