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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*10ā¶

Explanation:

Applying Newton version of Kepler's third law;

aĀ³ = (Mā‚ + Mā‚‚) Ɨ PĀ²

where a = semimajor axis and

Ā  Ā  Ā  Ā  Ā  Ā P = orbital period

Ā  Ā  Ā  Ā  Ā  Mā‚ = Sun's mass

Ā  Ā  Ā  Ā  Ā  Mā‚‚ = Planet's mass

We would assume that the factor Mā‚ + Mā‚‚ ā‰” Mā‚ (the mass around which the star moves in its Keplerian orbit) since the planetā€™s mass is so small by comparison

Therefore, Mā‚ = aĀ³ / PĀ²

Convert light days to Astronomical Unit (AU);

5.7 light days = 173*5.7AU=986.1AU

Mā‚ = 986.1Ā³ / 17.2Ā²

Mā‚ = 3.241*10ā¶

So the mass around which the star moves in its Keplerian orbit = 3.241*10ā¶

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