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chiddler

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Given U = GMeMo/r
Where Me = mass of earth
Mo = mass of object

A spacecraft goes a distance 3R(earth) from the surface of the earth where it stops, and allows itself to drop back onto the earth from that altitude. When it strikes the earth, the potential energy is:

A. zero
B. -GMeMo/R
C. mgRe
D. infinity

Answer: B.

I want to make sure I am reasoning through this correctly and did not just luck out. Because I suck with gravitational stuff.

The reason it is not zero is because the gravitational potential energy is 0 only when the craft is at the center of the mass, which is impossible. Therefore, things at the surface of any planet have gravitational potential energy and 0 G.P.E is impossible. Because you can't go inside the core of a planet.

Yes?
 

bajoneswadup

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idk, i think this is a silly question. all of those PE questions i did have all been wrong then b/c i used h as the height from the surface of earth instead of h+R?? i guess what ur sayin makes sense, but what i dont get is that expression is the force between 2 masses, not PE?

Ohh nvm I see now they told u it equals to U. ok makes sense
 

rjosh33

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The reason it is not zero is because the gravitational potential energy is 0 only when the craft is at the center of the mass, which is impossible. Therefore, things at the surface of any planet have gravitational potential energy and 0 G.P.E is impossible. Because you can't go inside the core of a planet.

Yes?

Check out the equation again. Because r is in the denominator, the potential energy will only be zero when r = infinity. So at a point closer than infinity, the potential energy will be -GMeMo/r. Also, note that gravitational potential energy has to be negative.
 

chiddler

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Check out the equation again. Because r is in the denominator, the potential energy will only be zero when r = infinity. So at a point closer than infinity, the potential energy will be -GMeMo/r. Also, note that gravitational potential energy has to be negative.

that's right. so it approaches infinity as it approaches the core.

i didn't think of that: why does it have to be negative? why is the answer to this question negative?
 
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rjosh33

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that's right. so it approaches infinity as it approaches the core.

i didn't think of that: why does it have to be negative? why is the answer to this question negative?

The best I've heard it explained is through conservation of energy. If you have just the Earth and a soccer ball (for example) that are both at rest and separated by an infinite distance, then the kinetic and potential energies are both zero. However, if the ball were to move closer to the Earth, then it would gain some kinetic energy (no matter how slight). Because the kinetic energy is positive and the total energy is zero, then the potential energy must be negative.

Additionally, you can refer to the equation U = -Gm1m2/r. G is a positive constant, mass can't be negative, and the distance will never be negative, so the resulting energy must be negative.
 

chiddler

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The best I've heard it explained is through conservation of energy. If you have just the Earth and a soccer ball (for example) that are both at rest and separated by an infinite distance, then the kinetic and potential energies are both zero. However, if the ball were to move closer to the Earth, then it would gain some kinetic energy (no matter how slight). Because the kinetic energy is positive and the total energy is zero, then the potential energy must be negative.

Additionally, you can refer to the equation U = -Gm1m2/r. G is a positive constant, mass can't be negative, and the distance will never be negative, so the resulting energy must be negative.

thanks. makes sense.
 

milski

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that's right. so it approaches infinity as it approaches the core.

i didn't think of that: why does it have to be negative? why is the answer to this question negative?

Potential energy is relative - you have to define your own zero since there is no place where you can say that it does not exist. For some problems, like dropping balls, etc, it's convenient to set the zero at the height from which you drop it or at which it falls. For the case you talk about planetary distances, the convention is to set the zero at infinity (which sort of makes sense - the further away you go, the less you care about how you interact with something). When you get closer to the object the potential energy decreases and since it's 0 at infinity, all real potential energies end up lower than that and thus negative.
 

milski

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Another thing - the potential energy in the center of the body (Earth, whatever) is not -infinity. Energy approaching infinity in general is "not a good thing" (tm). It is a finite number and you can calculate even though the formula does change once you are below the surface. In a way, you could say that U at the center of the body is a better zero, since it really cannot get any lower. But first, that's only the potential energy from that body - there is still potential energy from other bodies. Second, that means that you have to deal with some arbitrary large numbers any time you're away from the center of that body, which becomes really painful, really fast.
 
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