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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 ...
The weight of an object on Earth's surface is the downwards force on that object, given by Newton's second law of motion, or F = m a (force = mass × acceleration). Gravitational acceleration contributes to the total gravity acceleration, but other factors, such as the rotation of Earth, also contribute, and, therefore, affect the weight of the ...
The general formula for the escape velocity of an object at a distance r from the center of a planet with mass M is [12] = =, where G is the gravitational constant and g is the gravitational acceleration. The escape velocity from Earth's surface is about 11 200 m/s, and is irrespective of the direction of the object.
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.
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 ...
The standard gravitational parameter μ of a celestial body is the product of the gravitational constant G and the mass M of that body. For two bodies, the parameter may be expressed as G ( m 1 + m 2 ) , or as GM when one body is much larger than the other: μ = G ( M + m ) ≈ G M . {\displaystyle \mu =G(M+m)\approx GM.}
For instance, gravity loss on a rocket of mass m would reduce a 3mg thrust directed upward to an acceleration of 2g. However, the same 3 mg thrust could be directed at such an angle that it had a 1 mg upward component, completely canceled by gravity, and a horizontal component of mg× 3 2 − 1 2 {\displaystyle {\sqrt {3^{2}-1^{2}}}} = 2.8 mg ...
At the Earth's surface this becomes: = ^ where g is the downward 9.8 m/s 2 acceleration due to gravity, and ^ is a unit vector in the radially outward direction from the center of the gravitating body. Thus here an outward proper force of mg is needed to keep one from accelerating downward.