An object is released from rest at an altitude h above the surface of the Earth. (a) Show that its speed at a distance r from the Earth's center, where RE < r < RE + h, given by v = squareroot (2GME (1/r - 1/(RE + h)) Assume the release altitude is 500 km. perform the integral: (Delta) t = (integral from i to f) dt = - (integral from i to f) dr/v to find the time of fall as the object moves from the release point to the Earth's surface. The negative sign appears because the object is moving opposite to the radial direction, so its speed is v = -dr/dt. Perform the integral numerically.
An object is released from rest at an altitude h above the surface of the Earth. (a) Show that its speed at a distance r from the Earth's center, where RE < r < RE + h, given by v = squareroot (2GME (1/r - 1/(RE + h)) Assume the release altitude is 500 km. perform the integral: (Delta) t = (integral from i to f) dt = - (integral from i to f) dr/v to find the time of fall as the object moves from the release point to the Earth's surface. The negative sign appears because the object is moving opposite to the radial direction, so its speed is v = -dr/dt. Perform the integral numerically.
Classical Dynamics of Particles and Systems
5th Edition
ISBN:9780534408961
Author:Stephen T. Thornton, Jerry B. Marion
Publisher:Stephen T. Thornton, Jerry B. Marion
Chapter10: Motion In A Noninertial Reference Frame
Section: Chapter Questions
Problem 10.12P
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An object is released from rest at an altitude h above the surface of the Earth. (a) Show that its speed at a distance r from the Earth's center, where RE < r < RE + h, given by v = squareroot (2GME (1/r - 1/(RE + h)) Assume the release altitude is 500 km. perform the integral: (Delta) t = (integral from i to f) dt = - (integral from i to f) dr/v to find the time of fall as the object moves from the release point to the Earth's surface. The negative sign appears because the object is moving opposite to the radial direction, so its speed is v = -dr/dt. Perform the integral numerically.
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