Question

Assume the a star orbits the center of the Galaxy in 340 million years at a distance of 40,000 light-years. Given that D3=(M1+M2)×P2D3=(M1+M2)×P2, where DD is the orbital distance (in AUs) and PP is the orbital period (in years), what is the mass of the Galaxy within the star's orbit? (Hint: 1 light-year = 63,240 AU).

Answer 1

We know that ,

From generalised Kepler's Third Law , we have ,

MP² = a³ ,

where , P is the period of orbiting , ( in earth years )

M is the mass of the galaxy ( in solar mass )

and 'a' is the orbital distance , ( in AU )

So , here we have ,

P = 340 million years = 340,000,000 years

a = 40,000 light years = 40,000 * 63,240 AU = 2,529,600,000 AU

therefore,

Thus , the mass of the galaxy ,

Where , 1 solar mass = 2 10³⁰ kg

Hence , M = 2.8 10⁴¹ Kg ( approx.)