Question

Planet geos in orbit a distance of 1

Answer 1

Answer:

8 years

Explanation:

Kepler's third law states that the ratio between the cube of the distance of a planet from its star and the square of its orbital period is constant for all the planets orbiting around that star:

$\frac{d^3}{T^2}=const.$

where d is the distance of the planet from the star and T is the orbital period.

By applying this law to the two planets of this problem, we can write

$\frac{d_g^3}{T_g^2}=\frac{d_L^3}{T_L^2}$

where $d_g=1 AU$ is the distance of geos from the star, $T_g=1 y$ is its orbital period, $d_L=4 AU$ is the distance of logos from the star. Re-arranging the equation , we can find $T_L$, the orbital period of logos around the star:

$T_L=\sqrt{\frac{T_g^2 d_L^3}{d_Lg^3}}=\sqrt{\frac{(1 y)^2 (4 AU)^3}{(1 AU)^3}}=\sqrt{4^3}=8 years$

Answer 2

Planet Geos in orbit a distance of 1 A.U. (astronomical unit) from the star Astra has an orbital period of 1 "year." If planet Logos is 4 A.U. from Astra, how long does Logos require for a complete orbit?

TB = 8 years