To calculate the orbital period of a satellite in a circular orbit around the Earth, you can use Kepler's third law, which is simplified for circular orbits. The formula for the orbital period \( T \) is given by:
\[ T = 2\pi \sqrt{\frac{r^3}{GM}} \]
where:
- \( T \) is the orbital period,
- \( r \) is the distance from the center of the Earth to the satellite (which is equal to Earth's radius plus the altitude of the satellite),
- \( G \) is the universal gravitational constant (approximately \( 6.674 \times 10^{-11} , \text{m}^3, \text{kg}^{-1}, \text{s}^{-2} \)),
- \( M \) is the mass of the Earth (approximately \( 5.972 \times 10^{24} , \text{kg} \)).
Steps to Calculate the Orbital Period:
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Determine the Variables:
- Identify the mass of the Earth \( M \).
- Identify the radius of the Earth \( R_E \) (approximately \( 6.371 \times 10^{6} , \text{m} \)).
- Determine the altitude \( h \) of the satellite above the Earth's surface.
- Calculate the total distance from the center of the Earth to the satellite: \[ r = R_E + h \]
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Plug Values into the Formula:
- Use \( r \) in the formula for \( T \).
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Perform the Calculation:
- Compute \( r^3 \),
- Multiply by \( GM \),
- Take the square root, and multiply by \( 2\pi \) to find \( T \).
Example Calculation
Assume a satellite is in orbit at an altitude of 300 km above the Earth's surface.
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Calculate \( r \):
- \( R_E \approx 6.371 \times 10^{6} , \text{m} \)
- \( h = 300 \times 10^{3} , \text{m} \)
- \( r = R_E + h = 6.371 \times 10^{6} , \text{m} + 300 \times 10^{3} , \text{m} = 6.671 \times 10^{6} , \text{m} \)
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Calculate \( T \):
- \( G = 6.674 \times 10^{-11} , \text{m}^3, \text{kg}^{-1}, \text{s}^{-2} \)
- \( M = 5.972 \times 10^{24} , \text{kg} \)
- Compute \( T \): \[ T = 2\pi \sqrt{\frac{(6.671 \times 10^{6})^3}{(6.674 \times 10^{-11})(5.972 \times 10^{24})}} \]
- Calculate \( r^3 \): \[ r^3 \approx 2.973 \times 10^{20} , \text{m}^3 \]
- Calculate \( GM \): \[ GM \approx 3.986 \times 10^{14} , \text{m}^3, \text{s}^{-2} \]
- Now plug in the values: \[ T = 2\pi \sqrt{\frac{2.973 \times 10^{20}}{3.986 \times 10^{14}}} \]
Finally, calculate \( T \) to find the orbital period in seconds. You can further convert this to minutes if needed (by dividing by 60).
Conclusion:
This approach will give you the orbital period of a satellite based on its altitude and the parameters of the Earth.