EK Physics Lesson test 3, discrete

Discussion in 'MCAT Discussions' started by aa4, Jul 27, 2006.

1. aa4 Member

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can someone please explain to me why the answer is b. i would guess it to be d, g decreases to 0, wouldn't the PE have to as well (PE=mgh)

thanks

A rocket is launched from earth to explore our solar system and beyond. As the rocket moves out of the earth's atm and into deep space the gravatational constant g decreases and approaches 0, and the gravatational potential E of the rocket:
a) also decreases and approaches 0
b) continually increases
c) remains constant
d) increases at first and then decreases and approaches 0

3. moocow012 New Member

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PE = mgh refers only to the PE within the earth's gravitational pull. This PE is commonly used because many questions refer to objects within a gravitational field. But In general, PE increases with increasing height, so even if you left earth's gravitational pull, PE will continue to increase. Hope that helps.

4. estairella Senior Member

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To add to that, technically, you can never leave earth's gravitational pull, until you are infinitely away from it.

Like he said, PE = mgh is an approximation. Imagine an isolated universe with 2 balls of some mass. No matter how much you move them away, there will always be the attractive force of gravity. And because there is always a net attractive force, no matter how small, the farther away you separate them, the more potential energy they have.

5. DrBowtie Final Countdown Moderator Emeritus

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The formula is U=-Gmm/r
As r-> infinity it gets closer to zero aka increases.

Enjoy!

6. aa4 Member

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thanks for the help. makes more sense now

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