Physics
posted by Steve .
Take into account the Earth's rotational speed (1 rev/day) and determine the necessary speed, with respect to the Earth, for a rocket to escape if fired from the Earth at the equator in a direction a) eastward; b) westward; c) vertically upward.
The answers are 1.07 x10^4 m/s, 1.17 x10^4 m/s and 1.12 x10^4 m/s. I got c but I don't understand what to do for a and b.

Escape velocity is Ve = sqrt(2µ/r) where µ = the gravitational constant of the earth = 3.9863x10^14m^3/sec.^2 and r = the radius of the earth = 3963 miles = 6.377519x10^6 meters.
The earth's surface velocity at the equator is [3963(5280)2(3.1416)/3.281]/[86,400sec. = .0463m/s.
The vertical escape velocity is therefore 1.1180m/s.
Launching eastward, the escape velocity for the rockets 1.1180  .0463 = 1.0707m/s.
Launchung westward, the escape velocity is 1.1180 + .0463 = 1.1643m/s.
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