Spacecraft could create a type of artificial gravity by spinning fast enough to create a radial acceleration equal to 9.8 m/s2. If a giant space station with a mass of 2.8 x 1017 kg and a radius of 5.05 x 106 m wanted to create that type of artificial gravity, what angular velocity would be required? O 1.39 x 103 rad/s 7030 rad/s 4.95 x 107 rad/s O 1.94 x 10-6 rad/s

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### Artificial Gravity in Spacecraft

Spacecraft could create a type of artificial gravity by spinning fast enough to create a radial acceleration equal to 9.8 m/s². 

#### Problem Statement:
If a giant space station with a mass of \(2.8 \times 10^{17}\) kg and a radius of \(5.05 \times 10^6\) m wanted to create that type of artificial gravity, what angular velocity would be required?

#### Answer Choices:
- \(1.39 \times 10^{-3} \text{ rad/s}\)
- \(7030 \text{ rad/s}\)
- \(4.95 \times 10^7 \text{ rad/s}\)
- \(1.94 \times 10^{-6} \text{ rad/s}\)

To solve this, we can use the formula for radial (centripetal) acceleration \(a_r = \omega^2 r\), where \(\omega\) is the angular velocity and \(r\) is the radius.

Given:
- Radial acceleration, \(a_r = 9.8 \text{ m/s}^2\)
- Radius, \(r = 5.05 \times 10^6 \text{ m}\)

Rearranging the formula to solve for \(\omega\):
\[ \omega = \sqrt{\frac{a_r}{r}} \]

Substituting the given values:
\[ \omega = \sqrt{\frac{9.8}{5.05 \times 10^6}} \]

Performing the calculation:
\[ \omega \approx 1.39 \times 10^{-3} \text{ rad/s} \]

Thus, the correct answer is:
- \(1.39 \times 10^{-3} \text{ rad/s}\)
Transcribed Image Text:### Artificial Gravity in Spacecraft Spacecraft could create a type of artificial gravity by spinning fast enough to create a radial acceleration equal to 9.8 m/s². #### Problem Statement: If a giant space station with a mass of \(2.8 \times 10^{17}\) kg and a radius of \(5.05 \times 10^6\) m wanted to create that type of artificial gravity, what angular velocity would be required? #### Answer Choices: - \(1.39 \times 10^{-3} \text{ rad/s}\) - \(7030 \text{ rad/s}\) - \(4.95 \times 10^7 \text{ rad/s}\) - \(1.94 \times 10^{-6} \text{ rad/s}\) To solve this, we can use the formula for radial (centripetal) acceleration \(a_r = \omega^2 r\), where \(\omega\) is the angular velocity and \(r\) is the radius. Given: - Radial acceleration, \(a_r = 9.8 \text{ m/s}^2\) - Radius, \(r = 5.05 \times 10^6 \text{ m}\) Rearranging the formula to solve for \(\omega\): \[ \omega = \sqrt{\frac{a_r}{r}} \] Substituting the given values: \[ \omega = \sqrt{\frac{9.8}{5.05 \times 10^6}} \] Performing the calculation: \[ \omega \approx 1.39 \times 10^{-3} \text{ rad/s} \] Thus, the correct answer is: - \(1.39 \times 10^{-3} \text{ rad/s}\)
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