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Hitman42 [59]
2 years ago
14

A planet has been observed orbiting a nearby star. This star has a mass of 3.5 solar masses, and the planet is 4.2 AU from the s

tar. What is the planet’s orbital period in Earth years? Round your answer to the nearest whole number. __ Earth years
Physics
2 answers:
nadezda [96]2 years ago
7 1

Answer:

16 years just got it wrong for choosing 4.6 believe me if u want

Explanation:

kondaur [170]2 years ago
4 1

Answer:

4.6 years

Explanation:

This is solved using Kepler's third law which says:

T^2=\frac{4\pi ^2}{GM} a^2

Where

T = Orbital period of the planet (in seconds)

a = Distance from the star (in meters)

G = Gravitational constant

M = Mass of the parent star (in kg)

From the information given

M = 3.5M_{sun} = 6.96*10^{30} kg

a=4.2AU = 6.28*10^{11} meters

G = 6.67*10^{-11}m^3kg^{-1}s^{-2}

We put this into Kepler's law and get:

T=\sqrt{\frac{4\pi ^2}{6.67*10^{-11}*6.96*10^{30}} (6.28*10^{11})^3}=145,128,196 seconds.

This when converted to years is 4.6 years.

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Explanation:

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Answer:

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Good question

Keep thinking like this :)

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