Gravitational potential energy near the earth

  • #1

Main Question or Discussion Point

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?
 

Answers and Replies

  • #2
Simon Bridge
Science Advisor
Homework Helper
17,841
1,642
Do you have a higher PE on the ground or higher up?
 
  • #3
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?
 
  • #4
1,924
223
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.
 
  • #5
Simon Bridge
Science Advisor
Homework Helper
17,841
1,642
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.
 
  • #6
Doc Al
Mentor
44,865
1,112
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.
 
  • #7
Simon Bridge
Science Advisor
Homework Helper
17,841
1,642
@demonelite: this sound good to you?
 
  • #8
@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.
 
  • #9
jtbell
Mentor
15,484
3,249
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.
 
  • #10
Simon Bridge
Science Advisor
Homework Helper
17,841
1,642
Thank you jtbell :)
@deminelite123 - since y > y0, in order to go down you have to start at y.
 
  • #11
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!
 
  • #12
Like everyone said, you missed a minus sign. :)
 
  • #13
Simon Bridge
Science Advisor
Homework Helper
17,841
1,642
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.
 
  • #14
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?
 
  • #15
Simon Bridge
Science Advisor
Homework Helper
17,841
1,642
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.
 
  • #16
Thanks a lot
 
  • #17
Thanks!
 
  • #18
Simon Bridge
Science Advisor
Homework Helper
17,841
1,642
No worries.
Tracking the minus signs can be tricky - especially if things are slowing down as well as changing direction.
 

Related Threads for: Gravitational potential energy near the earth

Replies
3
Views
872
Replies
10
Views
515
Replies
2
Views
3K
Replies
2
Views
361
Replies
0
Views
1K
Replies
4
Views
23K
Top