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← Revision 20 as of 2016-06-08 23:37:34 ⇥
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| Deletions are marked like this. | Additions are marked like this. |
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| = A Bad Gravitational Approximation = | = An Inaccurate Gravitational Approximation = |
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| $ F_N ~ = ~ \large { { G m_1 m_2 } \over { r ^ 2 } } $ | 1) $ F_N ~ = ~ \Large { { G m_1 m_2 } \over { r ^ 2 } } $ |
| Line 10: | Line 10: |
| 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 $, | If $ r_1 $ is the distance from $ m_1 $ to the center of mass of the system, and $ r_2 $ the distance from the center of mass to $ m_2 $, |
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| $ r_1 m_1 ~ = ~ r_2 m_2 ~ ~ ~ $ and $ ~ ~ ~ r_1 + r_2 ~ = ~ r $ | 2a) $ r_1 m_1 ~ = ~ r_2 m_2 ~ ~ ~ $ and 2b) $ ~ ~ ~ r_1 + r_2 ~ = ~ r $ |
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| $ r_1 = \large { r \over { 1 + m_2 / m_1 } } ~ ~ ~ $ and $ ~ ~ ~ r_2 = \large { r \over { 1 + m_1 / m_2 } } $ | 3) $ 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 \large { \left( { m_1 \over r_1 } \right) \left( { m_2 \over r_2 } \right) } $ . . . ???? | 4) $ F_? = G \Large { \left( { m_1 \over r_1 } \right) \left( { m_2 \over r_2 } \right) } $ . . . ???? |
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| $ 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) } $ . . . ???? | 5) $ 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_? = { \large { \left( { { G m_1 m_2 } \over { r ^ 2 } } \right) } } ( 1 + m_2 / m_1 ) ( 1 + m_1 / m_2 ) $ . . . ???? | 6) $ 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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| $ b = m_1 / m_2 $ | 7) $ b = m_1 / m_2 $ |
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| $ E = ( 1 + m_2 / m_1 ) ( 1 + m_1 / m_2 ) $ | 8) $ E = ( 1 + m_2 / m_1 ) ( 1 + m_1 / m_2 ) $ |
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| $ F_? ~ = ~ F_N \times E $ | so that |
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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 $ | 9) $ F_? ~ = ~ F_N \times E $ |
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| || $ b $ || $ E $ || | If $ b = 1 $ then $ E = 4 $. If $ b = 2 $ or $ b = 0.5 $ then $ E = 4.5 $. For large $ b $, $ E \approx 2 + b \approx \approx b $, and for small $ b $, $ E \approx 2 + 1 / b \approx \approx 1/b $. || mass ratio $ b $ || force ratio $ E $ || |
| Line 49: | Line 55: |
| || 0.3 || 5.6333...|| | || 0.2 || 7.2 || || 0.5 || 4.5 || |
| Line 51: | Line 58: |
| || 3.0 || 5.3333...|| | || 2.0 || 4.5 || || 5.0 || 7.2 || |
| Line 55: | Line 63: |
| || 1047.4 || 1049.4 || Sun to Jupiter ratio || || 1e33 || 1e33 || Sun to sand grain ratio || || 1.2e47 || 1.2e47 || Sun to hydrogen atom ratio || |
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| $ F_? ~ \approx ~ G \left( { {m_2}^2 \over { r^2 } } \right) $ | 10) $ F_? \approx F_N / b \approx F_N m_2 / m_1 \approx G \left( \Large { {m_2}^2 \over { r^2 } } \right) $ |
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| $ a_? ~ = ~ F_? / m_1 ~ \approx ~ G \left( { {m_2}^2 \over { m_1 ~ r^2 } } \right) = a_N * { m_2 / m_1 } $ | 11) $ a_? = F_? / m_1 \approx G \left( \Large { {m_2}^2 \over { m_1 ~ r^2 } } \right) = a_N \times { m_2 / m_1 } $ |
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| The force difference between an electron and a proton could tear a hydrogen atom apart. Not very likely. | A circular orbit has a centripedal acceleration $ a = v^2 / r $, so the orbital velocity is proportional to the square root of acceleration. 1047 times the acceleration means 32.4 times the orbital velocity. The unrestricted 3 body problem is very difficult to solve - approximation and computers are needed, but are good enough to deliver space probes to other planets with parts-per-billion accuracy. My colleague's "approximation" is incorrect, yet more difficult to solve. |
An Inaccurate 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:
1) 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 the center of mass to m_2 , r_1 and r_2 can be calculated from
2a) r_1 m_1 ~ = ~ r_2 m_2 ~ ~ ~ and 2b) ~ ~ ~ r_1 + r_2 ~ = ~ r
With a little bit of algebra, we can solve for r_1 ~ ~ ~ and ~ ~ ~ r_2 :
3) 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:
4) 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:
5) 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:
6) 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
7) b = m_1 / m_2
Define the error factor E :
8) E = ( 1 + m_2 / m_1 ) ( 1 + m_1 / m_2 )
so that
9) F_? ~ = ~ F_N \times E
If b = 1 then E = 4 .
If b = 2 or b = 0.5 then E = 4.5 .
For large b , E \approx 2 + b \approx \approx b , and for small b , E \approx 2 + 1 / b \approx \approx 1/b .
mass ratio b |
force ratio E |
|
0.001 |
1002.001 |
|
0.01 |
102.01 |
|
0.1 |
12.1 |
|
0.2 |
7.2 |
|
0.5 |
4.5 |
|
1.0 |
4.0 |
|
2.0 |
4.5 |
|
5.0 |
7.2 |
|
10.0 |
12.1 |
|
100.0 |
102.01 |
|
1000.0 |
1002.001 |
|
1047.4 |
1049.4 |
Sun to Jupiter ratio |
1e33 |
1e33 |
Sun to sand grain ratio |
1.2e47 |
1.2e47 |
Sun to hydrogen atom ratio |
For very small m_1 compared to m_2 , the F_? "force" becomes:
10) F_? \approx F_N / b \approx F_N m_2 / m_1 \approx G \left( \Large { {m_2}^2 \over { r^2 } } \right)
and the acceleration of m_1 is
11) a_? = F_? / m_1 \approx G \left( \Large { {m_2}^2 \over { m_1 ~ r^2 } } \right) = a_N \times { m_2 / m_1 }
A circular orbit has a centripedal acceleration a = v^2 / r , so the orbital velocity is proportional to the square root of acceleration. 1047 times the acceleration means 32.4 times the orbital velocity.
The unrestricted 3 body problem is very difficult to solve - approximation and computers are needed, but are good enough to deliver space probes to other planets with parts-per-billion accuracy. My colleague's "approximation" is incorrect, yet more difficult to solve.