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Chapter 3: Problem 39

Estimate the acceleration of the Moon, which completes a nearly circular orbit of \(385,000 \mathrm{km}\) radius in 27 days.

Short Answer

Expert verified
The Moon's acceleration is estimated to be \(0.0028 \mathrm{m/s^2}\).

Step by step solution

01

Calculate the Circumference

The first step is to find the total distance covered in one orbit, which is the circumference of the circle. The formula for the circumference of a circle is \(C = 2\pi r\), where \(r\) is the radius. We know the radius is \(385,000 \mathrm{km}\), so \(C = 2\pi(385,000) = 2,420,000 \mathrm{km}\).
02

Calculate the Speed

Next, calculate the average speed over one orbit. The speed is distance divided by time. The distance is the circumference, and the time is 27 days. However, to keep the units consistent, the time should be converted into seconds: \(27 \mathrm{days} \times 24 \mathrm{hours/day} \times 60 \mathrm{minutes/hour} \times 60 \mathrm{seconds/minute} = 2,332,800 \mathrm{seconds}\). So, \(v = \frac{2,420,000,000 \mathrm{meters}}{2,332,800 \mathrm{seconds}} = 1,037 \mathrm{m/s}\).
03

Calculate the Acceleration

Finally, we can calculate the acceleration using the centripetal acceleration formula \(a = \frac{v^2}{r}\). With \(v = 1,037 \mathrm{m/s}\) and \(r = 385,000,000 \mathrm{meters}\), we find that \(a = \frac{(1,037 \mathrm{m/s})^2}{385,000,000 \mathrm{m}} = 0.0028 \mathrm{m/s^2}\).

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