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At this point, assuming the thrust fell below the force of gravity, velocity and its corresponding KE would begin to diminish with a corresponding increase in PE.

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- Thread starter Jackson Smith
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- #1

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At this point, assuming the thrust fell below the force of gravity, velocity and its corresponding KE would begin to diminish with a corresponding increase in PE.

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I would like confirmation of this concept of PE creation.

- #3

Chestermiller

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Oh yeah? What if you temporarily increased the thrust to raise the object to a new altitude, and then balanced the forces again (so the final KE were the equal to the initial KE)?

At this point, assuming the thrust fell below the force of gravity, velocity and its corresponding KE would begin to diminish with a corresponding increase in PE.

- #4

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You posit incorrectly. A weight doing work while being lowered at constant velocity has no change in KE but does have a change in PE.I posit that the change in PE never occurs UNTIL there is a change in KE.

This violates the conservation of energyThus, in the case of an object with a constant velocity UP, there is no gain in PE until the instant that the thrust no longer balances the force of gravity.

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Could someone then please explain this:

V = V0 + a*t. Therefore, when thrust = gravity, (Net "a" will be zero), V will be constant (V = V0). So, assuming we have thrust = gravity and no other forces (launch in a vacuum), how do we generate PE as the object ascends to a higher altitude without any change to velocity that could be used to convert KE to PE?

First, you need no change to thrust; we ARE gaining altitude since we have constant velocity equal to initial velocity (I didn't assume initial V was 0; let's make it 100M/s/s). Also, even if we change velocity, I think the PE would again return to zero as soon as thrust = gravity Because, if PE requires g, and you have thrust that offsets g exactly, then you have no Net g and therefore no PE, right?

V = V0 + a*t. Therefore, when thrust = gravity, (Net "a" will be zero), V will be constant (V = V0). So, assuming we have thrust = gravity and no other forces (launch in a vacuum), how do we generate PE as the object ascends to a higher altitude without any change to velocity that could be used to convert KE to PE?

Oh yeah? What if you temporarily increased the thrust to raise the object to a new altitude, and then balanced the forces again (so the final KE were the equal to the initial KE)?

First, you need no change to thrust; we ARE gaining altitude since we have constant velocity equal to initial velocity (I didn't assume initial V was 0; let's make it 100M/s/s). Also, even if we change velocity, I think the PE would again return to zero as soon as thrust = gravity Because, if PE requires g, and you have thrust that offsets g exactly, then you have no Net g and therefore no PE, right?

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- #6

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Energy is the capacity to do work. As the object goes up it has more capacity to do work by falling, regardless of its velocity.So, assuming we have thrust = gravity and no other forces (launch in a vacuum), how do we generate PE as the object ascends to a higher altitude

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You posit incorrectly. A weight doing work while being lowered at constant velocity has no change in KE but does have a change in PE.

This violates the conservation of energy

I don't see how there is ANY PE change or indeed, any PE at all because at constant velocity, thrust = gravity. Any existing velocity, UP or DOWN, remains. In addition, PE = m*g*h. If thrust = gravity, isn't the (Net) g in that formula ZERO?

- #8

DrClaude

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If thrust = gravity, it means that there is a force other than gravity acting on the object. It is that force that provides the additional potential energy. (In the absence of any gravitational potential, that same force would accelerate the object, leading to an increase of kinetic energy instead.)I don't see how there is ANY PE change or indeed, any PE at all because at constant velocity, thrust = gravity. Any existing velocity, UP or DOWN, remains.

In that equation, ##g## is the gravitational acceleration, not the net acceleration. It tells you how much energy you can recover if an object of mass ##m## travels a distance ##h## with a constant acceleration of ##g##.In addition, PE = m*g*h. If thrust = gravity, isn't the (Net) g in that formula ZERO?

- #9

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You posit incorrectly. A weight doing work while being lowered at constant velocity has no change in KE but does have a change in PE.

This violates the conservation of energy

I don't see a change in PE or any PE at all when we have constant velocity. In that case, thrust must equal gravity. So, to calculate PE, we have PE = m*g*h. What is the Net "g" we would use in this equation? Gravity less thrust = ZERO, so PE = 0. We have constant velocity (based on the equation, V = V0 + a*t). whenever Net "a" is zero, which occurs when thrust = gravity. Please explain how this violates the conservation of energy.

- #10

DrClaude

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If I drop a baseball above you, you don't mind if it falls on you from 1 cm above your head or 1 km above your head, provided it moved that km at constant velocity?I don't see a change in PE or any PE at all when we have constant velocity.

