Aplanet 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.

M = 1.38 10⁵⁹ kg

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

answer: its c lololo

radiation

explanation:

heat is transmitted by radiation