Answer
The asteroid's orbital radius is $4.37\times 10^{11}~m$
The asteroid's orbital speed is 17.4 km/s
Work Step by Step
First, we convert the orbital period to units of seconds.
$T = (5.0~years)[\frac{(365~days)(24~hours/day)(3600~s/hr)}{1~year}]$
$T = 1.577\times 10^8~s$
Then, we use the orbital period to find the orbital radius $R$. Let $M_s$ be the mass of the sun.
$T^2 = \frac{4\pi^2~R^3}{G~M_s}$
$R^3 = \frac{T^2~G~M_s}{4\pi^2}$
$R = (\frac{T^2~G~M_s}{4\pi^2})^{1/3}$
$R = (\frac{(1.577\times 10^8~s)^2(6.67\times 10^{-11}~m^3/kg~s^2)(1.99\times 10^{30}~kg}{4\pi^2})^{1/3}$
$R = 4.37\times 10^{11}~m$
The asteroid's orbital radius is $4.37\times 10^{11}~m$.
We then use the orbital radius to find the orbital speed.
$v = \sqrt{\frac{G~M_s}{R}}$
$v = \sqrt{\frac{(6.67\times 10^{-11}~m^3/kg~s^2)(1.99\times 10^{30}~kg)}{4.37\times 10^{11}~m}}$
$v = 1.74\times 10^4~m/s$
$v = 17.4~km/s$
The asteroid's orbital speed is 17.4 km/s.
