Question

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.

Answer 1

ANSWER:

The velocity is 29884.89 m/s

The avergae velocity is 0 m/s

STEP-BY-STEP 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:

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

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

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