Question

The earth makes one complete revolution around the sun in 365.24 days. Assuming that the orbit of the earth is circular and has a radius of 93,000,000 mi, determine the speedand magnitude of the acceleration of the earth, while treating earth as a point particle.[Hint: you can find the angular speed]Please be consistent with the units.

Answer 1

Answer:

The magnitudes of the speed and acceleration of the Earth are approximately 66,661.217 miles per hour and 47.782 miles per square hour, respectively.

Explanation:

Given that the Earth has a circular orbit and make a revolution at constant speed around the Sun. Then, the kinematic formulas for the speed and acceleration of the planet are, respectively:

Speed

$v = \frac{2\pi\cdot R}{T}$ (1)

Acceleration

$a = \frac{4\pi^{2}\cdot R}{T^{2}}$ (2)

Where:

$v$ - Speed of the planet, measured in miles per hour.

$a$ - Acceleration of the planet, measured in miles per square hour.

$R$ - Radius of the orbit, measured in miles.

$T$ - Period of rotation, measured in hours.

If we know that $R = 93,000,000\,mi$ and $T = 8,765.76\,h$, then the magnitudes of the speed and acceleration of the planet is:

$v = \frac{2\pi\cdot (93,000,000\,mi)}{8,765.76\,h}$

$v \approx 66,661.217\,\frac{mi}{h}$

$a = \frac{4\pi^{2}\cdot (93,000,000\,mi)}{(8,765.76\,h)^{2}}$

$a\approx 47.782\,\frac{mi}{h^{2}}$

The magnitudes of the speed and acceleration of the Earth are approximately 66,661.217 miles per hour and 47.782 miles per square hour, respectively.