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For an object of mass the energy required to escape the Earth's gravitational field is GMm / r, a function of the object's mass (where r is radius of the Earth, nominally 6,371 kilometres (3,959 mi), G is the gravitational constant, and M is the mass of the Earth, M = 5.9736 × 10 24 kg).
The InSight mission to Mars launched with a C 3 of 8.19 km 2 /s 2. [5] The Parker Solar Probe (via Venus) plans a maximum C 3 of 154 km 2 /s 2. [6] Typical ballistic C 3 (km 2 /s 2) to get from Earth to various planets: Mars 8-16, [7] Jupiter 80, Saturn or Uranus 147. [8] To Pluto (with its orbital inclination) needs about 160–164 km 2 /s 2. [9]
At a fixed point on the surface, the magnitude of Earth's gravity results from combined effect of gravitation and the centrifugal force from Earth's rotation. [ 2 ] [ 3 ] At different points on Earth's surface, the free fall acceleration ranges from 9.764 to 9.834 m/s 2 (32.03 to 32.26 ft/s 2 ), [ 4 ] depending on altitude , latitude , and ...
Based on wind resistance, for example, the terminal velocity of a skydiver in a belly-to-earth (i.e., face down) free-fall position is about 195 km/h (122 mph or 54 m/s). [3] This velocity is the asymptotic limiting value of the acceleration process, because the effective forces on the body balance each other more and more closely as the ...
Therefore, the movements of the celestial bodies should be modified in the order v/c, where v is the relative speed between the bodies and c is the speed of gravity. The effect of a finite speed of gravity goes to zero as c goes to infinity, but not as 1/c 2 as it does in modern theories.
Based on air resistance, for example, the terminal speed of a skydiver in a belly-to-earth (i.e., face down) free fall position is about 55 m/s (180 ft/s). [3] This speed is the asymptotic limiting value of the speed, and the forces acting on the body balance each other more and more closely as the terminal speed is approached. In this example ...
At latitudes nearer the Equator, the outward centrifugal force produced by Earth's rotation is larger than at polar latitudes. This counteracts the Earth's gravity to a small degree – up to a maximum of 0.3% at the Equator – and reduces the apparent downward acceleration of falling objects.
At any time the average speed from = is 1.5 times the current speed, i.e. 1.5 times the local escape velocity. To have t = 0 {\displaystyle t=0\!\,} at the surface, apply a time shift; for the Earth (and any other spherically symmetric body with the same average density) as central body this time shift is 6 minutes and 20 seconds; seven of ...