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Gravity decreases with altitude as one rises above the Earth's surface because greater altitude means greater distance from the Earth's centre. All other things being equal, an increase in altitude from sea level to 9,000 metres (30,000 ft) causes a weight decrease of about 0.29%.
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 ...
Escape speed at a distance d from the center of a spherically symmetric primary body (such as a star or a planet) with mass M is given by the formula [2] [3] = = where: G is the universal gravitational constant (G ≈ 6.67 × 10 −11 m 3 ⋅kg −1 ⋅s −2 [4])
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 ...
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 longitude.
The terminal velocity depends on many factors including mass, drag coefficient, and relative surface area and will only be achieved if the fall is from sufficient altitude. A typical skydiver in a spread-eagle position will reach terminal velocity after about 12 seconds, during which time they will have fallen around 450 m (1,500 ft). [4] Free ...
The actual Hill radius for the Earth-Moon pair is on the order of 60,000 km (i.e., extending less than one-sixth the distance of the 378,000 km between the Moon and the Earth). [ 9 ] In the Earth-Sun example, the Earth ( 5.97 × 10 24 kg ) orbits the Sun ( 1.99 × 10 30 kg ) at a distance of 149.6 million km, or one astronomical unit (AU).
The angular velocity of the Earth is defined to be ω = 72.921 15 × 10 −6 rad/s. [ 11 ] This leads to several computed parameters such as the polar semi-minor axis b which equals a × (1 − f ) = 6 356 752 .3142 m , and the first eccentricity squared, e 2 = 6.694 379 990 14 × 10 −3 .