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

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.

Step-by-step 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 = (2\pi\cdot R)/(T)` (1)

Acceleration

`a = (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 = (2\pi\cdot (93,000,000\,mi))/(8,765.76\,h)`

`v \approx 66,661.217\,(mi)/(h)`

`a = (4\pi^(2)\cdot (93,000,000\,mi))/((8,765.76\,h)^(2))`

`a\approx 47.782\,(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.