A satellite has an orbital period of 4 hours. what is the altitude?
thanks.
To determine the altitude of a satellite with a given orbital period, we can use Kepler's Third Law of Planetary Motion. This law states that the cube of the semi-major axis of an orbit is directly proportional to the square of the orbital period.
The semi-major axis of an orbit represents the average distance between the satellite and the center of the Earth. However, in the case of a circular orbit, like the one mentioned in the question, the semi-major axis is equal to the altitude of the satellite.
Let's use the formula for Kepler's Third Law to find the altitude:
T^2 = (4π^2 / GM) * a^3
Where:
T is the orbital period (in seconds).
G is the gravitational constant (approximately 6.67430 × 10^-11 m^3 kg^-1 s^-2).
M is the mass of the Earth (approximately 5.972 × 10^24 kg).
a is the semi-major axis (altitude) of the orbit.
First, we need to convert the orbital period to seconds. Since the given period is 4 hours, we multiply it by 60 minutes per hour and 60 seconds per minute:
T = 4 hours * 60 minutes/hour * 60 seconds/minute = 14,400 seconds
Now we can rearrange the formula to solve for a:
a = [(T^2 * GM) / (4π^2)]^(1/3)
Substituting the known values:
a = [(14,400^2 * 6.67430 × 10^-11 * 5.972 × 10^24) / (4π^2)]^(1/3)
Calculating this equation will give us the altitude of the satellite.