Gravitational potential energy

[Chenkel]

TL;DR Summary

I'm watching Walter Lewin's lectures on work and energy and have a couple questions about gravitational potential energy

Hello everyone! I noticed in the derivation of potential energy, Mr Lewin defined the gravitational potential energy of a mass m at point P relative to a much larger mass M. He says the potential energy of m at point P is equal to the work he would have to do to move the mass m from infinity to the point P. The point P is positioned R meters from mass M, and he defined r as the distance from M to m. He says the force that he would have to produce as he moves from infinity to point P is $F_{WL} = \frac {mMG} {r^2}$ he then defines the work he would have to do as $$ W_{WL} = \int_{\infty }^{R} \frac {mMG} {r^2} dr $$

This is confusing to me because I thought potential energy is equal to mass times g times height. I'm guessing this is a special case of the principle that he's expressing, but I fail to see how.

Also the lower limit of integration for his work is larger than the upper limit of integration, how am I supposed to interpret that?

Let me know what you guys think, thank you!

 

[Dale]

Summary: I'm watching Walter Lewin's lectures on work and energy and have a couple questions about gravitational potential energy

This is confusing to me because I thought potential energy is equal to mass times g times height

That is only for the case of a uniform gravitational field. When you start using large distances gravity is no longer uniform and so the uniform field equation doesn’t apply.

 

[kuruman]

The gravitational potential energy of mass $m$ at distance $r$ from the center of the Earth is $$U_g(r)=-\frac{GM_em}{r}$$Now let's say you move this mass from the surface of the Earth at $r=R_e$ to height $h$ above the surface of the Earth. The change in potential energy is $$\Delta U_g=-\frac{GM_em}{R_e+h}-\left(-\frac{GM_em}{R_e}\right)= GM_em\left( \frac{1}{R_e}-\frac{1}{R_e+h}\right)=\frac{GM_emh}{R_e(R_e+h)}.$$This expression is exact and applies to all heights $h$ above the surface of the Earth. Near the surface of the Earth where $h<<R_e$ one can approximate the denominator $R_e(R_e+h)\approx R_e^2$ in which case the change in potential energy becomes $$\Delta U_g=\frac{GM_e}{R_e^2}mh.$$The constant quantity $\frac{GM_e}{R_e^2}$ up front is the acceleration of gravity $g$. Look up the numbers and put them in. To three significant figures, $\frac{GM_e}{R_e^2}=9.78~$m/s². Thus, near the surface of the Earth $$\Delta U_g=mgh.$$This is an elaboration of @Dale's post.

On edit: To also elaborate on @Delta2's post, the potential energy is negative everywhere at finite distances from the center of the Earth. As the mass moves farther from the center of the Earth, the potential energy increases. That's because at infinity the potential energy is zero and a negative number increases as it gets closer to zero.

 

[Delta2]

Summary: I'm watching Walter Lewin's lectures on work and energy and have a couple questions about gravitational potential energy

Also the lower limit of integration for his work is larger than the upper limit of integration, how am I supposed to interpret that?

Yes the limits of integration are correct because we move from infinity to distance R, integral from a=infinity to b=R as we say for $$\int_a^b f(x)dx$$

$W_{WL}$ will come out negative because $G\frac{mM}{r^2}$ is always positive but $dr$ is negative (and dr is negative because the lower limit of integration which is $\infty$ is larger than the higher limit of integration which is $R$).

 

[Chenkel]

Yes the limits of integration are correct because we move from infinity to distance R, integral from a=infinity to b=R as we say for $$\int_a^b f(x)dx$$

$W_{WL}$ will come out negative because $G\frac{mM}{r^2}$ is always positive but $dr$ is negative (and dr is negative because the lower limit of integration which is $\infty$ is larger than the higher limit of integration which is $R$).

Would it be possible to integrate the work you would have to do in order to move the mass from the center of Earth to radius R where point P is in order to get the potential energy? Could we say the following: $$U = \int_{0}^{R} G{\frac{mM}{r^2}}dr$$

I can make some sense of the situation when the potential energy is calculated by totalling the work we perform as we move the object from the center of Earth to point P, but I still lack some intuition as to how we can get the potential energy by integrating from infinity to point P, why should these two methods be equivalent?

 

[Delta2]

Would it be possible to integrate the work you would have to do in order to move the mass from the center of Earth to radius R where point P is in order to get the potential energy? Could we say the following: $$U = \int_{0}^{R} G{\frac{mM}{r^2}}dr$$

I can make some sense of the situation when the potential energy is calculated by totalling the work we perform as we move the object from the center of Earth to point P, but I still lack some intuition as to how we can get the potential energy by integrating from infinity to point P, why should these two methods be equivalent?

