Final answer:
The tangential speed of the Earth as it orbits the Sun is approximately 29,855 m/s, 107,478 km/h, and 66,787 mph, calculated by dividing the circumference of the Earth's orbit by the period of revolution.
Step-by-step explanation:
The question involves calculating the tangential speed of the Earth as it orbits the Sun. Given that the Earth makes a approximately circular path as it travels around the Sun at a radius of 1.5 x 10^11 meters and the period of revolution of the Earth around the Sun (365 days), we can find the tangential speed using the formula for the circumference of a circle (C = 2πr) and the time it takes to complete one orbit.
To find the tangential speed in meters per second, we first calculate the circumference of Earth's orbit: C = 2π(1.5 x 10^11 m). The period, in seconds, is 365 days × 24 hours/day × 3600 seconds/hour. Dividing the circumference by the period gives us the speed in meters per second.
For kilometers per hour (km/h), we take the speed in meters per second and multiply by 3.6 (since 1 m/s = 3.6 km/h). Finally, to convert to miles per hour (mph), we multiply the km/h by 0.621371 (since 1 km = 0.621371 miles).
Now let's do the calculations:
- Circumference: C = 2π(1.5 x 10^11 m) = 9.42 x 10^11 m
- Period: T = 365 days × 24 h/day × 3600 s/h = 3.156 x 10^7 s
- Tangential speed (m/s): V = C/T = 9.42 x 10^11 m / 3.156 x 10^7 s ≈ 29,855 m/s
- Tangential speed (km/h): V × 3.6 ≈ 107,478 km/h
- Tangential speed (mph): 107,478 km/h × 0.621371 ≈ 66,787 mph