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It is believed that, at the center of our galaxy, is a "super-massive" black hole. One datum that leads to this conclusion is the important recent observation of stellar motion in the vicinity of the galactic center. If one such star moves in an elliptical orbit with a period of 17.2 years and has a semi-major axis of 5.7 light-days (the distance light travels in 5.7 days), what is the mass around which the star moves in its Keplerian orbit

Respuesta :

Answer:

3.241*10⁶

Explanation:

Applying Newton version of Kepler's third law;

a³ = (M₁ + M₂) × P²

where a = semimajor axis and

           P = orbital period

          M₁ = Sun's mass

          M₂ = Planet's mass

We would assume that the factor M₁ + M₂ ≡ M₁ (the mass around which the star moves in its Keplerian orbit) since the planet’s mass is so small by comparison

Therefore, M₁ = a³ / P²

Convert light days to Astronomical Unit (AU);

5.7 light days = 173*5.7AU=986.1AU

M₁ = 986.1³ / 17.2²

M₁ = 3.241*10⁶

So the mass around which the star moves in its Keplerian orbit = 3.241*10⁶

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