Question

An asteroid is discovered in a nearly circular orbit around the Sun, with an orbital radius that is 3.570 times Earth's. What is the
asteroid's orbital period, in terms of Earth years?
orbital period:
years

Answer 1

Explanation:

Given:

$r_a = 3.570R_E$

$R_E = 1.499×10^{11}\:\text{m}$

$M_S = 1.989×10^{30}\:\text{kg}$

$G = 6.674×10^{-11}\:\text{N-m}^2\text{/kg}^2$

Let $m_s$= mass of the asteroid and $r_a$ = orbital radius of the asteroid around the sun. The centripetal force $F_c$ is equal to the gravitational force $F_G:$

$F_c = F_G \Rightarrow m_a\dfrac{v_a^2}{r_a} = G\dfrac{m_aM_S}{r_a^2}$

or

$\dfrac{4\pi^2 r_a}{T^2} = G\dfrac{M_S}{r_a^2}$

where

$v = \dfrac{2\pi r_a}{T}$

with T = period of orbit. Rearranging the variables, we get

$T^2 = \dfrac{4\pi^2 r_a^3}{GM_S}$

Taking the square root,

$T = 2\pi \sqrt{\dfrac{r_a^3}{GM_S}}$

$\:\:\:\:=2\pi \sqrt{\dfrac{(3.57(1.499×10^{11}\:\text{m}))^3}{(6.674×10^{-11}\:\text{N-m}^2\text{/kg}^2)(1.989×10^{30}\:\text{kg})}}$

$\:\:\:\:= 2.13×10^8\:\text{s} = 6.75\:\text{years}$