Answer :
The period of the orbit is approximately 8.86 hours.
The period of an orbit is the time it takes for a satellite to complete one full orbit around the Earth. The period is determined by the distance of the satellite from the Earth's center and the gravitational force acting on it.
We can find the period of the orbit using Kepler's third law of motion:
T² = (4π² × (R+h)³) / (G × M)
where T is the period of the orbit in seconds, h is the altitude of the satellite (500 miles), R is the radius of the Earth (4000 miles), G is the gravitational constant (6.67 × 10⁻¹¹ N(m/kg)²) and M is the mass of the Earth (5.97 × 10²⁴ kg).
First, we need to convert the radius of the Earth and the altitude of the satellite from miles to meters.
1 mile = 1609.34 meters
R = 4000 miles × 1609.34 meters/mile = 6,371,600 meters
h = 500 miles × 1609.34 meters/mile = 804,670 meters
T² = (4π² × (6371600 + 804670)³) / (6.67 × 10⁻¹¹ × 5.97 × 10²⁴)
T² = 1.087 × 10¹⁶
T = √(1.087 × 10¹⁶)
T = 3.296 × 10⁸ seconds
To get the period in hours, we divide by the number of seconds in an hour:
T = 3.296 × 10⁸ seconds / (60 × 60) = approximately 8.86 hours
To learn more about Kepler's third law of motion visit: https://brainly.com/question/13532269
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Answer:
The orbital period of a satellite with circular orbit is given by
T
2
R
3
=
4
π
2
G
M
where
T
is the orbital period,
R
is the distance between the center of the planet and the satellite,
G
is the gravitational constant, and
M
is the mass of the Earth.
