I recently started studying a book about Orbital Mechanics. Its title is "Orbital Mechanics for Engineering students", 4th edition. I am currently at section 1.7 (page 30) trying to find the absolute velocity and acceleration of a point P "flying" around the Earth. The followig figure shows the Earth-centered Inertial frame (XYZ) as well the local ENU reference system (xyz):
I cannot understand the derivation of the angular velocity of the xyz frame (local ENU reference system). The book says that the angular velocity $\omega$ of the xyz frame relative to the non-rotating Earth is found in terms of rates of change of latitude $\phi$ and longitude $\Lambda$:
$$\omega = -\dot{\phi}\hat{i}+\dot{\Lambda}\cos \phi \hat{j} + \dot{\Lambda} \sin \phi \hat{k} \tag{1}$$
and then
$$ \frac{\partial \hat{k}}{\partial t} = \omega \times \hat{k} = \dot{\Lambda}\cos \phi \hat{i} + \dot{\phi}\hat{j} \tag{2}$$
I cannot understand the derivation of $\omega$ (Eq. 1) as well as the Eq.2. I tried to derive it by myself, but I could not figure that out. If answering with the derivation of $\omega$ is too much, having a link to some additional resources would be great too.
Best Answer
Their formula $$ \frac{\partial\hat{\boldsymbol{k}}}{\partial t}=\boldsymbol{\omega}\times\hat{\boldsymbol{k}} $$ indicates that $\boldsymbol{\omega}$ is the spin angular velocity of the radial unit vector $\hat{\boldsymbol{k}}$ that moves when latitude and longitude relative to earth change with time.
I prefer to use a more standard notation for spherical polar coordinates:
The conversion between the Cartesian and polar basis vectors is $$ \begin{bmatrix}\hat{\boldsymbol{r}}\\\hat{\boldsymbol{\theta}}\\\hat{\boldsymbol{\phi}}\end{bmatrix}= \begin{bmatrix} \sin\theta\cos\phi&\sin\theta\sin\phi&\cos\theta\\ \cos\theta\cos\phi&\cos\theta\sin\phi&-\sin\theta\\ -\sin\phi &\cos\phi & 0 \end{bmatrix} \begin{bmatrix}\hat{\boldsymbol{x}}\\\hat{\boldsymbol{y}}\\\hat{\boldsymbol{z}}\end{bmatrix}\,. $$ Therefore your $\displaystyle\frac{\partial\hat{\boldsymbol{k}}}{\partial t}$ is \begin{align} \dot{\hat{\boldsymbol{r}}}&=\Big(\dot\theta\cos\theta\cos\phi-\dot\phi\sin\theta\sin\phi\Big)\hat{\boldsymbol{x}}+\Big(\dot\theta\cos\theta\sin\phi+\dot\phi\sin\theta\cos\phi\Big)\hat{\boldsymbol{y}}-\dot\theta\sin\theta\,\hat{\boldsymbol{z}}\\ &=\dot\theta\,\hat{\boldsymbol{\theta}}+\dot\phi\,\sin\theta\,\hat{\boldsymbol{\phi}}\,. \end{align} Since $\sin\theta$ is $\cos(\theta-\pi/2)=\cos(\pi/2-\theta)$ this is the same equation that they have in their notation.
