#format jsmath
= 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.