Search results
Results from the WOW.Com Content Network
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 ...
The gravity g′ at depth d is given by g′ = g(1 − d/R) where g is acceleration due to gravity on the surface of the Earth, d is depth and R is the radius of the Earth. If the density decreased linearly with increasing radius from a density ρ 0 at the center to ρ 1 at the surface, then ρ ( r ) = ρ 0 − ( ρ 0 − ρ 1 ) r / R , and the ...
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.
Saturn V Moon rocket just after launch and the gravity of Neptune where atmospheric pressure is about Earth's 1.14 g: Bugatti Veyron from 0 to 100 km/h in 2.4 s 1.55 g [b] Gravitron amusement ride 2.5-3 g: Gravity of Jupiter at its mid-latitudes and where atmospheric pressure is about Earth's 2.528 g: Uninhibited sneeze after sniffing ground ...
For example, as the Earth's rotational velocity is 465 m/s at the equator, a rocket launched tangentially from the Earth's equator to the east requires an initial velocity of about 10.735 km/s relative to the moving surface at the point of launch to escape whereas a rocket launched tangentially from the Earth's equator to the west requires an ...
So if a comet approaching Earth (effective radius ~6400 km) with a velocity of 12.5 km/s (the approximate minimum approach speed of a body coming from the outer Solar System) is to avoid a collision with Earth, the impact parameter will need to be at least 8600 km, or 34% more than the Earth's radius.
Non-zero coefficients C n m, S n m correspond to a lack of rotational symmetry around the polar axis for the mass distribution of Earth, i.e. to a "tri-axiality" of Earth. For large values of n the coefficients above (that are divided by r ( n + 1) in ( 9 )) take very large values when for example kilometers and seconds are used as units.