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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).
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 ...
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 ...
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 ...
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 ...
The average speed is 7.7 km/s, the net delta-v to reach this orbit is 8.1 km/s (the actual delta-v is typically 1.5–2.0 km/s more for atmospheric drag and gravity drag). The increase per meter would be 4.4 J/kg; this rate corresponds to one half of the local gravity of 8.8 m/s 2. For an altitude of 100 km (radius is 6471 km):
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.
Putting the Sun immobile at the origin, when the Earth is moving in an orbit of radius R with velocity v presuming that the gravitational influence moves with velocity c, moves the Sun's true position ahead of its optical position, by an amount equal to vR/c, which is the travel time of gravity from the sun to the Earth times the relative ...