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?

1 Answer
Nov 14, 2016

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#


=(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, 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.