The Bully Mnemonic

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Bully Mnemonic Topics:

• The Bully Mnemonic
• The Bully Mnemonic & Calendar Variations

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The Bully Mnemonic is a technique for remembering the exact number of seconds that occur in Earth's sidereal year and tropical year; a good approximation of the Earth's Great Year; and a rough approximation of the Solar System's galactic year.

The following relationships are encoded in the Bully Mnemonic:

$$1SiderealYear=31,558,150Seconds$$

$$1TropicalYear=31,556,926Seconds$$

$$1GreatYear\approx 25,824SiderealYears\approx 25,825TropicalYears$$

$$1GalacticYear\approx 213,417,800TropicalYears$$

The following additional relationships are closely related to values used in the Bully Mnemonic

$$Eart{h}^{\prime }sStandardGravitationalParameter=398,600,441,800,000\frac{m^3}{s^2}$$

$$Eart{h}^{\prime }sSchwarzschildRadius=8.87005606mm$$

Bully Mnemonic Steps

Initial Definitions

Step 1

The first step is to write down the first five digits:

$$\begin{array}{ccccc}1& 2& 3& 4& 5\end{array}$$

Step 2

The second step is to select odd digits and intersperse them with zeros to form integers a) and b) as shown below: (important to remember that the first integer ends with 33 followed by a 0, whereas the second integer ends with 55 with no trailing 0)

$$\begin{array}{ccccc}1& \text{2}& 3& \text{4}& 5\end{array}$$

$$a)10330$$ $$b)3055$$

Step 3

The third step is to select even digits and define numbers c) and d) as shown below:

$$\begin{array}{ccccc}\text{1}& 2& \text{3}& 4& \text{5}\end{array}$$

$$c)2$$ $$d)0.40$$

Sidereal & Tropical Years

Step 4

Multiply integers a) and b) from Step 2 to get the total number of seconds in a sidereal year.

$$10330\times 3055=31558150=\frac{1SiderealYear}{1Second}$$

Using Long Multiplication:

       3055
×     10330
————————————
   3055
    0000
     9165
      9165
+      0000
————————————
   31558150

Step 5

The tropical year has a slightly shorter duration than the sidereal year. The approximate number of seconds in a tropical year is obtained by reducing integer a) by amount d), and then multiplying by b).

$$(10330-0.40)\times 3055=31556928\approx \frac{1TropicalYear}{1Second}$$

The exact number of seconds in a tropical year is obtained by reducing integer a) by amount d), multiplying by b), and then reducing by c).

$$((10330-0.40)\times 3055)-2=31556926=\frac{1TropicalYear}{1Second}$$

Using the Distributive Property of Multiplication:

(10330 - 0.40) × 3055 = (10330 × 3055) - (0.40 × 3055)
                            =    31558150    -     1222
                            =    31556928

Great Years

Step 6

The Great Year is, by definition, a least common multiple of the sidereal year and the tropical year. From steps 4 and 5 above, we have that the ratio of tropical years to sidereal years is:

$$\frac{1TropicalYear}{1SiderealYear}\approx \frac{(10330-0.40)\times 3055sec}{10330\times 3055sec}$$

Divide top and bottom by amount d) and use the Distributive Property of Multiplication to obtain:

$$\frac{1TropicalYear}{1SiderealYear}\approx \frac{(\frac{10330}{0.40}-\frac{0.40}{0.40})\times 3055sec}{(\frac{10330}{0.40})\times 3055sec}$$

From whence:

$$\frac{1TropicalYear}{1SiderealYear}\approx \frac{(25825-1)\times 3055sec}{(25825)\times 3055sec}$$

Consequently:

$$\frac{25825TropicalYear}{25824SiderealYear}\approx \frac{25825\times (25824)\times 3055sec}{25824\times (25825)\times 3055sec}=1$$

Finally:

$$1GreatYear\approx 25825TropicalYears\approx 25824SiderealYears$$

In terms of Long Multiplication; 0.40, 25825, and 10330 are related as follows:

       0.40
×  25825
————————————
   080
    200
     320
      08.0
+      2.00
————————————
   10330.00

Galactic Years

Step 7

Multiply integer c) by the square of integer a) to get a rough approximate galactic year (the number of tropical years required for the Solar System to orbit once around the galactic center).

$$2\times {10330}^2=213417800\approx \frac{1GalacticYear}{1TropicalYear}$$

Using Long Multiplication:

       10330
×      10330
——————————————
   10330
    00000
     30990
      30990
+      00000
——————————————
   106708900

And finally:

106708900 × 2 = 213417800

Additional Relationships

Step 8

Divide integer b) seconds by the product of integer c) and integer a). The resulting value will be roughly ten orders of magnitude bigger than earth's standard gravitational parameter (μ = MG) divided by the speed of light (c) cubed.

$$10^{10}\times \frac{\mu }{c^3}=0.147936611505s$$

$$\frac{3055s}{2\times 10330}=0.147870280736s$$

Step 9

A more accurate approximation is obtained by reducing a) by 4.6316922:

$$\frac{3055}{2\times (10330-4.6316922)}=0.147936611505s$$

Step 10

The value of an objects Schwarzschild radius can obtained from the standard gravitational parameter by multiplying by two and dividing by the speed of light squared.

$$10^{10}\times \frac{R_S}{c}=0.147936611505s$$

$$\frac{3055}{(10330-4.6316922)}=0.295873223010s$$