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The planet Earth orbits the sun with a period of one year. In this problem, you may assume the radius of the Earth's orbit is 1.50 × 10^8 km. what is the average velocity of relative to the sun over the period of one year in meters per second.

User Yue Jiang
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1 Answer

14 votes
14 votes

ANSWER:

The velocity is 29884.89 m/s

The avergae velocity is 0 m/s

Explanation:

We know that the speed is given by the distance for a certain time (in this case a year).

We know that the distance traveled would be equal to the circumference of the circle formed, which can be calculated by means of the radius.

We have the following equivalences:

1 km = 1000 m

1 year = 365 days

1 day = 24 hours

1 hour = 60 minutes

1 minute = 60 seconds

Therefore, we calculate the velocity as follows:


\begin{gathered} v=(d)/(t) \\ d=2\cdot\pi\cdot r=2\cdot3.1415\cdot1.5\cdot10^8\operatorname{km}\cdot\frac{1000\text{ m}}{1\text{ km}}=942450000000 \\ t=1\text{ yr}*\frac{365\text{ days}}{1\text{ yr}}*\frac{24\text{ hr}}{1\text{ day}}*\frac{60\text{ min}}{1\text{ hr}}*\frac{60\text{ sec}}{1\text{ min}}=31536000 \\ \text{ replacing} \\ v=(942450000000)/(31536000) \\ v=29884.89\text{ m/s } \end{gathered}

Now, since the initial and final position are the same, we would have:


\bar{v}=(\Delta x)/(\Delta t)=(x_f-x_i)/(\Delta t)=(0)/(\Delta t)=0\text{ m/s}

User Boyko Perfanov
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