Astronomy | Silicon Valley Math University

By Steve Dinh, a.k.a. Võ Đức Diên

I am taking the opportunity to write a phenomenon about the moon as I have been asked by many readers on why the moon appears to be round at full-moon and at other times it does not.

So the question is that whether the moon is ever perfectly round? The roundness I am talking about here is its illumination at the “full” moon and not its physical shape as we have already known or assumed that it shape is already round. The perfect roundness, by mathematical and geometrical definition, is 100% perfectly round. The general understanding thus far by humans is that the moon is perfectly round at full moon and not at 0.098… % percent round as it is about to be proven mathematically. While proving the illumination is not 100% round at full-moon we assume that its shape is already 100% round. Astronomically speaking, the moon illumination is never round because its shape is never perfectly round physically.

The calculations below are based on the assumption that the moon is 100% round physically and that there are no craters or rough areas on its surface.

The moon, as we have known, gets the sunlight from the Sun and, assuming that it’s 100% physically round, half of its surface is always illuminated. Basically, its illumination is always round unless when the earth blocks the sunlight from getting to it.

But humans on earth are never in a perfect position to see the total sunlight on the moon. Even on a full-moon, our eyes see the illumination almost perfectly whereas our calculations suggest otherwise. As a matter of fact, in order to see the moon light being 100% round, one has to position himself on the line connecting the centers of the sun and the moon. If a human is able to be in this position, the moon will be totally eclipsed by the earth, and thus there is no light on the moon to be seen instead of the round moon illumination.

We are looking at a scenario above when the there is a single ray of sun light which tangents both the earth and the moon. This can happen in real life although it is extremely rare. In such a situation if there is a person at position J which is at a time when the sun is setting down or rising up, that person will see the moon at an angle which is largest and with the highest illumination. Indeed, you will see the illumination of the moon at a smaller angle and less illumination if you are somewhere else. Basically, the best scenario is for a person to position at J.

The area colored green on the moon is the area where the illumination can not be seen by the person even positioning at J on earth. That person can only see the illumination on arc from C to B. Now we can find the angle α, phase angle β and distance from C to B in this case.

Let r be the radius of the moon, R the radius of the earth, d is the shortest distance from the surface of the earth to that of the moon, we have:

Two triangles OBI and O’JI are similar, hence

$\frac{OI}{OI'}$ = $\frac{r}{R}$ = $\frac{BI}{JI}$ (1)

EF = d is the distance from the earth to the moon, we have

OO’ = OI + IO’ = r + d + R (2)

From (1) and (2), we have OI (1 + $\frac{R}{r}$ ) = r + d + R

Or OI = $\frac{r + d + R}{1 + \frac{R}{r}}$

From there BI = $\sqrt {OI^{2}-r^{2}}$

BI = $\sqrt{({\frac{r + d + R}{1 + \frac{R}{r}}})^{2}-r^{2}}$

Also from (1) JI = $\frac{R*BI}{r}$ ; therefore,

BJ = BI + JI = BI (1 + $\frac{R}{r}$ ) = (1 + $\frac{R}{r}$ ) $\sqrt{({\frac{r + d + R}{1 + \frac{R}{r}}})^{2}-r^{2}}$

We also have OJ orthogonal to CB and BJ orthogonal to AB, and therefore the two triangles OBJ and ACB also similar, and we have

$\frac{CB}{AB}$ = $\frac{BJ}{OJ}$

Therefore, CB = $\frac{AB*BJ}{OJ}$ (3)

But OJ² = r² + BJ²

Or OJ² = r² + (1 + $\frac{R}{r}$ )²[ $(\frac{r + d + R}{1 + \frac{R}{r}})^{2}-r^{2}$ ]

AB = 2r, from (3) we have

CB = 2r(1 + $\frac{R}{r}$ ) $\sqrt{({\frac{r + d + R}{1 + \frac{R}{r}}})^{2}-r^{2}}$ /

$\sqrt{r^{2} + (1 + \frac{R}{r})^{2}[(\frac{r + d + R}{1 + \frac{R}{r}})^{2}-r^{2}]}$

Substituting in their values to get

r = 1737 km

R = 6371 km

d = 384000 km

into the equation of CB we have CB = 3473.97 km.

And we have found the distance CB of the area of illumination which a person at position J can see. The remaining illuminated area colored green is the area we can not see has the distance AC.

Now let’s find the phase angle β.

