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Yes.

A version of Kepler's 3rd Law is: The square of the period of revolution of

a planet (or asteroid) around the Sun (in Earth years) is equal to the cube of its average distance (in AU from) the Sun.

In this case that means 8 squared (64) should equal 4 cubed (64).

It does.

So, the asteroid is keeping to the Law. Thank goodness for that.

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Q: If an asteroid is 4 AU from the sun and its period of evolution around the sun is 8 years does it obey Kepler's law?
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