No you can't say this, I want to hear your view on why you think this integral from the center of the Earth to R would give the potential energy. It just is not the standard way potential energy is defined.

Btw the gravitational field "inside" the Earth at radius $r<R_e$ is $$G\frac{M_r}{r^2}$$ where $M_r$ the mass of sphere of only radius r (and not the whole mass of the sphere of radius $R_e$). This is a consequence of Gauss's law for gravity or simply a result known as shell theorem which took Isaac Newton a cumbersome calculation to prove, cause he didn't know that his law of inverse square for gravity is equivalent to Gauss's law for gravity.

 

[Delta2]

On second thought I think you might be right, we can put the zero for potential anywhere we want and not necessarily at infinity. Hold on while I expand on this.

 

[Chenkel]

No you can't say this, I want to hear your view on why you think this integral from the center of the Earth to R would give the potential energy. It just is not the standard way potential energy is defined.

Btw the gravitational field "inside" the Earth at radius $r<R_e$ is $$G\frac{M_r}{r^2}$$ where $M_r$ the mass of sphere of only radius r (and not the whole mass of the sphere of radius $R_e$). This is a consequence of Gauss's law for gravity or simply a result known as shell theorem which took Isaac Newton a cumbersome calculation to prove, cause he didn't know that his law of inverse square for gravity is equivalent to Gauss's law for gravity.

It comes from my intuition about movement of objects in a gravitational field, my intuition is that the potential energy increase is proportional to the height I lift it. It's hard for me to see it another way.

I do see a problem with my integral, if I total the work that I have to do to move mass m from center of earth, it would give me an infinite number as the lower limit of integration approaches zero. I wonder if there's a way around this seeming approach of an infinite number.

 

[Delta2]

I wonder if there's a way around this seeming approach of an infinite number.

yes the solution for this problem is as I said that we consider only the mass of sphere of radius $r$ for $r<R_e$, so the gravitation field is actually ($\rho$ the density of Earth which we can consider approximately constant) $$g(r)=G\frac{M_r}{r^2}=G\frac{\rho\frac{4}{3}\pi r^3}{r^2}=G\frac{4\rho\pi}{3}r$$ , so for $r=0$ it is $g(r)=0$.

 

[hutchphd]

I do see a problem with my integral, if I total the work that I have to do to move mass m from center of earth, it would give me an infinite number as the lower limit of integration approaches zero. I wonder if there's a way around this seeming approach of an infinite number.

Yes. You should take the time to actually read the previous answers because your formula for force is only good for r>R.
The rest of your issues stem from the fact that the work supplied by a force is $\vec F \cdot \vec x$ and so in this case $$dW= \vec F \cdot d\vec x$$ but$$d\vec x=-\hat r dr$$

 

[Delta2]

Yes. You should take the time to actually read the previous answers because your formula for force is only good for r>R.
The rest of your issues stem from the fact that the work supplied by a force is $\vec F \cdot \vec x$ and so in this case $$dW= \vec F \cdot d\vec x$$ but$$d\vec x=-\hat r dr$$

I think it is actually $d\vec{x}=\hat r dr$ but $dr$ is negative as we move from infinity to position R.

 

[Chenkel]

I think it is actually $d\vec{x}=\hat r dr$ but $dr$ is negative as we move from infinity to position R.

When you write $d\vec{x}=\hat r dr$ are you saying a little bit of displacement along a unit vector $\hat r$?

 

[Delta2]

When you write $d\vec{x}=\hat r dr$ are you saying a little bit of displacement along a unit vector $\hat r$?

Yes, along the unit vector $\hat r$ but in opposite direction to it because $dr$ is negative. In general if we have an integral $$\int_a^b f(r)dr$$ where $a>b$ (like in this case where $\infty>R$ ) then $dr$ is negative and the whole integral is negative if $f(r)\geq 0$ for $r\in [a,b]$.

 

[vanhees71]

It's much easier explained in terms of the potential of the force:
$$\vec{F}=-\frac{G mM}{r^3} \vec{r}=-\vec{\nabla} V.$$
It's easy to see that by symmetry $V$ is only a function of $r=|\vec{r}|$, i.e.,
$$\vec{F}=-\vec{\nabla} V(r)=-V'(r) \vec{\nabla} r = -V'(r) \frac{\vec{r}}{r} \; \Rightarrow \; -V'(r)=-\frac{G m M}{r^2} \; \Rightarrow \; V(r)=-\frac{G mM}{r}+C,$$
where $C$ is a constant. Usually one makes $V(r) \rightarrow 0$ for $r \rightarrow \infty$, i.e., sets $C=0$.

