explanation to kaplan question

[thebillsfan]

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Question: is the escape velocity of a particle higher at the equator or at the poles? Answer: the poles.

Explanation:
Remember, the centripetal force is the force required to keep a particle traveling in a circu-
lar orbit of radius r. In this case, the force of gravity is the centripetal force. Since the force of
gravity is approximately the same for all particles on the surface of the Earth and since the par-
ticles at the poles require a smaller attractive force to keep them at the surface of the Earth than
the particles at the equator, the velocity required to escape from the surface of the Earth will be
greater at the poles than at the equator

It seems like Kaplan is contradicting itself. If it's moving faster at the equator (since it's farther from the axis at the equator than at the poles), than it has a greater centripetal force keeping it on earth. this would make it seem to me that it would be HARDER for it to escape, requiring an even higher velocity.

 

[ezsanche]

i did this problem too. I dont get it. Can some one explain it in laymen terms.

 

[kentavr]

I'll try.

From 2nd Newton law

Let say we have rocket engine that can move it up with force F. Then to start movement
1. on pole : mg-F = 0 => F = mg.
2. on equator: mg-F = mv^2/R (where V - speed of earth rotation, R is radius of Earth) => F = mg-mv^2/R
In case 2. F is less then in case 1. So we need less power in equator. Interesting. It should also mean that weight on equator for the same mass is less the on pole.

p.S. It looks like that Kaplan ignores that Earth is not exact sphere, and radius for Pole and Equator are different. That should contarbalance this effect at some degree.

 

[ezsanche]

thanks i got it!!!

 

[thebillsfan]

thanks i got it!!!

okay, assuming the earth is spherical, i still dont get it. i understand that there is more centrip force at the equator. however, since gravity is the agent of this centripetal force, wouldnt it require MORE force by the engine since it has to counteract not just its own weight but also the centripetal force?