Escape velocity and radius (TBR question)

[ilovemedi]

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Passage 2 of chapter 4 in physics 1.

Equation in passage: U= (-GM(earth)*mass spacecraft)/ r . r is = distance away from center of Earth

Question: The magnitude of the escape velocity of a rocket is necessarily larger if:
A) The rocket has a larger gravitational PE upon liftoff.
B) The rocket has a smaller gravitational PE upon liftoff.
Answer: B. States that a greates escape vel is needed when rocket has small Grav PE (close to earth).

I'm confused as to how to approach this. They gave NO info on escape velocity. Just that conservation of energy is related to this. I thought that the on earth the spaceship has a large PE (since small r=large PE based on the above equation). It's CLOSE TO EARTH so has a low R, and thus high PE?

 

[brood910]

For escape velocity question, just use Gmm/r = 1/2mv^2

So, smaller r = larger v.

Note the negative sign in front of Gmm/r.
That's why higher the number, lower the PE is.

Just think about this as mgh.
Higher, h, higher PE.
In fact, Gmm/r is derived from mgh, where g = Gm/r^2 and h = r.

 

[Odi]

I'm confused as to how to approach this. They gave NO info on escape velocity. Just that conservation of energy is related to this. I thought that the on earth the spaceship has a large PE (since small r=large PE based on the above equation). It's CLOSE TO EARTH so has a low R, and thus high PE?

You're forgetting about the negative sign.

As R decreases, the negative number gets larger--meaning the actual value of that number is decreasing. Think of a number line.

 

[ilovemedi]

Thanks everyone. I did miss the negative sign!