Physics: Potential Energy Beyond Earth's Surface

• nhmllr
In summary: Earth. As discussed earlier, this is an approximation that does not take into account the changing gravitational force. The equation -mMG/r, on the other hand, is a more accurate representation of potential energy that considers the changing force. However, when h is much smaller than the radius of the Earth, the two equations become equivalent and both represent the potential energy of the object.
nhmllr
Sorry to post two PE posts in the same day...
Anyway, I understand how near the surface of the earth, PE = mgh,
becuase E = F*d, mg = F, h = d.
When the object hits the ground after height h, it has had the force mg work on it for h
But "g" changes with the distance from earth
Fgravity = mMG/r^2
Fgravity * h = mMG/r
(M = Earth's mass)
So should PE = mMG/r (when not near the surface, such as in orbits) because that's how I've seen it been written.
Should it be something different because the force working on it increases as it falls?
Thanks

It is exactly as you described. The equation U = -mgh is an approximation for objects near the surface of the Earth. The equation you described comes from

$$U = \int^{x}_{\infty}\frac{GMm}{r^{2}}dr$$
$$U = -\frac{GMm}{x}$$

which takes into account the changing gravitational force. In physics, it is often convinient to define potential energy as the amount of work needed to bring an object from infinity to the current location. This formula follows from that convention.

Yuqing said:
It is exactly as you described. The equation U = -mgh is an approximation for objects near the surface of the Earth. The equation you described comes from

$$U = \int^{x}_{\infty}\frac{GMm}{r^{2}}dr$$
$$U = -\frac{GMm}{x}$$

which takes into account the changing gravitational force. In physics, it is often convinient to define potential energy as the amount of work needed to bring an object from infinity to the current location. This formula follows from that convention.
Hm... that's interesting how you got it from the sums of all of the forces through a distance...
But how do you know that the work needed to get it from infinity to x is the same as from x to 0? In fact, when you plug 0 and x into the integral, you get infinity!

They are not the same. The force at the point 0 contains a singularity, it makes no real sense to define potentials from 0.

Yuqing said:
They are not the same. The force at the point 0 contains a singularity, it makes no real sense to define potentials from 0.

What I mean is, you defined PE as the work done to get the object from infinity to x. mgh means that PE is work that will be done when you get it from x to 0. However, that will be infinite. So I'm wondering how the work to done to get it from infinity to x = work done to get it from x to 0.

nhmllr said:
What I mean is, you defined PE as the work done to get the object from infinity to x. mgh means that PE is work that will be done when you get it from x to 0. However, that will be infinite. So I'm wondering how the work to done to get it from infinity to x = work done to get it from x to 0.

The work done to get it from infinity to X DOES NOT EQUAL the work done to get it from X to 0.

An object at infinity has zero gravitational potential energy (just set x=infinity in Yuqing's equation above). And at some point X close to the surface of Earth, the GPE is finite and non-zero. To the work done in moving the object from infinity to the new location X is FINITE, not infinite.

Now, the gravitational force at r=0 is NOT given by F=GMm/r^2. In fact, there's a delta function at r=0; all the gravitational flux emanates from r=0. When integrating to find the amount of work required in moving an object from the center of the Earth to location X, the limits of integration are from some infinitesimal distance from the center of the Earth--which we can call epsilon-- all the way up to the location X. Again, when you follow Yuqing's equation, plugging in these limits (epsilon and X), you'll get a FINITE amount of work done.

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Geezer said:
The work done to get it from infinity to X DOES NOT EQUAL the work done to get it from X to 0.

An object at infinity has zero gravitational potential energy (just set x=infinity in Yuqing's equation above). And at some point X close to the surface of Earth, the GPE is finite and non-zero. To the work done in moving the object from infinity to the new location X is FINITE, not infinite.

Now, the gravitational force at r=0 is NOT given by F=GMm/r^2. In fact, there's a delta function at r=0; all the gravitational flux emanates from r=0. When integrating to find the amount of work required in moving an object from the center of the Earth to location X, the limits of integration are from some infinitesimal distance from the center of the Earth--which we can call epsilon-- all the way up to the location X. Again, when you follow Yuqing's equation, plugging in these limits (epsilon and X), you'll get a FINITE amount of work done.

What I don't understand is that mgh is taught that it's the energy that the object will have when it reaches the ground. -mMG/r has a completely different definition, yet it shares the same name! How are they the same?

