Kepler's Third law in Newton's form helps us here. It states that for an object (a star in our case) orbiting another much more massive object (the galactic center)
`P^2/R^3 = (4pi^2)/(G M),`
where `P` is the orbital period, `R` is the radius of orbit, `M` is the mass of a central object and `G approx 6.7*10^(-11) (m^3)/(kg*s^2)` is the gravitational constant.
It is easy to find `M` from this equation, `M = ((2pi)/P)^2 R^3/G.` We know the speed `V,` it is obvious that `V = (2pi R)/P,` so `(2pi)/P = V/R` and the final formula is
`M = (V^2 * R)/G.`
To find the numeric answer we have to convert the given radius from light-days into meters. It is `22*2.6*10^(13) m` and the answer is
`((1200000)^2*22*2.6*10^(13)) / (6.7*10^(-11)) approx 12.3*10^(36) = 1.23*10^(37) (kg).`
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