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If thrust = gravity, it means that there is a force other than gravity acting on the object. It is that force that provides the additional potential energy. (In the absence of any gravitational potential, that same force would accelerate the object, leading to an increase of kinetic energy instead.)

In that equation, ##g## is the gravitational acceleration, not the net acceleration. It tells you how much energy you can recover if an object of mass ##m## travels a distance ##h## with a constant acceleration of ##g##.

First, there need be no additional force, assuming we started with an existing velocity (say 100 M/s) when thrust was initiated to balance gravity. But based on V = V0 + a*t, velocity cannot change when thrust = gravity since the "a" when be zero. So from where would we get PE?

I also can't agree since our object does not experience g as long as thrust = gravity. It only experiences the difference. To prove this, I offer: ASSUME an initial velocity of 100 M/Sec and the ability to maintain thrust equal to gravity indefinitely. What happens to the velocity as the rocket ascends? Does it DECREASE? Based upon V = V0 + a*t, that can't happen. ("a" would be thrust - gravity or ZERO). Hence, with no change to velocity and therefore no change to KE, there is no means to increase PE.

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Agree; as long as the velocity is constant, momentum will also be constant. Perhaps you were thinking of constant ACCELERATION? Now I WOULD prefer that the constant velocity also be a LOW velocity! :)If I drop a baseball above you, you don't mind if it falls on you from 1 cm above your head or 1 km above your head, provided it moved that km at constant velocity?

- #13

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No, g in that formula is not the net acceleration. It is only the acceleration due to gravity. The net acceleration would be indicated by a, not by g.In addition, PE = m*g*h. If thrust = gravity, isn't the (Net) g in that formula ZERO?

The expression mgh is nonzero even if a is zero.

- #14

DrClaude

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Then where does all the energy from the thrust go?Hence, with no change to velocity and therefore no change to KE, there is no means to increase PE.

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No, g in that formula is not the net acceleration. It is only the acceleration due to gravity. The net acceleration would be indicated by a, not by g.

The expression mgh is nonzero even if a is zero.

So, please help me understand how the PE is created, keeping in mind the equation V = V0 + a*t. If thrust = gravity, this equation results in constant velocity, right? And constant velocity is the scenario we have with our rocket. If it have constant velocity with thrust = gravity, where does the KE come from to be converted to PE. This is where I can't make ends meet.

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Then where does all the energy from the thrust go?

The thrust is used to offset (balance) the force of gravity. From V = V0 + a*t, we know that constant velocity will result when thrust = gravity since acceleration will then be ZERO.

- #17

DrClaude

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I think you misunderstand the concept of potential energy. There is a book on my bookshelf. It is at rest. If I move it to a higher shelf, and it is now at rest there, the book has gained potential energy.Agree; as long as the velocity is constant, momentum will also be constant. Perhaps you were thinking of constant ACCELERATION? Now I WOULD prefer that the constant velocity also be a LOW velocity! :)

KE is not converted to PE. The energy from the thrust goes directly into PE.where does the KE come from to be converted to PE

That doesn't answer my question. If you do not agree with what I wrote just above, you have to tell me where the energy of the thrust is going.The thrust is used to offset (balance) the force of gravity.

Come back to what I mentioned in post #8. Take two identical rockets, one close to Earth, the other in the middle of empty space. The same engines provide the same thrust. The rocket in the middle of nowhere accelerates from the thrust, gaining KE. The rocket near Earth gains no KE because the thrust is completely compensated by the force of gravity. Where does the energy of the thrust go?

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If thrust = gravity, it means that there is a force other than gravity acting on the object. It is that force that provides the additional potential energy. (In the absence of any gravitational potential, that same force would accelerate the object, leading to an increase of kinetic energy instead.)

In that equation, ##g## is the gravitational acceleration, not the net acceleration. It tells you how much energy you can recover if an object of mass ##m## travels a distance ##h## with a constant acceleration of ##g##.

I think you misunderstand the concept of potential energy. There is a book on my bookshelf. It is at rest. If I move it to a higher shelf, and it is now at rest there, the book has gained potential energy.

KE is not converted to PE. The energy from the thrust goes directly into PE.

That doesn't answer my question. If you do not agree with what I wrote just above, you have to tell me where the energy of the thrust is going.

Come back to what I mentioned in post #8. Take two identical rockets, one close to Earth, the other in the middle of empty space. The same engines provide the same thrust. The rocket in the middle of nowhere accelerates from the thrust, gaining KE. The rocket near Earth gains no KE because the thrust is completely compensated by the force of gravity. Where does the energy of the thrust go?

