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The tangential speed of Earth's rotation at a point on Earth can be approximated by multiplying the speed at the equator by the cosine of the latitude. [42] For example, the Kennedy Space Center is located at latitude 28.59° N, which yields a speed of: cos(28.59°) × 1,674.4 km/h = 1,470.2 km/h.
In most situations it is impractical to achieve escape velocity almost instantly, because of the acceleration implied, and also because if there is an atmosphere, the hypersonic speeds involved (on Earth a speed of 11.2 km/s, or 40,320 km/h) would cause most objects to burn up due to aerodynamic heating or be torn apart by atmospheric drag. For ...
A space vehicle's flight is determined by application of Newton's second law of motion: =, where F is the vector sum of all forces exerted on the vehicle, m is its current mass, and a is the acceleration vector, the instantaneous rate of change of velocity (v), which in turn is the instantaneous rate of change of displacement.
Then, Earth’s rotation itself causes a measurable change within the photons. It may seem silly to spend so much time and so many resources on clocking the speed of Earth’s rotation ...
The above example can be used to explain why the Eötvös effect starts diminishing when an object is traveling westward as its tangential speed increases above Earth's rotation (465 m/s). If the westward train in the above example increases speed, part of the force of gravity that pushes against the track accounts for the centripetal force ...
With orbital radius 42,000 km from the Earth's center, the satellite's tangential speed through space is thus v = 42,000 km × 0.26/h ≈ 11,000 km/h. The angular velocity is positive since the satellite travels prograde with the Earth's rotation (the same direction as the rotation of Earth).
Without applying force (such as firing a rocket engine), the period and shape of the satellite's orbit will not change. A satellite in a low orbit (or a low part of an elliptical orbit) moves more quickly with respect to the surface of the planet than a satellite in a higher orbit (or a high part of an elliptical orbit), due to the stronger ...
Near Earth's surface, the acceleration due to gravity, accurate to 2 significant figures, is 9.8 m/s 2 (32 ft/s 2). This means that, ignoring the effects of air resistance , the speed of an object falling freely will increase by about 9.8 metres per second (32 ft/s) every second.