Hello. The number on the right doesn't look correct. How did you get that value?barryj said:T^2/R^3 for Earth = 2.97E-19 (sec^2/m^3)
barryj said:T^2/R^3 for Earth = 2.97E-19 (sec^2/m^3)
The period of a low Earth satellite can be calculated using Kepler's third law, which states that the square of a satellite's orbital period is proportional to the cube of its semi-major axis. This can be expressed as P^2 = (4π^2 / GM) * a^3, where P is the period, G is the gravitational constant, M is the mass of the central body, and a is the semi-major axis of the satellite's orbit.
The semi-major axis is the average distance between the satellite and the center of the Earth. It is related to the period of a low Earth satellite through Kepler's third law, where the cube of the semi-major axis is directly proportional to the square of the orbital period.
Yes, the period of a low Earth satellite can change over time due to factors such as atmospheric drag, gravitational perturbations from other bodies, and changes in the satellite's altitude or orbit. These changes can be calculated and predicted using mathematical models.
The mass of the central body, in this case the Earth, affects the period of a low Earth satellite through Kepler's third law. As the mass of the central body increases, the period of the satellite will also increase, assuming all other factors remain constant.
The period of a low Earth satellite is inversely proportional to its altitude. This means that as the altitude of the satellite increases, its period will decrease. This is because a higher altitude will result in a larger semi-major axis, which in turn increases the orbital period according to Kepler's third law.