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 From Kepler's third law, an asteroid with an orbital period of 8 years lies at an average distance from the Sun equal to 16 astronomical units. 2 astronomical units. 8 astronomical units. 4 astronomical units. It depends on the asteroid's mass.

Answer 1

Answer: 4 AU

Kepler's third law states that square of orbital period of a planet is proportional to the cube of the average distance from the Sun.

$T^2 \propto r^3$

$\Rightarrow T^2=\frac {4\pi}{GM}r^3$

where, T is the orbital period, r is the average distance from Sun, G is the gravitational constant and M is the mass of the Sun.

If T is in Earth years, M is in Solar mass (1 Solar mass = mass of sun), r is in AU (astronomical unit) then:

$T^2=r^3$

$\Rightarrow r=T^{2/3}$

$\Rightarrow r=8^{2/3}=4 AU$