Gravitational potential energy near the earth

[demonelite123]

I am a bit confused about gravitational potential energy near the earth, namely the formula given by mgh.

I know that potential energy is defined as U(x) = -W(x₀ to x) where x₀ is our chosen reference point. Let's take the Earth's surface as the zero point and let's travel upwards to a point x. Since mg is pointing down and my displacement is pointing up, the dot product is negative so W(x₀ to x) = -mgx. Therefore, U(x) = -(-mgh) = mgh.

however, let's say i want to travel downwards instead. then since mg is pointing down and my displacement is also pointing down, the dot product is positive this time so W(x₀ to x) = mgx (x is negative). but then U(x) = -(mgx) = -mgx. If i say travel downwards from x = -2 to x = -4, i get U(-4) - U(-2) = mg(4) - mg(2) > 0, which doesn't make sense since if i am going downwards, shouldn't my change in potential energy be negative?

 

[Simon Bridge]

Do you have a higher PE on the ground or higher up?

 

[demonelite123]

Using the formula U(x) = -mgx derived from taking the work going from the zero point (ground) to a point below the surface, it seems that points above the surface have negative potential.

what i don't understand is why both approaches don't give me the same U(x). for U(x) = mgx, i traveled upwards so i can only use positive values of x (above the surface). If i want to use negative values of x, I travel downwards but i come up with U(x) = -mgx which seems to contradict with the function i got from traveling upwards since taking points above the surface, one gives me a positive PE and the other gives me a negative PE.

is there a way to reconcile them?

 

[willem2]

I am a bit confused about gravitational potential energy near the earth, namely the formula given by mgh.

I know that potential energy is defined as U(x) = -W(x₀ to x) where x₀ is our chosen reference point. Let's take the Earth's surface as the zero point and let's travel upwards to a point x. Since mg is pointing down and my displacement is pointing up, the dot product is negative so W(x₀ to x) = -mgx. Therefore, U(x) = -(-mgh) = mgh.

however, let's say i want to travel downwards instead. then since mg is pointing down and my displacement is also pointing down, the dot product is positive this time so W(x₀ to x) = mgx (x is negative)

You lost a minus sign here.

 

[Simon Bridge]

U=-W, W=F.d

F=-(mg)j
d=(y-y0)j if we go "up" (final minus initial)
... since y0<y, d>0 - for later reference, put h=|d|
U= -(-mg)[+(y-y0)] = mgh

if you go "down" the d=(y0-y)j = - (y-y0)j
U=-(-mg)[-(y-y0)] = -mgh

Keeping track of the minus signs is, indeed, the trick to keeping things consistent.

 

[Doc Al]

however, let's say i want to travel downwards instead. then since mg is pointing down and my displacement is also pointing down, the dot product is positive this time so W(x₀ to x) = mgx (x is negative). but then U(x) = -(mgx) = -mgx.

As others have already explained, you are essentially double counting the minus sign.

ΔU = -(F)*(Δx) = -(-mg)*(Δx) = (mg)*(Δx)

When Δx is positive, ΔU is positive; when Δx is negative, ΔU is negative.

 

[Simon Bridge]

@demonelite: this sound good to you?

 

[demonelite123]

@demonelite: this sound good to you?

sorry about my late response. I'm still a bit confused. for the case "going up" i agree that F = (-mg)j and d = (y - y₀)j since as you said it was final point (y) - initial point (y₀).

but for the "going down" case, F = (-mg)j but how come d = (y₀ - y)j? isn't final point - initial point still y - y₀?

thanks for all the replies so far everyone.

 

[jtbell]

sorry about my late response. I'm still a bit confused. for the case "going up" i agree that F = (-mg)j and d = (y - y₀)j since as you said it was final point (y) - initial point (y₀).

but for the "going down" case, F = (-mg)j but how come d = (y₀ - y)j? isn't final point - initial point still y - y₀?

In the "going down" case the object is going "from" y "to" y₀, so final point - initial point is y₀ - y which is negative.

 

[Simon Bridge]

Thank you jtbell :)
@deminelite123 - since y > y₀, in order to go down you have to start at y.

 

[demonelite123]

Thank you jtbell :)
@deminelite123 - since y > y₀, in order to go down you have to start at y.

ah ok. what you said earlier makes sense now. thanks!

 

[CrazyNeutrino]

Like everyone said, you missed a minus sign. :)

 

[Simon Bridge]

ah ok. what you said earlier makes sense now. thanks!

No worries, happens to the best of us.

In general, it is a good discipline to have the same labels refer to the same things. In this case: locations - when you were going down, you forgot to account for your y0>y so that y-y0 < 0. Try it with a diagram.

 

[CrazyNeutrino]

Wait a second... The professor at MIT said that when I do positive work, gravity does negative work and when I do negative work, gravity does positive work. How does this make sense. When I do negative work from A to B then gravity is also doing negative work right? since it is -mg?

 

[Simon Bridge]

Keep track of the directions ...

If you lower a mass at constant speed, then you are exerting a force in the opposite direction to the displacement, and gravity is acting in the same direction as displacement.

In your example, A must be higher than B so the displacement from A to B is negative.

 

[CrazyNeutrino]

Thanks a lot

 

[CrazyNeutrino]

Thanks!

 

[Simon Bridge]

No worries.
Tracking the minus signs can be tricky - especially if things are slowing down as well as changing direction.