3 The time period \(T\) of a satellite in circular orbit around a planet satisfies the equation
\(G M T ^ { 2 } = 4 \pi ^ { 2 } R ^ { 3 }\),
where

• \(G\) is the universal gravitational constant,
• \(M\) is the mass of the planet,
• \(\quad R\) is the radius of the orbital circle.
  1. Find the dimensions of \(G\).

A student suggests the following formula to model the approach speed between two orbiting bodies.
\(v = k G { } ^ { \alpha } { } ^ { \beta } { } _ { r } \gamma _ { m _ { 1 } } m _ { 2 } \left( m _ { 1 } + m _ { 2 } \right)\),
where

• \(\quad v\) is the approach speed of the two bodies,
• \(k\) is a dimensionless constant,
• \(\quad c\) is the speed of light,
• \(\quad r\) is the distance between the two bodies,
• \(\quad m _ { 1 }\) and \(m _ { 2 }\) are the masses of the bodies.
• Use dimensional analysis to determine the values of \(\alpha , \beta\) and \(\gamma\).
• Calculate, according to the student's model, how many times greater the approach speed is between a pair of stars which are 6.13 light-years apart and the same pair of stars if they were 8.64 light-years apart. (A light-year is a unit of distance.)