- #1
prabhjyot
- 10
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need help please urgent
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, is given by
v = sqrt(2GME (1/r -1/ (RE + h) )
(b) 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, is given by
v = sqrt(2GME (1/r -1/ (RE + h) )
(b) 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.
