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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 ...
Near the surface of the Earth, the acceleration due to gravity g = 9.807 m/s 2 (metres per second squared, which might be thought of as "metres per second, per second"; or 32.18 ft/s 2 as "feet per second per second") approximately. A coherent set of units for g, d, t and v is essential.
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.
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 ...
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.
After reducing the problem to the relative motion of the bodies in the plane, he defines the constant of the motion c 3 by the equation ẋ 2 + ẏ 2 = 2k 2 M/r + c 3, where M is the total mass of the two bodies and k 2 is Moulton's notation for the gravitational constant. He defines c 1, c 2, and c 4 to be other constants of the
One g is the force per unit mass due to gravity at the Earth's surface and is the standard gravity (symbol: g n), defined as 9.806 65 metres per second squared, [5] or equivalently 9.806 65 newtons of force per kilogram of mass. The unit definition does not vary with location—the g-force when standing on the Moon is almost exactly 1 ⁄ 6 that