 

[kuruman]

To @Chenkel : To complete the picture the gravitational potential energy function is given by the integral $$V(r)-V(\text{r=ref})=V(r)-0=-\int_{\text{r=ref}}^r \mathbf{g}\cdot d\mathbf{r}$$where "ref" is the reference point where one chooses the potential to be zero. It can be at infinity, at the center of the sphere or wherever one pleases. With $~\mathbf{g}=-\dfrac{4\pi G \rho}{3}\mathbf{r}$, the potential function at a point $r$ inside the sphere ($r<R$) relative to the center, is
$$V(r)=+\frac{4\pi G \rho}{3}\int_{\text{0}}^r rdr=\frac{2\pi G \rho}{3}r^2=\frac{GM_e r^2}{2R_e^3}.$$If you want the potential at a point above the surface of he Earth ($r>R_e$), then you have to take into account that the field outside changes form and is $\mathbf{g}=-\dfrac{GM_e}{r^2}\mathbf{\hat r}$. Thus, the potential relative to the center at a point above the surface is $$V(r)=\frac{4\pi G \rho}{3}\int_{\text{0}}^{R_e} rdr+GM_e\int_{R_e}^r\frac{dr}{r^2}=\frac{GM_eR_e^2}{2R_e^3}-\frac{GM_e}{r}+\frac{GM_e}{R_e}=-\frac{GM_e}{r}+\frac{3GM_e}{2R_e}.$$ For points above the surface of the Earth, where most calculations are done, it is more convenient to take the zero of potential at infinity instead of the center of the Earth and not have to remember the constant term.

On edit: Corrected typo as indicated in post #16.

 

[Chenkel]

To @Chenkel : To complete the picture the gravitational potential energy function is given by the integral $$V(r)-V(\text{r=ref})=V(r)-0=-\int_{\text{r=ref}}^r \mathbf{g}\cdot d\mathbf{r}$$where "ref" is the reference point where one chooses the potential to be zero. It can be at infinity, at the center of the sphere or wherever one pleases. With $~\mathbf{g}=-\dfrac{4\pi G \rho}{3}\mathbf{r}$, the potential function at a point $r$ inside the sphere ($r<R$) relative to the center, is
$$V(r)=+\frac{4\pi G \rho}{3}\int_{\text{0}}^r rdr=\frac{2\pi G \rho}{3}r^2=\frac{GM_e r^2}{R_e^3}.$$If you want the potential at a point above the surface of he Earth ($r>R_e$), then you have to take into account that the field outside changes form and is $\mathbf{g}=-\dfrac{GM_e}{r^2}\mathbf{\hat r}$. Thus, the potential relative to the center at a point above the surface is $$V(r)=\frac{4\pi G \rho}{3}\int_{\text{0}}^{R_e} rdr+GM_e\int_{R_e}^r\frac{dr}{r^2}=\frac{GM_eR_e^2}{R_e^3}-\frac{GM_e}{r}+\frac{GM_e}{R_e}=-\frac{GM_e}{r}+\frac{2GM_e}{R_e}.$$ For points above the surface of the Earth, where most calculations are done, it is more convenient to take the zero of potential at infinity instead of the center of the Earth and not have to remember the constant term.

You wrote $$\frac{2\pi G \rho}{3}r^2=\frac{GM_e r^2}{R_e^3}$$ if I'm not mistaken, I think density is $$\rho = \frac {M_e} {{\frac 4 3}\pi{R_e}^3}$$So I think the correct equation is
$$\frac{2\pi G \rho}{3}r^2=\frac{GM_e r^2}{2R_e^3}$$

 

[Chenkel]

I believe I have a little intuition now of how the gravitational potential energy is calculated relative to 0 potential energy being at infinity. If we move from infinity to point P, in a closed system, the mechanical energy at infinity is equal to the mechanical energy at point P, but say we hold the object as we move it from infinity to point P, now we would have to slow it down with our body by the amount of joules equaling the change in potential energy, because the magnitude of change in kinetic energy is equal to the magnitude of change in potential energy. This object has lost potential energy as we move from infinity to point P in proportion to how much work we have done moving it.

