Explanation:
Given:
[tex]r_a = 3.570R_E[/tex]
[tex]R_E = 1.499×10^{11}\:\text{m}[/tex]
[tex]M_S = 1.989×10^{30}\:\text{kg}[/tex]
[tex]G = 6.674×10^{-11}\:\text{N-m}^2\text{/kg}^2[/tex]
Let [tex]m_s[/tex]= mass of the asteroid and [tex]r_a[/tex] = orbital radius of the asteroid around the sun. The centripetal force [tex]F_c[/tex] is equal to the gravitational force [tex]F_G:[/tex]
[tex]F_c = F_G \Rightarrow m_a\dfrac{v_a^2}{r_a} = G\dfrac{m_aM_S}{r_a^2}[/tex]
or
[tex]\dfrac{4\pi^2 r_a}{T^2} = G\dfrac{M_S}{r_a^2}[/tex]
where
[tex]v = \dfrac{2\pi r_a}{T}[/tex]
with T = period of orbit. Rearranging the variables, we get
[tex]T^2 = \dfrac{4\pi^2 r_a^3}{GM_S}[/tex]
Taking the square root,
[tex]T = 2\pi \sqrt{\dfrac{r_a^3}{GM_S}}[/tex]
[tex]\:\:\:\:=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})}}[/tex]
[tex]\:\:\:\:= 2.13×10^8\:\text{s} = 6.75\:\text{years}[/tex]
