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To find the angle giving the maximum height for a given speed calculate the derivative of the maximum height = / with respect to , that is = / which is zero when = / =. So the maximum height H m a x = v 2 2 g {\displaystyle H_{\mathrm {max} }={v^{2} \over 2g}} is obtained when the projectile is fired straight up.
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 ...
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 escape velocity at a given height is times the speed in a circular orbit at the same height, (compare this with the velocity equation in circular orbit). This corresponds to the fact that the potential energy with respect to infinity of an object in such an orbit is minus two times its kinetic energy, while to escape the sum of potential ...
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 ...
A hyperbolic trajectory is depicted in the bottom-right quadrant of this diagram, where the gravitational potential well of the central mass shows potential energy, and the kinetic energy of the hyperbolic trajectory is shown in red. The height of the kinetic energy decreases as the speed decreases and distance increases according to Kepler's laws.
The formula for an escape velocity is derived as follows. The specific energy (energy per unit mass) of any space vehicle is composed of two components, the specific potential energy and the specific kinetic energy. The specific potential energy associated with a planet of mass M is given by =
Compared with the potential energy at the surface, which is −62.6 MJ/kg., the extra potential energy is 3.4 MJ/kg, and the total extra energy is 33.0 MJ/kg. The average speed is 7.7 km/s, the net delta-v to reach this orbit is 8.1 km/s (the actual delta-v is typically 1.5–2.0 km/s more for atmospheric drag and gravity drag).