Question

From Kepler's third law, an asteroid with an orbital period of 8 years lies at an average distance from the Sun equal to:

Answer 1

Answer:

The correct option is (B).

Explanation:

The Kepler's third law of motion gives the relationship between the orbital time period and the distance from the semi major axis such that,

$T^2\propto a^3\\\\T^2=ka^3$

It is mentioned that, an asteroid with an orbital period of 8 years. So,

$(8)^2=ka^3\\\\64=ka^3\\\\a=(64)^{\dfrac{1}{3}}\\\\a=4\ AU$

So, an asteroid with an orbital period of 8 years lies at an average distance from the Sun equal to 4 astronomical units.