Use Kepler's Laws and data about the Moon to determine the orbital period (hours) of a satellite that is 17512 km above the SURFACE of the Earth.
HINT: How does the radius of the Earth affect the problem?
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To determine the orbital period of a satellite above the Earth's surface, we can use Kepler's third law, also known as the law of periods. This law states that the square of the period of any planet or satellite is proportional to the cube of its average distance from the center of the planet it is orbiting.
To use this law, we need to consider the average distance from the center of the Earth to the satellite. Since we are given the height above the Earth's surface, we need to account for the Earth's radius.
The radius of the Earth is approximately 6,371 km. So, to find the average distance from the center of the Earth to the satellite, we add the satellite's height to the radius of the Earth:
Average distance = Radius of the Earth + Height above the surface
Average distance = 6,371 km + 17,512 km = 23,883 km
Now, we can use Kepler's third law to find the orbital period (T):
T^2 = (R^3) / (G * M)
Where:
T = Orbital period of the satellite (in hours)
R = Average distance from the center of the Earth to the satellite (in km)
G = Gravitational constant (approximately 6.674e-11 m^3/kg/s^2)
M = Mass of the Earth (approximately 5.972e24 kg)
First, let's convert the average distance from kilometers to meters for consistency with the units of the gravitational constant:
Average distance = 23,883 km = 23,883,000 m
Now, we can substitute the values into the formula:
T^2 = (23,883,000^3) / (6.674e-11 * 5.972e24)
Simplifying the equation gives:
T^2 = 17.849 x 10^25
Taking the square root of both sides:
T ≈ √(17.849 x 10^25)
T ≈ 1,335,972 hours
Therefore, the orbital period of the satellite that is 17,512 km above the surface of the Earth is approximately 1,335,972 hours.