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⇤ ← Revision 1 as of 2016-06-07 19:17:09
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| Deletions are marked like this. | Additions are marked like this. |
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| A colleague proposes a simplification to the gravitational equation, using the system center of mass. | A colleague proposes a simplification to the gravitational equation, using the system center of mass. |
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| $ F_N ~ = ~ { G m_1 m_2 } \over { r ^ 2 } $ | $ F_N ~ = ~ \large { { G m_1 m_2 } \over { r ^ 2 } } $ |
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| $ r_1 $ and $ r_2 $ can be calculated from | $ r_1 $ and $ r_2 $ can be calculated from |
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| $ r_1 m_1 ~ = ~ r_2 m_2 $ and $ r_1 + r_2 ~ = ~ r $ | $ r_1 m_1 ~ = ~ r_2 m_2 ~ ~ ~ $ and $ ~ ~ ~ r_1 + r_2 ~ = ~ r $ |
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| With a little bit of algebra, we can solve for $ r_1 $ and $ r_2 $ : | With a little bit of algebra, we can solve for $ r_1 ~ ~ ~ $ and $ ~ ~ ~ r_2 $ : |
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| $ r_1 ~ = ~ r \over { 1 + m_2 / m_1 } $ and $ r_2 ~ = ~ r \over { 1 + m_1 / m_2 } $ | $ r_1 = \large { r \over { 1 + m_2 / m_1 } } ~ ~ ~ $ and $ ~ ~ ~ r_2 = \large { r \over { 1 + m_1 / m_2 } } $ |
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| $ F_? ~ = ~ G left( { m_1 \over r_1 } right) left( { m_2 \over r_2 } right) $ . . . ???? | $ F_? = G \large { \left( { m_1 \over r_1 } \right) \left( { m_2 \over r_2 } \right) } $ . . . ???? |
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| $ F_? ~ = ~ { G left( { m_1 ( 1 + m_2 / m_1 ) \over r } right) left( { m_2 ( 1 + m_1 / m_2 ) \over r } right) $ . . . ???? | $ F_? = G \large { \left( { m_1 ( 1 + m_2 / m_1 ) \over r } \right) \left( { m_2 ( 1 + m_1 / m_2 ) \over r } \right) } $ . . . ???? |
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| $ F_? ~ = ~ { left( { G m_1 m_2 } \over { r ^ 2 } } right) ( 1 + m_2 / m_1 ) ( 1 + m_1 / m_2 ) $ . . . ???? | $ F_? = { \large { \left( { { G m_1 m_2 } \over { r ^ 2 } } \right) } } ( 1 + m_2 / m_1 ) ( 1 + m_1 / m_2 ) $ . . . ???? |
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| Define $ b $, the ratio of the masses, as | Define $ b $, the ratio of the masses, as |
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| $ b ~ == ~ m_1 / m_2 $ | $ b = m_1 / m_2 $ |
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| $ E ~ = ~ ( 1 + m_2 / m_1 ) ( 1 + m_1 / m_2 ) $ | $ E = ( 1 + m_2 / m_1 ) ( 1 + m_1 / m_2 ) $ |
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| If $ b = 1 $, then $ E ~ = ~ 4 $ if $ b = 2 $ or $ b = 0.5 $, then E ~ = ~ 4.5 $. For very large $ b $, $ E ~ \approx ~ 2 + b $, and for very small $ b $, $ E ~ \approx ~ 2 + 1 / b $ | If $ b = 1 $ then $ E = 4 $, if $ b = 2 $ or $ b = 0.5 $ then $ E = 4.5 $. For very large $ b $, $ E \approx 2 + b $, and for very small $ b $, $ E \approx 2 + 1 / b $ |
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| || $ b $ || $ E $ || | || $ b $ || $ E $ || |
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| and the acceleration of $ m_1 $ is | and the acceleration of $ m_1 $ is |
A Bad Gravitational Approximation
A colleague proposes a simplification to the gravitational equation, using the system center of mass.
Assume two masses m_1 and m_2 separated by distance r. According to standard Newtonian physics, the force between them is:
F_N ~ = ~ \large { { G m_1 m_2 } \over { r ^ 2 } }
If r_1 is the distance from m_1 to the center of mass of the system, and r_2 the distance from COM to m_2 , r_1 and r_2 can be calculated from
r_1 m_1 ~ = ~ r_2 m_2 ~ ~ ~ and ~ ~ ~ r_1 + r_2 ~ = ~ r
With a little bit of algebra, we can solve for r_1 ~ ~ ~ and ~ ~ ~ r_2 :
r_1 = \large { r \over { 1 + m_2 / m_1 } } ~ ~ ~ and ~ ~ ~ r_2 = \large { r \over { 1 + m_1 / m_2 } }
My colleague (incorrectly) claims that the force can be calculated with:
F_? = G \large { \left( { m_1 \over r_1 } \right) \left( { m_2 \over r_2 } \right) } . . . ????
Substituting the equations for r_1 and r_2 we get:
F_? = G \large { \left( { m_1 ( 1 + m_2 / m_1 ) \over r } \right) \left( { m_2 ( 1 + m_1 / m_2 ) \over r } \right) } . . . ????
Simplifying:
F_? = { \large { \left( { { G m_1 m_2 } \over { r ^ 2 } } \right) } } ( 1 + m_2 / m_1 ) ( 1 + m_1 / m_2 ) . . . ????
... which is never less than 4 times the actual Newtonian gravitational force.
Define b , the ratio of the masses, as
b = m_1 / m_2
Define the error factor E :
E = ( 1 + m_2 / m_1 ) ( 1 + m_1 / m_2 )
F_? ~ = ~ F_N \times E
If b = 1 then E = 4 , if b = 2 or b = 0.5 then E = 4.5 . For very large b , E \approx 2 + b , and for very small b , E \approx 2 + 1 / b
b |
E |
0.001 |
1002.001 |
0.01 |
102.01 |
0.1 |
12.1 |
0.3 |
5.6333... |
1.0 |
4.0 |
3.0 |
5.3333... |
10.0 |
12.1 |
100.0 |
102.01 |
1000.0 |
1002.001 |
For very small m_1 compared to m_2 , the F_? "force" becomes:
F_? ~ \approx ~ G \left( { {m_2}^2 \over { r^2 } } \right)
and the acceleration of m_1 is
a_? ~ = ~ F_? / m_1 ~ \approx ~ G \left( { {m_2}^2 \over { m_1 ~ r^2 } } \right) = a_N * { m_2 / m_1 }
The force difference between an electron and a proton could tear a hydrogen atom apart. Not very likely.