The Bully Mnemonic is a technique for remembering the exact (eight digits) number of seconds that occur in Earth's sidereal year and tropical year; a good approximation (four digits) 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:

$${\displaystyle {1\,Sidereal\,Year}={31,558,150\,Seconds}}$$

$${\displaystyle {1\,Tropical\,Year}={31,556,926\,Seconds}}$$

$${\displaystyle 1\,Great\,Year\approx 25,824\,Sidereal\,Years\approx 25,825\,Tropical\,Years}$$

$${\displaystyle {1\,Galactic\,Year}\approx 213,417,800\,Tropical\,Years}$$

A few additional approximations (four digits) can be obtained from values used in the Bully Mnemonic. These include an approximate relationship of the speed of light to the Earth's radius (r ≈ 6371), Schwarzschild radius (R), standard gravitational parameter (μ = MG ≈ 3.984e14), and a typical gravitational acceleration on earth's surface (g ≈ 9.813 ).

$${\displaystyle r_{earth}\approx {10^{-3}}\times c\times {\frac {{\color {Red}3}0{\color {Red}55}\,s}{\sqrt {{\color {Red}2}\times {{\color {Red}1}0{\color {Red}33}0}}}}=6371\,km}$$

$${\displaystyle R_{earth}={\frac {2\times GM_{earth}}{c^{2}}}\approx {10^{-10}}\times c\times {\frac {{\color {Red}3}0{\color {Red}55}\,s}{{\color {Red}1}0{\color {Red}33}0}}\approx {8.866\,mm}}$$

$${\displaystyle {\mu }_{earth}=GM_{earth}\approx {10^{-10}}\times c^{3}\times {\frac {{\color {Red}3}0{\color {Red}55}\,s}{{\color {Red}2}\times {{\color {Red}1}0{\color {Red}33}0}}}\approx {398,400,000,000,000\,{\frac {m^{3}}{s^{2}}}}}$$

$${\displaystyle g_{earth}={\frac {{\mu }_{earth}}{{r_{earth}}^{2}}}\approx {10^{-4}}\times c\times {\frac {1}{{\color {Red}3}0{\color {Red}55}\,s}}\approx {9.813\,{\frac {m}{s^{2}}}}}$$

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

$${\displaystyle {\begin{matrix}1&2&3&4&5\end{matrix}}}$$

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)

$${\displaystyle {\begin{matrix}{\color {Red}1}&\scriptstyle {\text{2}}&{\color {Red}3}&\scriptstyle {\text{4}}&{\color {Red}5}\end{matrix}}}$$

$${\displaystyle a)\,{\color {Red}1}0{\color {Red}33}0}$$ $${\displaystyle b)\,{\color {Red}3}0{\color {Red}55}}$$

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

$${\displaystyle {\begin{matrix}\scriptstyle {\text{1}}&{\color {Red}2}&\scriptstyle {\text{3}}&{\color {Red}4}&\scriptstyle {\text{5}}\end{matrix}}}$$

$${\displaystyle c)\,{\color {Red}2}}$$ $${\displaystyle d)\,0.{\color {Red}4}0}$$

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

$${\displaystyle {\color {Red}1}0{\color {Red}33}0\times {\color {Red}3}0{\color {Red}55}=31558150={\frac {1\,Sidereal\,Year}{1\,Second}}}$$

Using Long Multiplication:

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

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).

$${\displaystyle ({\color {Red}1}0{\color {Red}33}0-0.{\color {Red}4}0)\times {\color {Red}3}0{\color {Red}55}=31556928\approx {\frac {1\,Tropical\,Year}{1\,Second}}}$$

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).

$${\displaystyle (({\color {Red}1}0{\color {Red}33}0-0.{\color {Red}4}0)\times {\color {Red}3}0{\color {Red}55})-{\color {Red}2}=31556926={\frac {1\,Tropical\,Year}{1\,Second}}}$$

Using the Distributive Property of Multiplication:

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

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:

$${\displaystyle {\frac {1\,Tropical\,Year}{1\,Sidereal\,Year}}\approx {\frac {(10330-0.40)\times 3055\,sec}{10330\times 3055\,sec}}}$$

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

$${\displaystyle {\frac {1\,Tropical\,Year}{1\,Sidereal\,Year}}\approx {\frac {({\frac {10330}{0.40}}-{\frac {0.40}{0.40}})\times 3055\,sec}{({\frac {10330}{0.40}})\times 3055\,sec}}}$$

From whence:

$${\displaystyle {\frac {1\,Tropical\,Year}{1\,Sidereal\,Year}}\approx {\frac {(25825-1)\times 3055\,sec}{(25825)\times 3055\,sec}}}$$

Consequently:

$${\displaystyle {\frac {25825\,Tropical\,Year}{25824\,Sidereal\,Year}}\approx {\frac {25825\times (25824)\times 3055\,sec}{25824\times (25825)\times 3055\,sec}}=1}$$

Finally:

$${\displaystyle 1\,Great\,Year\approx 25825\,Tropical\,Years\approx 25824\,Sidereal\,Years}$$

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

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

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).

$${\displaystyle {\color {Red}2}\times {{\color {Red}1}0{\color {Red}33}0}^{2}=213417800\approx {\frac {1\,Galactic\,Year}{1\,Tropical\,Year}}}$$

Using Long Multiplication:

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

And finally:

106708900 × 2 = 213417800

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

$${\displaystyle 10^{10}\times {\frac {{\mu }_{earth}}{c^{3}}}=0.147\,936\,611\,505s}$$

$${\displaystyle {\frac {{\color {Red}3}0{\color {Red}55}\,s}{{\color {Red}2}\times {{\color {Red}1}0{\color {Red}33}0}}}=0.147\,870\,280\,736s}$$

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

$${\displaystyle {\frac {{\color {Red}3}0{\color {Red}55}}{{\color {Red}2}\times ({{\color {Red}1}0{\color {Red}33}0}-4.6316922)}}=0.147\,936\,611\,505s}$$

The value of an object's Schwarzschild radius (R) is obtained from the standard gravitational parameter by multiplying by two and dividing by the speed of light squared. Comparing with steps 8 and 9 above, one obtains:

$${\displaystyle 10^{10}\times {\frac {R_{earth}}{c}}=0.295\,873\,223\,010s}$$

$${\displaystyle {\frac {{\color {Red}3}0{\color {Red}55}}{({{\color {Red}1}0{\color {Red}33}0}-4.6316922)}}=0.295\,873\,223\,010s}$$

The Earth is not a perfect sphere. The radius and gravitational acceleration at the earth's surface are not constant values. Approximations can be obtained as follows:

$${\displaystyle r_{earth}\approx {10^{-3}}\times c\times {\frac {{\color {Red}3}0{\color {Red}55}\,s}{\sqrt {{\color {Red}2}\times {{\color {Red}1}0{\color {Red}33}0}}}}=6371\,km}$$

$${\displaystyle g_{earth}={\frac {{\mu }_{earth}}{{r_{earth}}^{2}}}\approx {10^{-4}}\times c\times {\frac {1}{{\color {Red}3}0{\color {Red}55}\,s}}\approx {9.813\,{\frac {m}{s^{2}}}}}$$