OJ is the bisector of angle CJB and the phase angle β on triangle ABC equals angle OJB and equals half angle α,

β = $\frac {1}{2}$ α

sinβ = $\frac {r}{OJ}$

From there we find angle β = arcsine $\frac {r}{OJ}$

β = arcsine{r/ $\sqrt{r^{2} + (1 + \frac{R}{r})^{2}[(\frac{r + d + R}{1 + \frac{R}{r}})^{2}-r^{2}]}$ }

Again substituting in their values and we get β = 0.28°.

The area we can not see k has the distance equals the diameter of the moon subtracted from CB, or

k = 2r – 2r(1 + $\frac{R}{r}$ ) $\sqrt{(\frac{r + d + R}{1 + \frac{R}{r}})^{2}-r^{2}}$ / $\sqrt{r^{2} + (1 + \frac{R}{r})^{2}[(\frac{r + d + R}{1 + \frac{R}{r}})^{2}-r^{2}]}$

or k = 34.10 m.

Therefore, we can conclude that the illumination we can see on a full-moon is not perfectly round, and its phase angle is 0.28°. Its percentage compared to the diameter of the moon is $\frac{34.10}{3474000}$ = 0.098 %

Attention: Wikipedia.com also gives data about the phase angle of the moon but did not offer a calculation or an explanation.

I guess it’s very interesting to find out what day of the week the human ancestors picked to start out the current solar calendar almost 2010 years ago. I assume that the current solar calendar started on January 01, year 1, abbreviated as 01/01/0001. I will leave out why they picked seven days for a week. The debates of why seven days was selected for a week have been widely circulated in the Internet.

So let’s find out what day of the week January 01, 0001 was: Monday, Tuesday, Wednesday, Thursday, Friday, Saturday or Sunday.

You all agree with me that if today August 13, 2009 is Thursday, seven days ago August 6, 2009 was also a Thursday. Let’s expand on that idea:

Use today as a reference point. Let N be the number of days it has been since 01/01/0001, n and k be positive integers.

We have: N = 7 n + k (i)

We can easily see that n is the number of weeks that has elapsed since that first day (01/01/0001), and k is the day of the week, and it must be less than 7.

Now let’s find out what N is.

We know that a year divisible by 4 is a leap year except that year divisible to 100 is not leap and divisible to 400 is leap.

We can write a little software program to find the numbers of days. Following are the only basic syntaxes of a complete program:

static char daytab[2][13] =

{

{ 0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31},

{ 0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}

};

// The leap year argument above is expressed in this for loop

// day variable is the number of days already into this current year

// daystotal is N

for ( year = 1; year <= 2010; year++ ) // year = 1 is year 0001

{

leap = year%4 == 0;

if ( year%100 == 0 && year%400 != 0 )

leap = 0;

for ( month = 1; month <= 12; month++ )

{

day += daytab[leap][month – 1]; daystotal = 365*(year-1) + (int)floor((year-1)/4) – (int)floor((year-1)/100)

+ (int)floor((year-1)/400) + day;

}

We found that daystotal which is N = 733632

To verify the total days we can do a simple calculation:

There are already 2008 full years since 01/01/0001. The total numbers of days in 2008 years not counting the leap days are 365 × 2008 = 732920 days. Assume that all years divisible by 4 are leap years, add 2008 / 4 = 502 to the total; then subtract the 15 leap days because the years divisible by 100 are leap and divisible by 400 are not; then add to the new total the numbers of days in 2009 that has passed which are 31 + 28 + 31 + 30 + 31 + 30 + 31 = 212 days and the 13 days of August, we have:

N = 732920 + 502 – 20 + 5 + 212 + 13 = 733632

What this means is that it has been 733632 days since that first day January 01, 0001.

Substitute the value of N into the above equation (i), we have:

733632 = 7 × 104804 + 4

we find n = 104804 and k = 4.

What those mean is that we are 4 days into the 104805^th week since 01/01/0001. The fourth day of the week is Thursday; therefore, the first day of the week must be Monday, and we conclude that the first day 01/01/0001 of the calendar is Monday.

Expand this idea by expanding the software program, we find that on year 3000 we will have a Friday the 13^th falling on June since the number of days will be 1095526.

More hypothetical, if we assume that this earth and people exist to year 1000000 (year one millionth), there will be a Friday the 13^th on October of that year because the number of days to Friday the 13^th of year 1000000 N will be 365242421.

Note: You can cross-check my calculation with the results from the website:

http://www.searchforancestors.com/utility/dayofweek.html

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Category Archives: Astronomy

Has the illumination on the moon at full-moon ever been perfectly round?

What day of the week was January 01, year 0001?