Confusion about Gravitational Potential

AI Thread Summary
Gravitational potential energy can be expressed in two ways: U(h) = mgh for uniform gravitational fields and U = -GMm/r for spherical bodies. The first formula assumes the zero potential is at ground level, while the second sets it at infinity, leading to the negative sign. Changes in potential energy are the focus, as constants can be added without affecting physical outcomes. The mgh approximation is useful for small height changes compared to the radius of the spherical body, simplifying calculations. Understanding these distinctions clarifies the concepts of gravitational potential energy.
dcl
Messages
54
Reaction score
0
Heya's
Im a bit confused reguarding gravitational potential energy.
I've seen 2 different calculations

U(h) = mgh
where h is the height above the ground.

and
U = -GMm/r (off the top of my head, havnt got my notes here with me right now)

Could anyone help clear this up for me? Why does one expression have a negative potential?
 
Physics news on Phys.org
dcl said:
Heya's
Im a bit confused reguarding gravitational potential energy.
I've seen 2 different calculations

U(h) = mgh
where h is the height above the ground.

and
U = -GMm/r (off the top of my head, havnt got my notes here with me right now)

Could anyone help clear this up for me? Why does one expression have a negative potential?
The former, U(h), applies to a uniform gravitational field where the zero of potential is at h = 0. The later, U = -GMm/r, applies to the gravitational field of a spherical body where the zero potential is at r = infinity. Plot each on a graph and you'll notice that each is a decreasing function of r, only the shape of the curve is different.

Keep in mind that only changes in potential are physically meaningful. You can always add a constant without changing the physics. The constant is usually chosen to make the math as simple as possible.

Pete
 
I see, makes sense. Thanks for clearing that up.
 
Just to add to what pmb_phy said, we use mgh much of the time in calculations because it is a good approximation to the change in the gravitational potential (given by U = -GMm/r) for small distances. i.e. when h << r.

Matt
 
r=R+h

So,

U =\frac {-GMm} {r} = \frac {-GMm} {R+h} = \frac {-GMm} {R} (1 + h/R)^{-1}

For h<<R, you can expand the last term binomially, and neglect terms of second order and up. So,

U = \frac {-GMm} {R} (1 - h/R) =\frac {-GMm} {R} + \frac {GMmh} {R^2}
= Constant + (m)*(\frac {GM} {R^2})*h = Constant + mgh

Since we are interested only in changes in potential, we can throw away the constant term.
 
The rope is tied into the person (the load of 200 pounds) and the rope goes up from the person to a fixed pulley and back down to his hands. He hauls the rope to suspend himself in the air. What is the mechanical advantage of the system? The person will indeed only have to lift half of his body weight (roughly 100 pounds) because he now lessened the load by that same amount. This APPEARS to be a 2:1 because he can hold himself with half the force, but my question is: is that mechanical...
Some physics textbook writer told me that Newton's first law applies only on bodies that feel no interactions at all. He said that if a body is on rest or moves in constant velocity, there is no external force acting on it. But I have heard another form of the law that says the net force acting on a body must be zero. This means there is interactions involved after all. So which one is correct?
Thread 'Beam on an inclined plane'
Hello! I have a question regarding a beam on an inclined plane. I was considering a beam resting on two supports attached to an inclined plane. I was almost sure that the lower support must be more loaded. My imagination about this problem is shown in the picture below. Here is how I wrote the condition of equilibrium forces: $$ \begin{cases} F_{g\parallel}=F_{t1}+F_{t2}, \\ F_{g\perp}=F_{r1}+F_{r2} \end{cases}. $$ On the other hand...
Back
Top