Closest Luna Looks Super. From the data on record over centuries, how do you prove that the so-far-largest Super Moon was nearly 1.14 times the so-far-smallest, in diameter, and 1.30 times, in area of the disc?

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Answer 1 · 2016-11-14T08:23:28

See explanation.

I have use (diameter)=(distance)X(angular spacing in radian; of the

lunar disc, at that distance), $r X alpha= D$ .

The diameter of the lunar disc, D = 3474 km.

So far, farthest 3-sd apogee $= 4.07 X 10^5 km$ .

So far, nearest 3-sd perigee $= 3.57 X 10^5 km.$

Let the angular spacing of the lunar discs at apogee = $alpha_a$

and that at perigee = $alpha_p$ .

Using the approximation formula for long distance and small angular

spacing of discs,

$Delta$ (angular spacing )/alpha_a#

$=(alpha_p-alpha_a)/alpha_a$

=(1/perigee-1/apogee)/(apogee), nearly

#=(407/357-1)=0.140, nearly

So, apparently, the diameter increases by 14%

Apparently, the area increases by about (1.14^2-1)X100=30%.

As reported in http://earthsky.org/astronomy-essentials/close-and-far-moons-in-2016, the last October 31 apogee = 406662 km and this Nov 14 perigee = 356509 km.

The reader can compute the current values (14.1% and 30.1%), from my answer.