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Chapter 5: Problem 56

A space station is shaped like a ring and rotates to simulate gravity. If the radius of the space station is \(120 \mathrm{m},\) at what frequency must it rotate so that it simulates Earth's gravity? [Hint: The apparent weight of the astronauts must be the same as their weight on Earth.]

Short Answer

Expert verified
Answer: The space station should rotate at a frequency of approximately 0.0125 Hz to simulate Earth's gravity.

Step by step solution

01

Determine the centripetal acceleration formula

The formula for centripetal acceleration (a_c) is given by: \(a_c = \frac{v^2}{r}\), where v is the linear speed of rotation and r is the radius of the circular path.
02

Relate centripetal acceleration with gravitational acceleration

We need the centripetal acceleration to be equal to Earth's gravitational acceleration (g) to simulate Earth's gravity. Therefore, we set the centripetal acceleration equal to g: \(a_c = g \Rightarrow \frac{v^2}{r} = g\)
03

Relate linear speed (v) to frequency (f)

The linear speed (v) can be expressed in terms of frequency (f) using the formula \(v = 2\pi rf\). Substitute this into our previous equation: \(\frac{(2\pi rf)^2}{r} = g\)
04

Solve the equation for frequency (f)

We need to find the frequency, so we will solve the above equation for f: \(\frac{(2\pi rf)^2}{r} = g \Rightarrow (2\pi rf)^2 = rg \Rightarrow f^2 = \frac{g}{(2\pi r)^2}\) Now take the square root: \(f = \sqrt{\frac{g}{(2\pi r)^2}}\)
05

Plug in the values and calculate the frequency

We are given the radius of the space station, \(r = 120 \,\text{m}\). Earth's gravitational acceleration, \(g = 9.81 \,\text{m/s}^2\). Substitute these values into the frequency equation: \(f = \sqrt{\frac{9.81}{(2\pi \cdot 120)^2}}\) Now, calculate the frequency: \(f \approx \sqrt{\frac{9.81}{(753.98)^2}} \approx 0.0125 \,\text{Hz}\) The space station must rotate at a frequency of approximately \(0.0125\) Hz to simulate Earth's gravity.

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