Integral for work done leading to potential energy sign confusion

[Ebby]

Homework Statement

What is the work done by gravity in moving the particle from a distance of infinity to a distance R from the centre of the Earth (where R > the radius of the Earth)?

Relevant Equations

W = F * r

What is the work done by gravity in moving the particle from a distance of $\infty$ to a distance $R$ from the centre of the Earth (where $R$ > the radius of the Earth)?

The answer is obvious, since the displacement and the force of gravity are in the same direction. Therefore, gravity does positive work in the amount $W_{grav} = GMm/R$.

However, I want to show you how I am tempted to formulate my answer wrongly, which leads to an answer of the wrong sign. I am hoping someone can explain clearly why this is wrong. I keep coming back to problems like this, and am plagued by the same confusion every time. I am hoping finally to resolve this issue in my mind!

So, we're going to use the equation generally given with the form: $W = \int_a^b \vec F \cdot \, d\vec r$. In this particular case we have: $$W_{grav} = \int_\infty^R -\frac {GMm} {x^2} \, {\hat x} \cdot -dx \, \hat x$$
$$= \int_\infty^R \frac {GMm} {x^2} \, dx$$
$$= GMm \int_\infty^R \frac 1 {x^2} \, dx$$
$$= \frac {-GMm} {x} \left. \right |_{\infty}^R$$
$$= \frac {-GMm} {R}$$

Which is of course wrong.

Additionally, I have difficulty thinking of a force (gravity in this case) doing positive work on an object, and yet the object having less P.E. afterwards, not more. It's more intuitive for me to think of an object being "worked on" and consequently having more energy. What is a better way to think about this?

EDIT: I see that my sketch is slightly wrong. There's a $\hat x$ that needs to be the numerator, not the denominator. Anyway, all the LaTeX is fine.

 

[Hill]

There should be dx rather than -dx in the first integral. dx goes in the x-axis direction, the force is negative as it is in the opposite direction, and the direction of motion is determined by the limits of the integral.
The work is converted to the kinetic energy of the falling body.

 

[kuruman]

Your problem is using the equation for work $W = \int_a^b \vec F \cdot \, d\vec r$ incorrectly. You can (a) use the "cosine" definition of work $dW=\vec F \cdot \, d\vec r=F\,dr\,\cos\theta$, in which case you have to worry about the angle between the force and the displacement or (b) use unit vector notation but not a mix of both which is what you did. The second method is foolproof if you set it up right. Here is how:

First write the two vectors in unit vector form. I will be using the radial direction but you can easily change symbols $r$ and $\mathbf{\hat {r}}$ to $x$ and $\mathbf{\hat {x}}$). By definition $\mathbf{\hat {r}}$ points radially out and $r$ increases in that direction which means $d\mathbf{r}=dr~\mathbf{\hat {r}}$ always. This is where you went wrong.

Next write the force and form the dot product
$\mathbf{F}=\dfrac{GmM}{r^2}(-\mathbf{\hat {r}}).$
$\mathbf{F}\cdot d\mathbf{r}=\dfrac{GmM}{r^2}(-\mathbf{\hat {r}})\cdot dr~\mathbf{\hat {r}}=-\dfrac{GmM}{r^2}dr.$
You now have the integrand, so put it under the integral sign, set the limits and integrate $$W=\int_{\infty}^R \left(-\frac{GmM}{r^2}dr\right)=\frac{GmM}{R}.$$ Note that if you were integrating the opposite way (radially out), all you would have to do is flip the limits of integration which flips the overall sign of the integral. When applied correctly, the unit vector method automatically takes care of all signs regardless of vector direction and sense of integration.

 

[kuruman]

It's more intuitive for me to think of an object being "worked on" and consequently having more energy.

What kind of energy? Answer: Kinetic not potential. The work energy theorem says that the change in kinetic energy is equal to the total work done on the object, $\Delta K=W_{tot}$. In this case $$\Delta K =W_g=\frac{GMm}{R}$$ and the kinetic energy increases. The gravitational potential energy change is, by definition, the negative of the work done by gravity, $$\Delta U = - W_g=-\frac{GMm}{R}.$$ The two equations can be combined to a statement of mechanical energy conservation: $$\Delta K+\Delta U=0$$ which says that the sum of energy changes is zero, i.e. one kind of energy increases at the expense of the other. The conversion from one form to another is mediated by the conservative force which does work on the object that is either positive (increase of kinetic and decrease of potential) or negative (decrease of kinetic and increase of potential).

 

[PhDeezNutz]

you don’t assign a direction to dx in the integral, that is taken care of by the order of the integration bounds. So don’t negate it twice.

 

[Ebby]

Thank you. This is very helpful, especially @kuruman's explanation.