Questions: A satellite going around the earth has an altitude (height above the earth's surface) of 3.5 × 10^6 m. How long would it take this satellite to go one full revolution around the earth? Remember, earth has a radius of 6.37 × 10^6 m and a mass of 6 × 10^24 kg. 0.007 hr 2.7 hr 6300 hr 24 hr 0.57 hr

Solution Steps

The orbital radius \( r \) is the sum of the Earth's radius and the satellite's altitude: \[ r = 6.37 \times 10^6 \, \text{m} + 3.5 \times 10^6 \, \text{m} = 9.87 \times 10^6 \, \text{m} \]

Kepler's Third Law for circular orbits states: \[ T^2 = \frac{4 \pi^2 r^3}{G M} \] where:

• \( T \) is the orbital period,
• \( r \) is the orbital radius,
• \( G \) is the gravitational constant \( 6.67430 \times 10^{-11} \, \text{m}^3 \text{kg}^{-1} \text{s}^{-2} \),
• \( M \) is the mass of the Earth \( 6 \times 10^{24} \, \text{kg} \).

First, calculate \( r^3 \): \[ r^3 = (9.87 \times 10^6 \, \text{m})^3 = 9.617 \times 10^{20} \, \text{m}^3 \]

Next, plug in the values: \[ T^2 = \frac{4 \pi^2 (9.617 \times 10^{20} \, \text{m}^3)}{6.67430 \times 10^{-11} \, \text{m}^3 \text{kg}^{-1} \text{s}^{-2} \times 6 \times 10^{24} \, \text{kg}} \]

Simplify the denominator: \[ 6.67430 \times 10^{-11} \times 6 \times 10^{24} = 4.00458 \times 10^{14} \]

Now, calculate \( T^2 \): \[ T^2 = \frac{4 \pi^2 \times 9.617 \times 10^{20}}{4.00458 \times 10^{14}} = \frac{3.802 \times 10^{22}}{4.00458 \times 10^{14}} = 9.49 \times 10^7 \, \text{s}^2 \]

Finally, take the square root to find \( T \): \[ T = \sqrt{9.49 \times 10^7} = 9740 \, \text{s} \]

Convert seconds to hours: \[ T = \frac{9740 \, \text{s}}{3600 \, \text{s/hr}} = 2.7056 \, \text{hr} \]

Final Answer