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"
Comments
This helped a ton! Thanks so much!! :)
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