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 .

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.