Well, mgh actually follows from the integral definition by an approximation.

Let us consider the difference in potential between two points of distance r and r+h.

We have

$$\frac{-GMm}{r+h} - \frac{-GMm}{r} = \frac{GMmh}{r(r+h)}$$

now let us take r as the radius of the Earth and restrict h << r (as is required by mgh).

Since h << r we have r + h $$r+h \approx r$$ where our equation becomes
$$\frac{GMmh}{r^{2}}$$
the term $$\frac{GM}{r^{2}}$$ if you recall is precisely g. Thus the expression reduces to mgh.

nhmllr said:
What I don't understand is that mgh is taught that it's the energy that the object will have when it reaches the ground. -mMG/r has a completely different definition, yet it shares the same name! How are they the same?

If you recall, "mgh" is NOT the energy that the object will have when it hits the ground unless the object falls from rest (and we neglect frictional forces, drag, and other outside forces). The change in potential energy equals the change in kinetic energy. If the object falls from rest (kinetic energy is initially zero), then that gravitational potential energy is converted into kinetic energy. The first sentence of this paragraph and this last statement may seem like they're equivalent, but they're not.

nhmllr said:
mgh means that PE is work that will be done when you get it from x to 0.
mgh is an approximation for relatively short distances from the earth.

mgh would be the formula for an infinite plane with enough mass per unit area to produce an acceleration of g on any mass. The force from such a plane is constant regardless of distance from the plane, and the maximum potential energy from such a plane mgh as h -> ∞ is infinite. For a infinite plane the usual "reference" distance is 0 distance from the plane.

For a finite size source of gravity the maximum potential is finite, so a distance of ∞ is the usual "reference" distance.

Although not common for gravity, an infinite line source for a field is sometimes used in electrical field problems. In this case for gravity from an infinite line, the potential energy would be m g ln(h), where h is distance from the line. In this case some non-zero but finite distance is normally used as the reference points (note that ln(1) = 0).

Yuqing said:
Well, mgh actually follows from the integral definition by an approximation.

Let us consider the difference in potential between two points of distance r and r+h.

We have

$$\frac{-GMm}{r+h} - \frac{-GMm}{r} = \frac{GMmh}{r(r+h)}$$

now let us take r as the radius of the Earth and restrict h << r (as is required by mgh).

Since h << r we have r + h $$r+h \approx r$$ where our equation becomes
$$\frac{GMmh}{r^{2}}$$
the term $$\frac{GM}{r^{2}}$$ if you recall is precisely g. Thus the expression reduces to mgh.

I agree, Yuqing.

Yuqing said:
Well, mgh actually follows from the integral definition by an approximation.

Let us consider the difference in potential between two points of distance r and r+h.

We have

$$\frac{-GMm}{r+h} - \frac{-GMm}{r} = \frac{GMmh}{r(r+h)}$$

now let us take r as the radius of the Earth and restrict h << r (as is required by mgh).

Since h << r we have r + h $$r+h \approx r$$ where our equation becomes
$$\frac{GMmh}{r^{2}}$$
the term $$\frac{GM}{r^{2}}$$ if you recall is precisely g. Thus the expression reduces to mgh.

That's a great explanation. I understand it now! Thanks everybody for your patience

1. What is potential energy?

Potential energy is the energy that an object possesses due to its position or configuration. It is the potential for an object to do work based on its position relative to other objects or forces.

2. How does potential energy differ from kinetic energy?

Kinetic energy is the energy an object possesses due to its motion, while potential energy is based on an object's position or configuration. They are both forms of energy that can be converted back and forth.

3. How does potential energy change with respect to distance from Earth's surface?

Potential energy increases as the distance from Earth's surface increases. This is because the gravitational force between two objects decreases as the distance between them increases, resulting in an increase in potential energy.

4. Can potential energy be negative?

Yes, potential energy can be negative. This occurs when the object's position is below the reference point, such as an object falling below the ground level. In this case, the potential energy is considered to be negative because it requires work to move the object back to the reference point.

5. How is potential energy utilized in space exploration?

Potential energy is essential for spacecraft to escape Earth's gravitational pull and travel beyond Earth's surface. By utilizing the potential energy of an object in orbit, spacecraft can achieve escape velocity and travel to other planets or even leave our solar system.

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