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A rocket's required mass ratio as a function of effective exhaust velocity ratio. The classical rocket equation, or ideal rocket equation is a mathematical equation that describes the motion of vehicles that follow the basic principle of a rocket: a device that can apply acceleration to itself using thrust by expelling part of its mass with high velocity and can thereby move due to 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 ...
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 ...
There are two main descriptions of motion: dynamics and kinematics.Dynamics is general, since the momenta, forces and energy of the particles are taken into account. In this instance, sometimes the term dynamics refers to the differential equations that the system satisfies (e.g., Newton's second law or Euler–Lagrange equations), and sometimes to the solutions to those equations.
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.
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 ...
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.
To escape the Solar System from a location at a distance from the Sun equal to the distance Sun–Earth, but not close to the Earth, requires around 42 km/s velocity, but there will be "partial credit" for the Earth's orbital velocity for spacecraft launched from Earth, if their further acceleration (due to the propulsion system) carries them ...
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