An asteroid is discovered in a nearly circular orbit around the sun, with an orbital radius that is 2.47 times earth's. what is the asteroid's orbital period, its "year," in terms of earth years?

Explanation:

This problem can be solved using the Third Kepler’s Law of Planetary motion:

“The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”.

This law states a relation between the orbital period of a body (the asteroid in this case) orbiting a greater body in space (the Sun in this case) with the size of its orbit:

   (1)

Where;

is the Gravitational Constant and its value is

is the mass of the Sun

 is orbital radius of the orbit the asteroid describes around the Sun.

Now, if we want to find the period, we have to express equation (1) as written below and substitute all the values:

  (2)

  (3)

  (4) This is the period in seconds, but we were asked to find it in Earth years. So, we have to make the conversion:

Finally:

 This is the period of the asteroid around the Sun in Earth years.

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