# Bully Mnemonic Extension

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The Bully Mnemonic Extension is a technique for remembering the exact number of meters that light travels in one second, and the approximate range of gravitational accelerations that occur on the surface of the Earth due to Newtonian gravity. The Bully Mnemonic Extension, when used in conjunction with the Bully Mnemonic, allows one to calculate a few physical quantities, including the number of meters in a light year.
The following relationships are encoded in the Bully Mnemonic Extension:
$Speed\,of\,light\,in\,vacuum={299,792,458{\frac {m}{s}}}$
$High\,end\,of\,gravity\,on\,Earth's\,surface\approx {9.86{\frac {m}{s^{2}}}}$
$Mid\,range\,gravity\,on\,Earth's\,surface\approx {9.81{\frac {m}{s^{2}}}}$
$Low\,end\,of\,gravity\,on\,Earth's\,surface\approx {9.76{\frac {m}{s^{2}}}}$
The following relationship can be derived using the Bully Mnemonic Extension in conjunction with the Bully Mnemonic:
$1\,light\,year\approx {9,460,528,400,000,000\,meters}$

# Bully Mnemonic Extension Steps
## Initial Definitions
### Step 1

Complete steps 1 and 2 of the The Bully Mnemonic to form integers a) and b) as shown below:
${\begin{matrix}{\color {Red}1}&\scriptstyle {\text{2}}&{\color {Red}3}&\scriptstyle {\text{4}}&{\color {Red}5}\end{matrix}}$
$a)\,{\color {Red}1}0{\color {Red}33}0$ $b)\,{\color {Red}3}0{\color {Red}55}$

### Step 2

The Bully Mnemonic Extension will use two variants of integer a). The first variant will have 33 removed and replaced with 00. The second variant will have 330 removed and replaced with 22:
$av1)\,{\color {Red}1}0000$ $av2)\,{\color {Red}1}0{\color {Red}22}$ $b)\,{\color {Red}3}0{\color {Red}55}$

## Speed of Light
### Step 3

Multiply integers av2) and b) from Step 2.
${\color {Red}1}0{\color {Red}22}\times {\color {Red}3}0{\color {Red}55}=3122210$
Using Long Multiplication:

### Step 4

Drop the zero from the integer obtained in step 3, swap each 2 with 3, and swap each 1 with 9, to obtain integer f) shown below:

### Step 5

Multiply integer av2) from Step 2, and integer f) from step 4, to get the total number of meters that light travels in one second.
${\color {Red}1}0{\color {Red}22}\times 293339=299792458$
Using Long Multiplication:

## Gravity on Earth
### Step 6

Divide the speed of light obtained in step 5, by integers av1) and b) from step 2, to obtain a value for Earth's gravity:
${\frac {299792458{\frac {m}{s}}}{{\color {Red}1}0000\times {\color {Red}3}0{\color {Red}55}\,s}}={\frac {299792458{\frac {m}{s}}}{30550000\,s}}\approx {9.81{\frac {m}{s^{2}}}}$
In terms of Long Multiplication, 30550000 and 9.81 are approximately related to 299792458 as follows:

### Step 7

The range of gravitational accelerations that occur on the surface of the Earth, due to Newtonian gravity, can be approximated by repeating step 6 with the denominator increased or decreased by half a percent:
${\frac {299792458{\frac {m}{s}}}{(30550000\,s)+(150000\,s)}}={\frac {299792458{\frac {m}{s}}}{30700000\,s}}\approx {9.76{\frac {m}{s^{2}}}}$
${\frac {299792458{\frac {m}{s}}}{(30550000\,s)-(150000\,s)}}={\frac {299792458{\frac {m}{s}}}{30400000\,s}}\approx {9.86{\frac {m}{s^{2}}}}$

## Additional Relationships
### Step 8

As shown in steps 8 and 9 of the Bully Mnemonic, the earth's standard gravitational parameter (μ = MG) divided by the speed of light cubed, can be approximated as follows:
$10^{10}\times {\frac {{\mu }_{earth}}{c^{3}}}={\frac {{\color {Red}3}0{\color {Red}55}\,s}{{\color {Red}2}\times ({{\color {Red}1}0{\color {Red}33}0}-4.6316922)}}\approx {\frac {{\color {Red}3}0{\color {Red}55}\,s}{{\color {Red}2}\times {{\color {Red}1}0{\color {Red}33}0}}}$
Rearranging terms:
${\mu }_{earth}=GM_{earth}={\frac {c^{3}\times 10^{-10}\times {\color {Red}3}0{\color {Red}55}\,s}{{\color {Red}2}\times ({{\color {Red}1}0{\color {Red}33}0}-4.6316922)}}\approx {\frac {c^{3}\times 10^{-10}\times {\color {Red}3}0{\color {Red}55}\,s}{{\color {Red}2}\times {{\color {Red}1}0{\color {Red}33}0}}}$
As shown in step 6 above, a typical gravitational acceleration on earth is:
$g_{earth}={\frac {c}{30550000\,s}}={\frac {299792458{\frac {m}{s}}}{30550000\,s}}\approx {9.81{\frac {m}{s^{2}}}}$
Taking a ratio of the standard gravitational parameter with the gravitational acceleration:
${\frac {{\mu }_{earth}}{g_{earth}}}={\frac {30550000\,s\times c^{3}\times 10^{-10}\times {\color {Red}3}0{\color {Red}55}\,s}{c\times {\color {Red}2}\times ({{\color {Red}1}0{\color {Red}33}0}-4.6316922)}}\approx {\frac {30550000\,s\times c^{3}\times 10^{-10}\times {\color {Red}3}0{\color {Red}55}\,s}{c\times {\color {Red}2}\times {{\color {Red}1}0{\color {Red}33}0}}}$
Simplifying Terms:
${\frac {{\mu }_{earth}}{g_{earth}}}={\frac {(c\times 3.055\,s)^{2}}{{\color {Red}2}\times ({{\color {Red}1}0{\color {Red}33}0}-4.6316922)}}\approx {\frac {(c\times 3.055\,s)^{2}}{{\color {Red}2}\times {{\color {Red}1}0{\color {Red}33}0}}}$

### Step 9

It turns out that the radius of the earth can be approximated as the square root of the ratio of standard gravitational parameter with the gravitational acceleration. Using the approximation obtained in step 8:
$r_{earth}\approx {\sqrt {\frac {{\mu }_{earth}}{g_{earth}}}}\approx {\sqrt {\frac {(c\times 3.055\,s)^{2}}{{\color {Red}2}\times {{\color {Red}1}0{\color {Red}33}0}}}}\approx {\frac {c\times 3.055\,s}{\sqrt {{\color {Red}2}\times {{\color {Red}1}0{\color {Red}33}0}}}}=6371\,km$