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The first equation shows that, after one second, an object will have fallen a distance of 1/2 × 9.8 × 1 2 = 4.9 m. After two seconds it will have fallen 1/2 × 9.8 × 2 2 = 19.6 m; and so on. The next-to-last equation becomes grossly inaccurate at great distances. If an object fell 10 000 m to Earth, then the results of both equations differ ...
Equation [3] involves the average velocity v + v 0 / 2 . Intuitively, the velocity increases linearly, so the average velocity multiplied by time is the distance traveled while increasing the velocity from v 0 to v, as can be illustrated graphically by plotting velocity against time as a straight line graph. Algebraically, it follows ...
L T−1. In kinematics, the speed (commonly referred to as v) of an object is the magnitude of the change of its position over time or the magnitude of the change of its position per unit of time; it is thus a scalar quantity. [ 1] The average speed of an object in an interval of time is the distance travelled by the object divided by 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.
Special relativity. Gravitational time dilation is a form of time dilation, an actual difference of elapsed time between two events, as measured by observers situated at varying distances from a gravitating mass. The lower the gravitational potential (the closer the clock is to the source of gravitation), the slower time passes, speeding up as ...
Lorentz factor. where and v is the relative velocity between two inertial frames . For two frames at rest, γ = 1, and increases with relative velocity between the two inertial frames. As the relative velocity approaches the speed of light, γ → ∞. Time dilation (different times t and t' at the same position x in same inertial frame)
Inversely, for calculating the distance where a body has to orbit in order to have a given orbital period T: a = G M T 2 4 π 2 3 {\displaystyle a={\sqrt[{3}]{\frac {GMT^{2}}{4\pi ^{2}}}}} For instance, for completing an orbit every 24 hours around a mass of 100 kg , a small body has to orbit at a distance of 1.08 meters from the central body's ...
Relative velocities between two particles in classical mechanics. The figure shows two objects A and B moving at constant velocity. The equations of motion are: = +, = +, where the subscript i refers to the initial displacement (at time t equal to zero).