I Line integral for work done by gravity

[bottle_shadow]

TL;DR Summary

Lack of understanding of the technique used to calculate the line integral.

Dear Physics Forums people,

My problem lies in understanding how the following line integral, which represents work done by the gravitational force, was calculated

Specifically, in the integral after the 2nd = sign, they implicitly used \hat{r}\cdot d\vec{s} = dr

I wish to understand what is the argumentation for that.

Following is the picture that ought to provide the argumentation (but doesn't do it for me).

 

[Gaussian97]

Gravitational force is $\vec{F}=-\frac{GMm}{r^2}\hat{r}$. So the integral is $$-GMm\int_{r_A}^{r_B} \frac{\hat{r}\cdot \text{d}\vec{s}}{r^2}$$. Now, you can think of $\text{d}\vec{s}$ as a very little change in position, since it's a vector you can write it in spherical coordinates: $\text{d}r\hat{r}+r\text{d}\theta \hat{\theta}+r\sin{\theta}\text{d}\phi\hat{\phi}$. When you do the scalar product $$\hat{r}\cdot \text{d}\vec{s}=\text{d}r$$.

Of course, this is not completely rigorous, you should parametrize the integral to reduce it to a Riemann integral, but at the end, this is a not completely wrong way to visualise it.

 

[bottle_shadow]

Gravitational force is $\vec{F}=-\frac{GMm}{r^2}\hat{r}$. So the integral is $$-GMm\int_{r_A}^{r_B} \frac{\hat{r}\cdot \text{d}\vec{s}}{r^2}$$. Now, you can think of $\text{d}\vec{s}$ as a very little change in position, since it's a vector you can write it in spherical coordinates: $\text{d}r\hat{r}+r\text{d}\theta \hat{\theta}+r\sin{\theta}\text{d}\phi\hat{\phi}$. When you do the scalar product $$\hat{r}\cdot \text{d}\vec{s}=\text{d}r$$.

Of course, this is not completely rigorous, you should parametrize the integral to reduce it to a Riemann integral, but at the end, this is a not completely wrong way to visualise it.

Thank you! I appreciate your answer :)

 

[jbriggs444]

Specifically, in the integral after the 2nd = sign, they implicitly used \hat{r}\cdot d\vec{s} = dr

I wish to understand what is the argumentation for that.

Static gravity is an example of a conservative vector field. The result of a path integral in a conservative field is independent of the path and is dependent only on the end points.

So use a path that swings from A over to a point at distance $|\vec{r_a}|$ along $\vec{r_S}$ (the integral over that part of the path is obviously zero since the path is always at right angles to the force) and then follows a path on out to B (that part is the simple scalar integral).