Question

The diameter of Earth is \(12,742\) kilometers; the diameter of the Moon is 3476 kilometers. a. If you flew around Earth by following the equator at a height of 10 kilometers, how many trips around the Moon could you take in the same amount of time, at the same height from the Moon, and at the same speed? Explain.

Short answer

Answer: Approximately 3.65 trips.

Step-by-step solution

Calculate the circumference of Earth and Moon's paths

First of all, we need to find the circumference of the path around the Earth and Moon, taking into account the 10 kilometers height. We'll use the formula for the circumference of a circle, which is: Circumference = \(2\pi r\) where "r" is the radius. For Earth: Diameter + 2 x Height = \(12742 + 2 \times 10 = 12762\) km Radius of Earth's path = \(\frac{12762}{2} = 6381\) km Circumference of Earth's path = \(2\pi \times 6381 \approx 40074.5\) km For the Moon: Diameter + 2 x Height = \(3476 + 2 \times 10=3496\) km Radius of Moon's path = \(\frac{3496}{2} = 1748\) km Circumference of Moon's path = \(2\pi \times 1748 \approx 10986.5\) km

Calculate the number of Moon trips

Now, we need to find how many times the Moon's path length fits into the Earth's path length, considering they are traveled at the same speed. To do that, we'll divide the circumference of Earth's path by the circumference of the Moon's path: Number of trips around the Moon = \(\frac{Circumference\, of\, Earth's\, path}{Circumference \,of\, Moon's\, path} = \frac{40074.5}{10986.5} \approx 3.65\) In the same amount of time, at the same speed and height, one can take approximately 3.65 trips around the Moon while taking one trip around the Earth following the equator.