 

[kuruman]

You wrote $$\frac{2\pi G \rho}{3}r^2=\frac{GM_e r^2}{R_e^3}$$ if I'm not mistaken, I think density is $$\rho = \frac {M_e} {{\frac 4 3}\pi{R_e}^3}$$So I think the correct equation is
$$\frac{2\pi G \rho}{3}r^2=\frac{GM_e r^2}{2R_e^3}$$

You are correct. Thank you for pointing this out. I fixed the earlier post.

 

[kuruman]

I believe I have a little intuition now of how the gravitational potential energy is calculated relative to 0 potential energy being at infinity. If we move from infinity to point P, in a closed system, the mechanical energy at infinity is equal to the mechanical energy at point P, but say we hold the object as we move it from infinity to point P, now we would have to slow it down with our body by the amount of joules equaling the change in potential energy, because the magnitude of change in kinetic energy is equal to the magnitude of change in potential energy. This object has lost potential energy as we move from infinity to point P in proportion to how much work we have done moving it.

You have to be careful how you say what you want to say. If you hold a mass and bring it in from infinity in the gravitational field of the Earth, you have to say how you do this. For example if you move the object at constant speed, there is no change in kinetic energy. In that case, the change in potential energy is negative and equal to the work done by your muscles on the mass. Nothing slows down when the mass is moved from point A to point B in the Earth's gravitational field.

 

[Delta2]

Nothing slows down when the mass is moved from point A to point B in the Earth's gravitational field.

Sorry what exactly do you mean here, If I have a ball and I throw it from point A on the surface of the Earth towards point B some altitude higher the ball will slow down due to the negative gravitational acceleration...

 

[kuruman]

Sorry what exactly do you mean here, If I have a ball and I throw it from point A on the surface of the Earth towards point B some altitude higher the ball will slow down due to the negative gravitational acceleration...

The entire picture is " For example if you move the object at constant speed, there is no change in kinetic energy. In that case, the change in potential energy is negative and equal to the work done by your muscles on the mass. Nothing slows down when the mass is moved from point A to point B in the Earth's gravitational field." The second sentence that you object to refers to the previous sentence where the mass in question is assumed to be moving at constant speed. I am sorry for not making it clearer.

 

[Delta2]

The entire picture is " For example if you move the object at constant speed, there is no change in kinetic energy. In that case, the change in potential energy is negative and equal to the work done by your muscles on the mass. Nothing slows down when the mass is moved from point A to point B in the Earth's gravitational field." The second sentence that you object to refers to the previous sentence where the mass in question is assumed to be moving at constant speed. I am sorry for not making it clearer.

Yes ok I see now thanks. Well the OP by slow down meant that the force by our hand has to counter balance the attractive gravitational force which would accelerate it.

 

[Chenkel]

You have to be careful how you say what you want to say. If you hold a mass and bring it in from infinity in the gravitational field of the Earth, you have to say how you do this. For example if you move the object at constant speed, there is no change in kinetic energy. In that case, the change in potential energy is negative and equal to the work done by your muscles on the mass. Nothing slows down when the mass is moved from point A to point B in the Earth's gravitational field.

So if you move an object at a constant speed you will have to overcome the work done by gravity along that line, and that work to overcome the work of gravity is the work your muscles are doing, it's interesting to see that if you move the object at a constant speed along a circular path where R is constant, I believe you won't need to do any work :)

 

[Delta2]

So if you move an object at a constant speed you will have to overcome the work done by gravity along that line, and that work to overcome the work of gravity is the work your muscles are doing, it's interesting to see that if you move the object at a constant speed along a circular path where R is constant, I believe you won't need to do any work :)

You would have to do work to keep it at constant speed during the circular orbit. This is because as gravity is a conservative force, its work does not depend on the path, so the work of gravity would be the same even if the body was moving on circular path. So you would have to do work to counter that work of gravity even in a circular path.

 

[Chenkel]

You would have to do work to keep it at constant speed during the circular orbit. This is because as gravity is a conservative force, its work does not depend on the path, so the work of gravity would be the same even if the body was moving on circular path. So you would have to do work to counter that work of gravity even in a circular path.

I'm trying to make sense of this, the displacement of the object would map out a circle around M, and the gravitational force would be perpendicular to this displacement vector, wouldn't the work done by gravity be 0?

 

[kuruman]

I'm trying to make sense of this, the displacement of the object would map out a circle around M, and the gravitational force would be perpendicular to this displacement vector, wouldn't the work done by gravity be 0?

