A satellite is arbithng a planet of a fathes of 5500 km from the planet's centen If Its ofbital veloelty \( 18250 \mathrm{~m} / \mathrm{s} \) deternine the inme pethat of the sotellite's motlon in hours

To determine the time period of the satellite's motion, we can use the formula for the orbital period \( T \) of a satellite, which is given by: \[ T = \frac{2\pi r}{v} \] where: - \( T \) is the orbital period, - \( r \) is the radius of the orbit (distance from the center of the planet to the satellite), - \( v \) is the orbital velocity. ### Step 1: Extract Known Conditions - The radius \( r \) is given as \( 5500 \) km, which we need to convert to meters: \[ r = 5500 \, \text{km} = 5500 \times 1000 \, \text{m} = 5,500,000 \, \text{m} \] - The orbital velocity \( v \) is given as \( 18250 \, \text{m/s} \). ### Step 2: Calculate the Orbital Period Now we can substitute the values into the formula: \[ T = \frac{2\pi (5,500,000)}{18250} \] Let's calculate this value. Calculate the value by following steps: - step0: Calculate: \(\frac{\left(2\pi \times 5500000\right)}{18250}\) - step1: Remove the parentheses: \(\frac{2\pi \times 5500000}{18250}\) - step2: Multiply the terms: \(\frac{11000000\pi }{18250}\) - step3: Reduce the fraction: \(\frac{44000\pi }{73}\) The calculated orbital period \( T \) is approximately \( 1893.56 \) seconds. ### Step 3: Convert Seconds to Hours To convert the period from seconds to hours, we use the conversion factor \( 1 \, \text{hour} = 3600 \, \text{seconds} \): \[ T_{\text{hours}} = \frac{T_{\text{seconds}}}{3600} = \frac{1893.56}{3600} \] Now, let's calculate this value. Calculate the value by following steps: - step0: Calculate: \(\frac{1893.56}{3600}\) - step1: Convert the expressions: \(\frac{\frac{47339}{25}}{3600}\) - step2: Multiply by the reciprocal: \(\frac{47339}{25}\times \frac{1}{3600}\) - step3: Multiply the fractions: \(\frac{47339}{25\times 3600}\) - step4: Multiply: \(\frac{47339}{90000}\) The orbital period of the satellite in hours is approximately \( 0.526 \) hours. ### Final Answer The time period of the satellite's motion is approximately \( 0.526 \) hours.