Question

Earth orbits once around the sun every 365.25 days at an average radius of 1.5X10^11 m. Earth's mass is 6X10^24 kg. What distance does the Earth travel in a year? what is Earth's average centripetal acceleration?

Answer 1

(a) The distance the Earth travels every year is the perimeter of the orbit; since we have the average radius of the orbit, we can calculate the perimeter:
$d=2\pi r = 2 \pi (1.5 \cdot 10^{11}m)=9.4 \cdot 10^{11}m$

(b) The average centripetal acceleration is given by
$a_c = \frac{v^2}{r}$
where v is the average speed of the Earth, that we can find by dividing the distance covered by the Earth in one year (=the perimeter of the orbit) by the the time corresponding to 1 year:
t=1 year=365.25 days=3.2 \cdot 10^7 s
So the velocity is: 
$v= \frac{d}{t}= \frac{9.4 \cdot 10^{11}m}{3.2 \cdot 10^7 s} =2.94 \cdot 10^4 m/s$
And so the centripetal acceleration is:
$a_c = \frac{v^2}{r}= \frac{(2.94 \cdot 10^4 m/s)^2}{9.4 \cdot 10^{11}m} =9.2 \cdot 10^{-4} m/s^2$