Potential Energy

[MedPR]

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You launch a rocket out of earth's atmosphere. As the rocket moves out of the atmosphere and into deep space, the gravitational constant g decreases and approaches 0, and the gravitational potential energy 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.

The answer is B.

Why isn't it D? As the rocket ascends within the atmosphere, its PE increases according to PE = mgh. Then, as g starts to decrease and approach 0, PE decreases and approaches 0, again according to PE = mgh.

How can PE continually increase if g is continually decreasing?

 

[ljc]

PE is the potential energy due to a separation of objects. When the rocket is flying, does it suddenly stop taking energy to escape earth's pull? No, so energy is still being put into the system, thus PE is increasing.

Another way to think of it: if there were only 2 objects in the whole universe, they would both attract each other via gravity. The further away these objects were, the more potential work they could do as they moved together. It wouldn't be a particularly large amount of work at first, as the gravitational force would be near zero, but it would still be positive.

Force necessary for the rocket to escape decreases as g shrinks, but most importantly, it never turns negative. F is always positive, so PE is always increasing.

 

[milski]

Choosing where the 0 is for any sort of potential energy is an arbitrary decision. You can say that it's at the surface of the Earth, if you are talking about throwing object out of your window, or the floor of your house if you are moving things between the floor and say a table.

When talking about gravitational potential energy, it's a convention to define the zero to be the potential energy at infinity. That gives you a nice formule for it for any distance being: U=-GMm/r. It also means that for all non-infinite distances the PE will be negative.

With that in mind, it should be easy to see that the PE does increase as you go further and further away from earth (after all you are putting work in the system, even if you're moving at constant velocity) until it eventually reaches 0.

 

[MedPR]

So PE is increasing, but at a decreasing rate?

 

[ljc]

Yep!

 

[MedPR]

Ok perfect, thank you!

 

[SaintJude]

Then why isn't it still B.? I understand, with U=-GMm/r , as you approach infinity, the U will approach 0.

I mean it's increasing continually. How is it increasing at a decreasing rate?

 

[ljc]

Then why isn't it still B.? I understand, with U=-GMm/r , as you approach infinity, the U will approach 0.

I mean it's increasing continually. How is it increasing at a decreasing rate?

It is B! ...?
And it increases at a decreasing rate because PE=mgh, and as G approaches zero, you need an increasingly large increase in H to gain the same change in PE.

 

[SaintJude]

Oh, I'm sorry haha. MedPr said it was B from the very first post. Maybe I need to take a break....