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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 ).

$r_{e a r t h} \approx 1 0^{- 3} \times c \times \frac{3055 s}{\sqrt{2 \times 10330}} = 6371 k m$

$R_{e a r t h} = \frac{2 \times G M_{e a r t h}}{c^{2}} \approx 1 0^{- 10} \times c \times \frac{3055 s}{10330} \approx 8.866 m m$

$μ_{e a r t h} = G M_{e a r t h} \approx 1 0^{- 10} \times c^{3} \times \frac{3055 s}{2 \times 10330} \approx 398, 400, 000, 000, 000 \frac{m^{3}}{s^{2}}$

$g_{e a r t h} = \frac{μ_{e a r t h}}{{r_{e a r t h}}^{2}} \approx 1 0^{- 4} \times c \times \frac{1}{3055 s} \approx 9.813 \frac{m}{s^{2}}$

Additional Relationships

Step 8

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.

$1 0^{10} \times \frac{μ_{e a r t h}}{c^{3}} = 0.147 936 611 505 s$

$\frac{3055 s}{2 \times 10330} = 0.147 870 280 736 s$

Step 9

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

$\frac{3055}{2 \times (10330 - 4.6316922)} = 0.147 936 611 505 s$

Step 10

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:

$1 0^{10} \times \frac{R_{e a r t h}}{c} = 0.295 873 223 010 s$

$\frac{3055}{(10330 - 4.6316922)} = 0.295 873 223 010 s$

Step 11

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

$r_{e a r t h} \approx 1 0^{- 3} \times c \times \frac{3055 s}{\sqrt{2 \times 10330}} = 6371 k m$

$g_{e a r t h} = \frac{μ_{e a r t h}}{{r_{e a r t h}}^{2}} \approx 1 0^{- 4} \times c \times \frac{1}{3055 s} \approx 9.813 \frac{m}{s^{2}}$

Bully Metric Main Page

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:

$S p e e d o f l i g h t i n v a c u u m = 299, 792, 458 \frac{m}{s}$

$H i g h e n d o f g r a v i t y o n E a r t h^{'} s s u r f a c e \approx 9.86 \frac{m}{s^{2}}$

$M i d r a n g e g r a v i t y o n E a r t h^{'} s s u r f a c e \approx 9.81 \frac{m}{s^{2}}$

$L o w e n d o f g r a v i t y o n E a r t h^{'} s s u r f a c e \approx 9.76 \frac{m}{s^{2}}$

The following relationship can be derived using the Bully Mnemonic Extension in conjunction with the Bully Mnemonic:

$1 l i g h t y e a r \approx 9, 460, 528, 400, 000, 000 m e t e r s$

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:

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

$a) 10330$ $b) 3055$

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:

$a v 1) 10000$ $a v 2) 1022$ $b) 3055$

Speed of Light

Step 3

Multiply integers av2) and b) from Step 2.

$1022 \times 3055 = 3122210$

Using Long Multiplication:

     3055
×    1022
————————————
     6110
    6110
   0000
  3055
————————————
  3122210

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:

   312221
f) 293339

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.

$1022 \times 293339 = 299792458$

Using Long Multiplication:

         1022
×      293339
——————————————
        9198
       3066
      3066
     3066
    9198
   2044
——————————————
   299792458

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:

$\frac{299792458 \frac{m}{s}}{10000 \times 3055 s} = \frac{299792458 \frac{m}{s}}{30550000 s} \approx 9.81 \frac{m}{s^{2}}$

In terms of Long Multiplication, 30550000 and 9.81 are approximately related to 299792458 as follows:

   30550000
×         9.81
————————————
     305500.00
   24440000.0
  274950000
————————————
  2997.....

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:

$1 0^{10} \times \frac{μ_{e a r t h}}{c^{3}} = \frac{3055 s}{2 \times (10330 - 4.6316922)} \approx \frac{3055 s}{2 \times 10330}$

Rearranging terms:

$μ_{e a r t h} = G M_{e a r t h} = \frac{c^{3} \times 1 0^{- 10} \times 3055 s}{2 \times (10330 - 4.6316922)} \approx \frac{c^{3} \times 1 0^{- 10} \times 3055 s}{2 \times 10330}$

As shown in step 6 above, a typical gravitational acceleration on earth is:

$g_{e a r t h} = \frac{c}{30550000 s} = \frac{299792458 \frac{m}{s}}{30550000 s} \approx 9.81 \frac{m}{s^{2}}$

Taking a ratio of the standard gravitational parameter with the gravitational acceleration:

$\frac{μ_{e a r t h}}{g_{e a r t h}} = \frac{30550000 s \times c^{3} \times 1 0^{- 10} \times 3055 s}{c \times 2 \times (10330 - 4.6316922)} \approx \frac{30550000 s \times c^{3} \times 1 0^{- 10} \times 3055 s}{c \times 2 \times 10330}$

Simplifying Terms:

$\frac{μ_{e a r t h}}{g_{e a r t h}} = \frac{(c \times 3.055 s)^{2}}{2 \times (10330 - 4.6316922)} \approx \frac{(c \times 3.055 s)^{2}}{2 \times 10330}$

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:

$r_{e a r t h} \approx \sqrt{\frac{μ_{e a r t h}}{g_{e a r t h}}} \approx \sqrt{\frac{(c \times 3.055 s)^{2}}{2 \times 10330}} \approx \frac{c \times 3.055 s}{\sqrt{2 \times 10330}} = 6371 k m$