Actually, if you review the initial scenario, the requirement is not that thrust remain constant, but that it BALANCE gravity. So, the rocket in the middle of nowhere would need minuscule thrust (gravity never goes to zero, at least theoretically). The energy from thrust is consumed in offsetting gravity. If you have two opposing forces of the same magnitude acting on the same object, there will be no change to the velocity of that object - it will remain stationary or maintain whatever initial velocity it had. Therefore, a rocket traveling directly up the line of gravitational force at 100M/s will maintain that velocity if thrust is managed to exactly balance gravity. Hence, I ask again, from where does the additional PE come in this scenario? There is no apparent source.

My posit makes sense. PE is not created unless the object is subjected to a NET force (gravitational in this case), which would occur the instant thrust no longer balances gravity.

If you believe the PE in PE = m*g*h is not affected by thrust, I have to disagree on another level. Calculate the KE that would result from your PE at a given height WITH and WITHOUT some thrust. You will find that as long as any thrust can be provided indefinitely, the PE available to be converted to KE is dependent on the NET g, not 9.8M/s/s.

The error lies in assuming thrust stops. And, as in my posit, THAT is exactly when/where the PE gets created. I still can't find any available source of energy to generate PE when thrust = gravity and we have any existing velocity (UP).

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And, uncomfortably, that means there is no PE accumulated when thrust = gravity and a constant (up) velocity. It also means that the constant velocity will remain until the thrust can't be supplied to balance gravity.

- #20

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PE is created by the position of the object in the gravitational field. By virtue of being at a height h an object has the capacity to do work mgh. Therefore, since energy is the capacity to do work, the object has an amount of energy equal to mgh which we call gravitational potential energy.So, please help me understand how the PE is created,

Right. That is correct, the velocity is constant.keeping in mind the equation V = V0 + a*t. If thrust = gravity, this equation results in constant velocity, right?

However, potential energy does not depend on velocity. It depends on position, specifically height.

There is no need for the energy to be KE before it is converted to PE. In the case of a rocket with thrust = gravity the energy comes from the chemical energy of the rocket fuel.If it have constant velocity with thrust = gravity, where does the KE come from to be converted to PE.

- #21

jbriggs444

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Hence, I ask again, from where does the additional PE come in this scenario?

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It is. If it included thrust then it would be a and not g.If in the equation PE = m*g*h, g is independent of thrust,

You don’t calculate KE from height. You calculate KE from v, specifically ##KE=\frac{1}{2} m v^2##the KE you calculate from the height

I have no idea how you think that this example supports your conclusion. The PE=0 on landing regardless of whether it is a crash or a gentle landing. The PE is irrelevant to the gentleness of the landing.We KNOW that is not true by watching the SpaceX booster rockets land. (They did not crash land; they landed very softly.) Hence, the PE from a given height is NOT calculated by m*g*h, but by m*(g - any opposing force)*h, or, the Net g.

No, it is not a proof. You have the PE the same and you vary the external work done and find that the KE varies. That implies that work can change your KE, not that the initial PE is different.With thrust at half gravity, the KE at ground contact will be half that of the KE achieved with unopposed gravity. That is the proof that the original PE must be reduced by any offsetting thrust.

- #23

Mister T

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If it have constant velocity with thrust = gravity, where does the KE come from to be converted to PE. This is where I can't make ends meet.

The energy comes from the fuel burned to produce the thrust.

What makes you think it has to be kinetic energy that's converted to potential energy? If that were the case conservation of energy would be a purely dynamical principle, but it is a much grander principle as it also encompasses thermodynamic considerations.

- #24

jbriggs444

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I think that I see a way to make sense of this within the conventional framework of potential energy.

At this point, assuming the thrust fell below the force of gravity, velocity and its corresponding KE would begin to diminish with a corresponding increase in PE.

Briefly, the idea is that if you have a thruster that is designed so that the thrust it produces is everwhere equal and opposite to the force of gravity in that location then:

1. Under this hypothetical, the force of the thruster amounts to a conservative vector field with its own associated potential.

2. You can compute a potential associated with the net force, gravity + thruster. The force is zero everywhere, so the potential is zero everywhere.

3. If you turn off the thruster, you now have a different potential field with a different potential at the object's current position.

We do not ordinarily consider thrusters to produce a potential because the force from a thruster does not normally amount to a conservative vector field. Gravitational potential energy is calculated based on gravity, not upon gravity plus some hypothetical thruster that may or may not be turned on.

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