a) The period of a satellite in orbit 6.7 x 10⁶ m from the center of Earth is approximately 3.12 days. b) The period of a satellite in orbit 6.7 x 10⁶ m from the center of Earth is approximately 270,720 seconds. c) The distance from the Earth's center to the satellite is 6.7 x 10⁶ m. d) the distance above Earth's surface is approximately 3.3 x 10⁵ m.
a) Find the period of a satellite in orbit 6.7 x 10⁶ m from the center of Earth
Kepler's third law states that the square of the period of a planet is proportional to the cube of its semi-major axis. In other words,
T² ∝ a³
where T is the period and a is the semi-major axis. We can use this relationship to find the period of a satellite in orbit 6.7 x 10⁶ m from the center of Earth.
First, we need to find the ratio of the semi-major axes of the two orbits. The semi-major axis of the moon's orbit is 3.9 x 10⁸ m, so the ratio of the semi-major axes is
a_satellite/a_moon = (6.7 x 10⁶ m)/(3.9 x 10⁸ m) ≈ 0.172
Next, we can use the ratio of semi-major axes to find the ratio of periods.
(T_satellite/T_moon)²= (a_satellite/a_moon)³
T_satellite/T_moon ≈
≈ 0.115
T_satellite ≈ 0.115 × T_moon ≈ 3.12 days
b) Convert the period found in part a) to hours, minutes, and seconds
There are 24 hours in a day, 60 minutes in an hour, and 60 seconds in a minute. Therefore,
3.12 days × (24 hours/day) × (60 minutes/hour) × (60 seconds/minute) ≈ 270,720 seconds
c) Find the distance from the Earth's center to the satellite
The distance from the Earth's center to the satellite is equal to the semi-major axis of the satellite's orbit. We already found that the semi-major axis is 6.7 x 10⁶ m.
d) Find the distance above Earth's surface
The distance above Earth's surface is equal to the difference between the distance from the Earth's center to the satellite and the Earth's radius. The Earth's radius is 6.37 x 10⁶ m. Therefore,
distance above surface = 6.7 x 10⁶ m - 6.37 x 10⁶ m ≈ 3.3 x 10⁵ m