That would be the case if the mass is moving in a horizontal plane. I think @Delta2 is talking about a vertical plane. A clever application of masses being raised and lowered at constant speed, not to mention the Archimedes principle, is the Falkirk wheel. Here is a fun and instructive video. There are many more on the web.

 

[Delta2]

I actually thought @Chenkel was talking about bringing the object from infinity by making it in a circular orbit of semi infinite radius.

 

[Delta2]

I'm trying to make sense of this, the displacement of the object would map out a circle around M, and the gravitational force would be perpendicular to this displacement vector, wouldn't the work done by gravity be 0?

Ah that kind of circle, well in this case the object remains at the same infinite distance from the mass M, the work of weight is indeed 0 but there is no change in its gravitational potential energy either.

 

[Chenkel]

I think one has to admit that the standard definition for gravitational potential energy is a little unintuitive, when we calculate the gravitational potential energy of a mass on the surface of earth, it is not 0 as you might expect, it's the amount of work you would have to do to move the mass from infinity far away in the sky to the surface of earth. So if you have a mass positioned at infinity, it will have 0 potential energy, and it can actually accelerate. Of course you could make 0 the surface of earth, in which case your potential energy would be mgh (a little more intuitive), but this only works when the change in g is relatively small. The change in potential energy says a lot more about the potential motion of an object than just the potential energy of mass at a point by itself.

 

[hutchphd]

when we calculate the gravitational potential energy of mass on the surface of Earth it is not 0 as you might expect, it's the amount of work you would have to do to move the mass from infinity far away in the sky to the surface of earth.

This is exactly the same logic as used by many geocentric natural philosophers and clergy. Of course its intuitive power would seem less to anyone not on earth. The more descriptive and less neutral term is prejudiced, not intuitive.

 

[Chenkel]

This is exactly the same logic as used by many geocentric natural philosophers and clergy. Of course its intuitive power would seem less to anyone not on earth. The more descriptive and less neutral term is prejudiced, not intuitive.

mgh is more local to earth, than say mars, so I can see your point about geocentric natural philosophy. I would say intuition of approach depends on the problem at hand, for objects near earth, mgh, at least to me, seems to be far more intuitive. I don't think I'm biased in which formula I use, but I think elegance comes with simplicity and common sense, and knowing which formula to use for the problem at hand. I think it's hard to say a method is prejudice without knowing what the specific physics problem is at hand. I'm not sure what you mean by less neutral, perhaps you mean less universal? And finally I wonder what you mean by more descriptive, what quality makes something more descriptive?

 

[kuruman]

$\dots$ it's the amount of work you would have to do to move the mass from infinity far away in the sky to the surface of earth.

The definition of potential energy is much more general than that and in any case it is not related to the work "you have to do." A correct definition of the potential energy of a system consisting of the Earth and a mass is the negative of the work done by the force of gravity on the mass as the mass moves from a reference point to another point. Note that
1. It is the conservative force of gravity that does the work, not "you" and infinity as a reference point is too restrictive.
2. It is not the mass that "has" potential energy. It is the configuration, i.e. the relative position, of the Earth-mass system that is associated with potential energy. A change in the relative position results in a change of potential energy. If you exclude the Earth from the system, the mass can only have kinetic energy. If gravity does work on it, the result can only be a change in this system's kinetic energy, nothing else.
3. The choice of the zero of potential energy is arbitrary. One can choose it anywhere one pleases, usually where it's most convenient. For masses near the surface of the Earth, it is convenient to use the surface of the Earth as zero. In fact all zeroes of energy are arbitrary and that includes kinetic energy. We say that an object at rest on the surface of the Earth has zero kinetic energy because we chose tit to be so. If we choose the center of the Earth as the zero, the same object will have quite a bit of kinetic energy that will vary with latitude.

 

[hutchphd]

Descriptive referred to the word "prejudiced". It better describes the situation than "intuitive" I think. Intuitive implies a universality of thought whereas prejudice implies a more parochial mindset. Not really important.

 

[Delta2]

2. It is not the mass that "has" potential energy. It is the configuration,

According to another interpretation it is the gravitational field that has the energy. See that thread of mine with the self potential energy of a sphere, posts by Orodruin.

 

[kuruman]

According to another interpretation it is the gravitational field that has the energy. See that thread of mine with the self potential energy of a sphere, posts by Orodruin.

Yes, indeed. However one must attempt to match the level of the answer to the level of the person who asked the question.