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Earth at seasonal points in its orbit (not to scale) Earth orbit (yellow) compared to a circle (gray) Earth orbits the Sun at an average distance of 149.60 million km (92.96 million mi), or 8.317 light-minutes, [1] in a counterclockwise direction as viewed from above the Northern Hemisphere.
The same (blue) area is swept out in a fixed time period. The green arrow is velocity. The purple arrow directed towards the Sun is the acceleration. The other two purple arrows are acceleration components parallel and perpendicular to the velocity. The orbital radius and angular velocity of the planet in the elliptical orbit will vary.
An orbit will be Sun-synchronous when the precession rate ρ = dΩ / dt equals the mean motion of the Earth about the Sun n E, which is 360° per sidereal year (1.990 968 71 × 10 −7 rad/s), so we must set n E = ΔΩ E / T E = ρ = ΔΩ / T , where T E is the Earth orbital period, while T is the period of the spacecraft ...
In gravitationally bound systems, the orbital speed of an astronomical body or object (e.g. planet, moon, artificial satellite, spacecraft, or star) is the speed at which it orbits around either the barycenter (the combined center of mass) or, if one body is much more massive than the other bodies of the system combined, its speed relative to the center of mass of the most massive body.
To escape the Solar System from a location at a distance from the Sun equal to the distance Sun–Earth, but not close to the Earth, requires around 42 km/s velocity, but there will be "partial credit" for the Earth's orbital velocity for spacecraft launched from Earth, if their further acceleration (due to the propulsion system) carries them ...
μ = Gm 1 + Gm 2 = μ 1 + μ 2, where m 1 and m 2 are the masses of the two bodies. Then: for circular orbits, rv 2 = r 3 ω 2 = 4π 2 r 3 /T 2 = μ; for elliptic orbits, 4π 2 a 3 /T 2 = μ (with a expressed in AU; T in years and M the total mass relative to that of the Sun, we get a 3 /T 2 = M) for parabolic trajectories, rv 2 is constant and ...
In the special case of perfectly circular orbits, the semimajor axis a is equal to the radius of the orbit, and the orbital velocity is constant and equal to = where: r is the circular orbit's radius in meters, This corresponds to 1 ⁄ √2 times (≈ 0.707 times) the escape velocity.
Kepler's second law states that a body in orbit traces equal areas over equal times; its orbital velocity is highest around perihelion and lowest around aphelion. [14] The Earth spends less time near perihelion and more time near aphelion. This means that the lengths of the seasons vary. [15]