Question

The planet Earth orbits the sun with a period of one year. In this problem, you may assume the radius of the Earths orbit is 1.50x10^8 km.

(A) Calculate the Earths average speed relative to the sun in meters per second.
(B) what is the earths average velocity relative to the sun over a period of one year in meters per second?

Answer 1

Solution (A): The earth's orbit around the sun insignificantly differs from a circular one. This means that its path is equal to the circumference of the circle $S=2\pi R=2\cdot 3.14\cdot 1.50\cdot 10^8km=9.42\cdot 10^{11}m$. This is the path the Earth travels in the year in which it is contained $t=1year=365 [day]\cdot 24 [hour]\cdot 60 [min]\cdot 60s=3.15\cdot 10^7 s$

By determining the average speed we write $v=\frac{S}{t}=\frac{9.42\cdot 10^{11}m}{3.15\cdot 10^7 s}=3.0\cdot 10^4m/s$ .

Solution (B): Relative to the sun the earth each year returns to the same point from which it started moving a year ago. Therefore relative to the sun it passes zero distance $\vec D=\vec 0 m$ per year. By determining the average velocity we write $\vec V=\frac {\vec D}{t}=\vec 0$ m/s.

Answers: (A)  the Earths average speed relative to the sun in meters per second is $3.0\cdot 10^4m/s$

(B) The earths average velocity relative to the sun over a period of one year is $0 m/s$