Use Kepler’s laws and the period of the Moon (27.4 d)
to determine the period of an artificial satellite orbiting
very near the Earth’s surface.
To determine the period of an artificial satellite orbiting very near the Earth's surface using Kepler's laws and the period of the Moon, we need to understand Kepler's laws and apply them appropriately.
Kepler's laws describe the motion of objects in orbit around a central mass, such as the Earth. The important laws for this calculation are Kepler's Third Law (the Law of Harmonies) and Kepler's Second Law (the Law of Areas).
1. Kepler's Third Law states that the square of the period of an orbiting object is directly proportional to the cube of its average distance from the central mass. Mathematically, it is expressed as:
T^2 = (4π^2 / GM) * r^3
Where T is the period, G is the gravitational constant, M is the mass of the central object (Earth in this case), and r is the distance from the center of the central object to the object in orbit.
2. Kepler's Second Law states that the line that connects an orbiting object to the central mass sweeps out equal areas in equal intervals of time. This means that an object moves faster when it is closer to the central mass and slower when it is farther away.
Now, let's use these laws to calculate the period of an artificial satellite orbiting near the Earth's surface:
1. We are given the period of the Moon (Tm) as 27.4 days. We can convert it to seconds for consistency:
Tm = 27.4 days * 24 hours * 60 minutes * 60 seconds
= 2,368,640 seconds
2. The average distance of the Moon from the Earth (rm) is approximately 384,400 km or 384,400,000 meters.
3. To calculate the period of the artificial satellite (Ts), we need to determine its average distance from the center of the Earth (rs). Since the satellite is orbiting very near the Earth's surface, we can assume rs is essentially equal to the Earth's radius (re), which is approximately 6,371 km or 6,371,000 meters.
4. We can now set up a proportion using Kepler's Third Law equation:
Tm^2 / Ts^2 = rm^3 / rs^3
Plugging in the known values:
(2,368,640 seconds)^2 / Ts^2 = (384,400,000 meters)^3 / (6,371,000 meters)^3
5. Rearranging the equation for Ts, we get:
Ts^2 = (2,368,640 seconds)^2 * (6,371,000 meters)^3 / (384,400,000 meters)^3
Taking the square root of both sides:
Ts = sqrt[(2,368,640 seconds)^2 * (6,371,000 meters)^3 / (384,400,000 meters)^3]
6. Calculating this expression will give us the period of the artificial satellite orbiting very near the Earth's surface.
It is important to note that this calculation assumes a circular orbit for the artificial satellite. For more precise calculations, eccentricity and other factors may need to be considered.