Question

Suppose that scientists discover an asteroid that is 2.0 AU from the sun. If 1 AU - 150 million km, how long would the asteroid's orbital period be in Earth years?

Answer 1

Answer:

2.84*10^-3 years = 0.00284*10^-3 years

Explanation:

To find the orbital period you use the following formula:

$T=\sqrt{\frac{4\pi^2r^3}{GM}}$

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

$T=\sqrt{\frac{4\pi^2(3*10^{11}m)^3}{(6.67*10^{-11}Nm^2/kg^2)(1.98*10^{30}kg)}}\\\\T=89839.27\ s$

Next, you calculate the time in seconds of one Earth's year:

$1\ year=365days*\frac{24\ h}{1\ day}*\frac{3600\ s}{1\ h}=31536000\ s$

thus, you use this value to find the orbital period of the asteroid in Earth's year:

$T=89839.27\ s*\frac{1\ year}{31536000\ s}=2.84*10^{-3}\ years$