Gravitational potential energy

[AG22]

Hey guys, maybe some of you could shed some light on this. I was always under the impression that Gravitational Potential Energy (GPE) tends to decrease the higher up you go, due to the inverse square law. Then why is it that the potential energy of a rocket, after it has exited the atmosphere is still LARGE...shouldn't it be smaller?

I'd appreciate your explanations

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[Fort]

Gravitational potential energy is directly proportional to the height above the plane we define to be zero.

So in the case of the rocket, we'll call the zero plane, or point, the ground level. As the height of the rocket increases, the gravitational potential energy of the rocket will also increase.

The equation for GPE is

GPE = mgh where m, g, and h are mass, the gravitational constant, and height, respectively.

I think you may be confusing GPE with the gravitational force between two objects, which does decrease with the square of the distance between them.

F = Gm1m2/r^2

Hope this helps.

 

[loveoforganic]

GPE = mgh wouldn't be applicable to a rocket though, would it? You've reached the point where g is no longer constant. I think your equation would be

GPE = (m1)^2*(m2)*(h) / (r)^2

or something of the kind. Not sure if that's valid really.

 

[Bernoull]

Hey guys, maybe some of you could shed some light on this. I was always under the impression that Gravitational Potential Energy (GPE) tends to decrease the higher up you go, due to the inverse square law. Then why is it that the potential energy of a rocket, after it has exited the atmosphere is still LARGE...shouldn't it be smaller?

I'd appreciate your explanations

This can be confusing.. but its an excellent question. I'll take a whack at it... If u don't like thorough answers, my apologies, I'm pretty sure someone will post a one-liner later...

First, force is a vector, typically we represent F=GMm/r^2 but technically it's F=-GMm/r^2 (usually with a unit vector to da right). The negative sign tells you that the force is ATTRACTIVE.

Second g=10m/s^2 is simply an approximation of gravitional force field, it's also a vector. A field is just "force density" in a specific direction. Therefore g= gravitational Force/unit mass and E= electric Force/unit charge.

g= Force/1kg = -GMm/mr^2 = -GM/r^2
(where m is a test mass =1kg, M=mass of Earth).

therefore F=mg=-GMm/r^2 (consider them equivalent)

Work/Energy = Fd(CosQ), GPE =Fr = GMm/r

Now to ur questions/statements:

"I was always under the impression that Gravitational Potential Energy (GPE) tends to decrease the higher up you go, due to the inverse square law"

Technically GPE INCREASES by the inverse law (not inverse squared law), r is NOT squared for GPE. But GPE INCREASES with r. Why?? See Below..

"Then why is it that the potential energy of a rocket, after it has exited the atmosphere is still LARGE...shouldn't it be smaller?"

1. Lets look at this form a perspective of work, W=Fr, what happens to F as r increases, it gets bigger... yes bigger.. F=-GMm/r^2 (a negative number gets bigger as it's magntude decreases, -1>-100).

As F increases with r, GPE = Fr increases as well. Most people get confused bcos they neglect the minus sign.

FYI, gravitational Force is maximum (F=0N) where r is infinite and minimum (F~ -infinity) when r is zero, If u want to prove this to urself, take the limits of F(r) =-k/r^2 as r approaches 0 and infinity... (I've substituted GmM as k since it's constant for rocket/earth system)

2. From a Field perspective, Work done on an object in a conservative field, is stored as potential energy, gravity is a conservative field so the rocket's engine does work against gravity to move it into space and this work is stored as GPE.. (Well, in reality there's a myriad of frictional forces etc... so not all the work is conserved.. regardless some of the work is stored as GPE therefore)


Hope this helps...

 

[Bernoull]

Gravitational potential energy is directly proportional to the height above the plane we define to be zero.

So in the case of the rocket, we'll call the zero plane, or point, the ground level. As the height of the rocket increases, the gravitational potential energy of the rocket will also increase.

The equation for GPE is

GPE = mgh where m, g, and h are mass, the gravitational constant, and height, respectively.

I think you may be confusing GPE with the gravitational force between two objects, which does decrease with the square of the distance between them.

F = Gm1m2/r^2

Hope this helps.

Just to clarify one point, MCAT questions usually ask you wat happens to the MAGNITUDE of F as r increases, in this case ur absolutely and [F] decreases with r, if they want just the value of F not absolute F, then that's not correct F will increase with r due to the negative sign...

I'll be the first to admit that I'm splitting hairs here, but it may earn/cost u a point if u dont appreciate these subleties..

 

[Fort]

GPE = mgh wouldn't be applicable to a rocket though, would it? You've reached the point where g is no longer constant. I think your equation would be

GPE = (m1)^2*(m2)*(h) / (r)^2

or something of the kind. Not sure if that's valid really.

For the rocket example, I was assuming that it wasn't very far from the Earth's surface. Gravity does decrease as you get higher, but I don't think the MCAT will take something like that into account, as h would need to be pretty big.

For objects near the earth surface you can just use GPE = mgh. But for objects in higher orbit you would have to use

U(r) = -G*m_object*M_earth/r
so I guess for objects way above the Earth's surface potential energy is inversely proportional to r but I wouldn't worry too much about this case. I would just know GPE = mgh for the MCAT.
Would make for an interesting passage, though.

 

[Fort]

Just to clarify one point, MCAT questions usually ask you wat happens to the MAGNITUDE of F as r increases, in this case ur absolutely and [F] decreases with r, if they want just the value of F not absolute F, then that's not correct F will increase with r due to the negative sign...

I'll be the first to admit that I'm splitting hairs here, but it may earn/cost u a point if u dont appreciate these subleties..

Yep, the r hat vector points opposite the direction of F, so you're right it would be negative. I think we're making this way too complicated, though.

 

[AG22]

Thanks Bernoull. I think I get it. The magnitude does decrease, however the work done to propel the rocket all the way away from the force of gravity increases with height. So I will take that into account according to what the question's asking!


Gracias

 

[Bernoull]

Thanks Bernoull. I think I get it. The magnitude does decrease, however the work done to propel the rocket all the way away from the force of gravity increases with height. So I will take that into account according to what the question's asking!


Gracias

Ur welcome, the simplest answer like FOrt said is GPE = mgh and u can clearly see that GPE is directly proportional to h..

I gave a thorough response to show that the prediction from "mgh" is consistent with that of "-GMm/r". Both predict that GPE increases with h. If you use "-GMm/r" remember the negative sign that's all!!!

 

[kwokkit]

First, force is a vector, typically we represent F=GMm/r^2 but technically it's F=-GMm/r^2 (usually with a unit vector to da right). The negative sign tells you that the force is ATTRACTIVE.

This is the key to understanding that GPE increases with r or h. Maybe this is getting complicated, but WHY is there a negative sign in that equation denoting an attractive force between m1 and m2? Just a convention so physics will agree with itself?

 

[Bernoull]

This is the key to understanding that GPE increases with r or h. Maybe this is getting complicated, but WHY is there a negative sign in that equation denoting an attractive force between m1 and m2? Just a convention so physics will agree with itself?

Definitely a sign convention, i think vector cross-product stuff... i leave it to someone else to expound..