Asteroid in a circular orbit about the Sun

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To determine the orbital period of an asteroid in a circular orbit around the Sun at a radius of 3.04 AU, Kepler's Third Law can be applied, which states that the square of the orbital period (T) is proportional to the cube of the semi-major axis (r) of its orbit. The relevant equation is T² = k * r³, where k is a constant that can be derived from Earth's known values. Given that Earth's period is 365.25 days at 1 AU, k can be calculated, and then used to find the period for the asteroid. By substituting r = 3.04 AU into the equation, the period can be calculated, yielding an approximate period of 5.25 years, or about 1920 days. This method effectively combines gravitational principles with Kepler's laws to solve for the orbital period.
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If the Earth's period is 365.25 days and its distance to the (center of the) Sun is one astronomical unit (AU), what is the period, to the nearest day, of an asteroid in a circular orbit about the Sun with a radius of 3.04 AU?

please show steps so that I can learn.

thanks a lot.
 
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What are the relevant equations that you think are needed here?
Newton's law of gravitation maybe?
 
I think the equation for the period of a planet is needed; but I'm not sure how to use it.
 
Kepler's equation relating period to radius.
If this isn't given, it can be derived by equating the equation for centripetal force to the equation for universal gravitational force.
 
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