Question

g question, An asteroid is discovered in a nearly circular orbit around the Sun, with an orbital radius that is 3.83 times Earth's. What is the asteroid's orbital period T , its "year," in terms of Earth years

Answer 1

Therefore the asteroids's orbital period is 7.45 years.

Step-by-step explanation:

Kapler's third law:

The orbital period of a planet squared is directly proportional to the average distance of the planet from the sun cubed.

`T^2\propto r^3`

T = orbital period of the planet

r = orbital radius of the planet

`\therefore (T_1^2)/(T_2^2)=(R_1^3)/(R_2^3)`

Here
`T_1` = Orbital period of the asteroid

`R_1`= Orbital radius of the asteroid

`T_2=` Orbital period of Earth = 1 year

`R_2` =Orbital radius of Earth

Given that the orbital radius of asteroid is 3.83 times of orbital radius of Earth.

`R_1 = 3.83R_2`

`(R_1)/(R_2)=3.83`

Therefore

`\therefore (T_1^2)/(T_2^2)=(R_1^3)/(R_2^3)`

`\Rightarrow ( (T_1)/(T_2))^2=((R_1)/(R_2))^3`

`\Rightarrow ( (T_1)/(T_2))^2=(3.83)^3`

`\Rightarrow (T_1)/(T_2)=(3.83)^\frac32`

`\Rightarrow {T_1=(3.83)^\frac32* T_2`

`\Rightarrow T_1= 7.45* (1 \ year)` [
`T_2=` 1 year]

`\Rightarrow T_1= 7.45 \ years`

Therefore the asteroids's orbital period is 7.45 years.