Their angular velocity is in my notation $$ \boldsymbol{\omega}=\dot\theta\hat{\boldsymbol{\phi}}\color{red}{-}\dot\phi\,\sin\theta\,\hat{\boldsymbol{\theta}} +\dot\phi\,\cos\theta\,\hat{\boldsymbol{r}}\,. $$ Note that their $-\dot\phi$ becomes $\dot\theta$ because they measure the zenith angle from the equator instead of the north pole. (They seem however to have a sign error in front of the $\hat{\boldsymbol{\theta}}$ term.) The verification that this angular velocity satisfies $$\tag{1} \dot{\hat{\boldsymbol{r}}}=\boldsymbol{\omega}\times \hat{\boldsymbol{r}} $$ is quite simple when one uses the relationships $$\boxed{\quad\phantom{\Bigg|} \hat{\boldsymbol{r}}\times\hat{\boldsymbol{\theta}}=\hat{\boldsymbol{\phi}}\,,\quad \hat{\boldsymbol{\theta}}\times\hat{\boldsymbol{\phi}}=\hat{\boldsymbol{r}}\,,\quad \hat{\boldsymbol{\phi}}\times\hat{\boldsymbol{r}}=\hat{\boldsymbol{\theta}}\,.\quad} $$ Since I have seen people being baffled about the correctness of those let's prove them: \begin{align}\require{cancel} \hat{\boldsymbol{r}}\times\hat{\boldsymbol{\theta}}&=\Big(\cancel{\sin\theta\cos\phi\cos\theta\sin\phi}-\cancel{\cos\theta\cos\phi\sin\theta\sin\phi}\Big)\,\underbrace{\hat{\boldsymbol{x}}\times\hat{\boldsymbol{y}}}_{\textstyle\hat{\boldsymbol{z}}}\\ &~-\Big(\sin^2\theta\sin\phi+\cos^2\theta\sin\phi\Big)\,\underbrace{\hat{\boldsymbol{y}}\times\hat{\boldsymbol{z}}}_{\textstyle\hat{\boldsymbol{x}}}\\ &~+\Big(\sin^2\theta\cos\phi+\cos^2\theta\cos\phi\Big)\, \underbrace{\hat{\boldsymbol{z}}\times\hat{\boldsymbol{x}}}_{\textstyle\hat{\boldsymbol{y}}}\\ &=-\sin\phi\,\hat{\boldsymbol{x}}+\cos\phi\,\hat{\boldsymbol{y}}\\ &=\hat{\boldsymbol{\phi}}\,,\\[2mm] \hat{\boldsymbol{\theta}}\times\hat{\boldsymbol{\phi}}&=\Big(\cos\theta\cos^2\phi+\cos\theta\sin^2\phi\Big)\,\underbrace{\hat{\boldsymbol{x}}\times\hat{\boldsymbol{y}}}_{\textstyle\hat{\boldsymbol{z}}}+\sin\phi\sin\theta\,\underbrace{\hat{\boldsymbol{z}}\times\hat{\boldsymbol{x}}}_{\textstyle\hat{\boldsymbol{y}}} -\sin\theta\cos\phi\,\underbrace{\hat{\boldsymbol{z}}\times\hat{\boldsymbol{y}}}_{\textstyle-\hat{\boldsymbol{x}}}\\ &=\hat{\boldsymbol{r}}\,,\\[2mm] \hat{\boldsymbol{\phi}}\times\hat{\boldsymbol{r}}&=-\Big(\sin\theta\cos^2\phi+\sin\theta\sin^2\phi\Big)\,\underbrace{\hat{\boldsymbol{x}}\times\hat{\boldsymbol{y}}}_{\textstyle\hat{\boldsymbol{z}}}-\sin\phi\cos\theta\,\underbrace{\hat{\boldsymbol{x}}\times\hat{\boldsymbol{z}}}_{\textstyle-\hat{\boldsymbol{y}}} +\cos\theta\cos\phi\,\underbrace{\hat{\boldsymbol{y}}\times\hat{\boldsymbol{z}}}_{\textstyle\hat{\boldsymbol{x}}}\\ &=\hat{\boldsymbol{\theta}}\,. \end{align}
Remark.
One thing I don't understand though is where the $\hat{\boldsymbol{r}}$-component in their angular velocity comes from. It has obviously no influence on the validity of the equation (1). Also, if we drop it it is easiy to see that the angular velocity becomes equal to the orbital angular velocity $$ \boldsymbol{\omega}=\hat{\boldsymbol{r}}\times \dot{\hat{\boldsymbol{r}}}\,. $$