QuestionFind the Earth's orbital speed given a radius of 1.49 x $10^8$ km and a period of 5.25 days.

Studdy Solution

STEP 1

Assumptions1. The Earth revolves around the Sun in a circular orbit. . The radius of the orbit is $1.49 \times10^8$ km.
3. The period of the orbit is365.25 days.
4. We are asked to find the orbital speed of the Earth.

STEP 2

The orbital speed can be calculated using the formula for the speed of an object moving in a circle, which is the circumference of the circle divided by the time it takes to travel that distance (the period). The formula is $v = \frac{2\pi r}{}$ where- $v$ is the orbital speed, - $r$ is the radius of the orbit, - $$ is the period of the orbit.

STEP 3

First, we need to convert the radius from kilometers to meters because the standard unit of speed is m/s. $1 \text{ km} =1000 \text{ m}$ So, $r =1.49 \times10^8 \text{ km} \times1000 =1.49 \times10^{11} \text{ m}$

STEP 4

Next, we need to convert the period from days to seconds because the standard unit of time in physics is seconds.
$1 \text{ day} =24 \times60 \times60 \text{ seconds}$ So,
$=365.25 \text{ days} \times24 \times60 \times60 =3.15576 \times10^7 \text{ seconds}$

STEP 5

Now we can substitute the values of $r$ and $$ into the formula to calculate the orbital speed.
$v = \frac{2\pi \times1.49 \times10^{11}}{3.15576 \times10^7}$

STEP 6

Calculate the orbital speed.
$v \approx29,785.55 \text{ m/s}$ The orbital speed of the Earth is approximately $29,785.55$ m/s.

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