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Norma-Jean [14]
3 years ago
14

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.
Physics
1 answer:
Anni [7]3 years ago
3 0

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.

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