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Size: 3262
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Size: 3070
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| Deletions are marked like this. | Additions are marked like this. |
| Line 62: | Line 62: |
| || 1000.0 || 1002.001 || | || 1000.0 || 1002.001 || Sun to Jupiter ratio, approximately || || 1.2e47 || 1.2e47 || Sun to hydrogen atom || |
| Line 72: | Line 73: |
| The force difference between an electron and a proton near a star could tear a hydrogen atom apart. Not very likely ... except oopsie, the $ F_? $ and center-of-mass calculation just got quite a bit more difficult, especially if the electron, proton, and star are not collinear. |
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:
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
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 )
so that
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 |
Sun to Jupiter ratio, approximately |
1.2e47 |
1.2e47 |
Sun to hydrogen atom |
For very small m_1 compared to m_2 , the F_? "force" becomes:
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
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. 1000 times the acceleration means 32 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.