Answer:
2.84*10^-3 years = 0.00284*10^-3 years
Explanation:
To find the orbital period you use the following formula:
[tex]T=\sqrt{\frac{4\pi^2r^3}{GM}}[/tex]
r: 2.0A = 2.0(1.5*10^11m)=3*10^11 m
G: Cavendish constant = 6.67*10^-11 Nm^2/kg^2
M: mass of the sun = 1.98*10^30 kg
[tex]T=\sqrt{\frac{4\pi^2(3*10^{11}m)^3}{(6.67*10^{-11}Nm^2/kg^2)(1.98*10^{30}kg)}}\\\\T=89839.27\ s[/tex]
Next, you calculate the time in seconds of one Earth's year:
[tex]1\ year=365days*\frac{24\ h}{1\ day}*\frac{3600\ s}{1\ h}=31536000\ s[/tex]
thus, you use this value to find the orbital period of the asteroid in Earth's year:
[tex]T=89839.27\ s*\frac{1\ year}{31536000\ s}=2.84*10^{-3}\ years[/tex]
