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A planet with a mass of 3.0 x 1023kg orbits 6.0 x 1010m from a star with a period of 3.4 x 1011seconds. Determine the mass of the star.

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Answer:

M = 1.38 10⁵⁹ kg

Step-by-step explanation:

For this problem we will use the law of universal gravitation

F = G m₁ m₂ / r²

Where G is the gravitation constant you are value 6.67 10⁻¹¹ N m2 / kg2, m are the masses and r the distance

In this case the mass of the planet is m = 3.0 10²³ kg and the mass of the start is M

Let's write Newton's second law

F = m a

The acceleration is centripetal

a = v² / r

The speed module is constant, so we can use the kinematic relationship

v = d / t

The distance remembered is the length of the circular orbit and the time in this case is called the period

d = 2π r

a = 2π r / T

Let's replace Newton's second law

G m M / r² = m (4π² r² / T²) / r

G M = 4 π² r³ / T²

M = 4 π² r³ / T² G

Let's calculate

M = 4 π² (3.0 10²³)³ / (3.4 10¹¹)² 6.67 10⁻¹¹

M = 13.82 10⁵⁸ kg

M = 1.38 10⁵⁹ kg

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