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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 terminal velocity is approached. In this example, a speed of 50 % of terminal velocity is reached after only about 3 seconds, while it takes 8 seconds to reach 90 %, 15 seconds to ...
In astrodynamics, an orbit equation defines the path of orbiting body around central body relative to , without specifying position as a function of time.Under standard assumptions, a body moving under the influence of a force, directed to a central body, with a magnitude inversely proportional to the square of the distance (such as gravity), has an orbit that is a conic section (i.e. circular ...
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 ...
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 ...
For example, the Schwarzschild radius r s of the Earth is roughly 9 mm (3 ⁄ 8 inch); at the surface of the Earth, the corrections to Newtonian gravity are only one part in a billion. The Schwarzschild radius of the Sun is much larger, roughly 2953 meters, but at its surface, the ratio r s / r is roughly 4 parts in a million.
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 principal reasoning is that Newton's law of gravitation yields an acceleration ¨ = /; if the product of gravitational constant and attractive mass at the center of the orbit are known, position and velocity are the initial values for that second order differential equation for () which has a unique solution.
In gravitationally bound systems, the orbital speed of an astronomical body or object (e.g. planet, moon, artificial satellite, spacecraft, or star) is the speed at which it orbits around either the barycenter (the combined center of mass) or, if one body is much more massive than the other bodies of the system combined, its speed relative to the center of mass of the most massive body.