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 partsperbillion accuracy. My colleague's "approximation" is incorrect, yet more difficult to solve.