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In orbital mechanics ... 6,378 km for Earth) is the geocentric distance is the oblateness ... Calculate the slant range, ...
[nb 1] Earth's orbital speed averages 29.78 km/s (19 mi/s; 107,208 km/h; 66,616 mph), which is fast enough to cover the planet's diameter in 7 minutes and the distance to the Moon in 4 hours. [3] The point towards which the Earth in its solar orbit is directed at any given instant is known as the "apex of the Earth's way". [4] [5]
Geosynchronous orbit (GSO): An orbit around the Earth with a period equal to one sidereal day, which is Earth's average rotational period of 23 hours, 56 minutes, 4.091 seconds. For a nearly circular orbit, this implies an altitude of approximately 35,786 kilometers (22,236 mi). The orbit's inclination and eccentricity may not necessarily be zero.
In astrodynamics, an orbit equation defines the path of orbiting body around central body relative to , without specifying position as a function of time.Under standard assumptions, a body moving under the influence of a force, directed to a central body, with a magnitude inversely proportional to the square of the distance (such as gravity), has an orbit that is a conic section (i.e. circular ...
For Earth, orbital eccentricity e ≈ 0.016 71, apoapsis is aphelion and periapsis is perihelion, relative to the Sun. For Earth's annual orbit path, the ratio of longest radius (r a) / shortest radius (r p) is = +
It is expressed as the angle between a reference plane and the orbital plane or axis of direction of the orbiting object. For a satellite orbiting the Earth directly above the Equator, the plane of the satellite's orbit is the same as the Earth's equatorial plane, and the satellite's orbital inclination is 0°. The general case for a circular ...
The International Space Station has an orbital period of 91.74 minutes (5504 s), hence by Kepler's Third Law the semi-major axis of its orbit is 6,738 km. [citation needed] The specific orbital energy associated with this orbit is −29.6 MJ/kg: the potential energy is −59.2 MJ/kg, and the kinetic energy 29.6 MJ/kg.
The basic orbit determination task is to determine the classical orbital elements or Keplerian elements, ,,,,, from the orbital state vectors [,], of an orbiting body with respect to the reference frame of its central body. The central bodies are the sources of the gravitational forces, like the Sun, Earth, Moon and other planets.