Action is defined as $A=\int_{\tau_1}^{\tau_2} L(\tau, q, \dot{q}) d \tau$where $\dot{q}$ means $\frac{d q}{d \tau}$

The word "path" is only spatial so here comes the word "world line" to describe the multi dimensional curve $(\tau, q(\tau))$. To find the world line that minimizes/maximize $A$. Assume $q(\tau)$ is the optimal and any small variation of $q(\tau)+\epsilon \cdot \eta(\tau)$ that keeps the end points intact, aka $\eta(\tau_1)=\eta(\tau_2)=0$, then the action $A$ becomes a function of $\epsilon$.

$\begin{aligned}A(\epsilon)&=\int_{\tau_1}^{\tau_2} L(\tau, q+\epsilon \cdot \eta, \dot{q}+\epsilon \cdot \dot{\eta}) d \tau\\\\0&=\frac{d A(\epsilon)}{d \epsilon} \vert_{\epsilon=0} \\\\&=\int_{\tau_1}^{\tau_2} \frac{\partial L}{\partial q} \cdot \eta +\frac{\partial L}{\partial \dot{q}} \cdot \dot{\eta} d \tau\\\\&=\int_{\tau_1}^{\tau_2} \frac{\partial L}{\partial q} \cdot \eta d \tau+\int_{\tau_1}^{\tau_2}\frac{\partial L}{\partial \dot{q}} \cdot \dot{\eta} d \tau\\\\&=\int_{\tau_1}^{\tau_2} \frac{\partial L}{\partial q} \cdot \eta d \tau+\frac{\partial L}{\partial \dot{q}}\vert_{\tau_2} \eta(\tau_2)-\frac{\partial L}{\partial \dot{q}}\vert_{\tau_1} \eta(\tau_1)-\int_{\tau_1}^{\tau_2}\eta \frac{d \frac{\partial L}{\partial \dot{q}}}{d \tau} d \tau\\\\&=\int_{\tau_1}^{\tau_2} \frac{\partial L}{\partial q} \cdot \eta d \tau -\int_{\tau_1}^{\tau_2}\eta \frac{d \frac{\partial L}{\partial \dot{q}}}{d \tau} d \tau\\\\&=\int_{\tau_1}^{\tau_2} \eta \cdot \left(\frac{\partial L}{\partial q}-\frac{d \frac{\partial L}{\partial \dot{q}}}{d \tau}\right) d \tau\end{aligned}$

Note. By the above math, one can see in variation problem the parameters of starting and end points are fixed so that taking derivative has nothing to do with the integration. This method was first introduced by Lagrange for classical physics where $L(q, \dot{q})=\text{energy}-\text{pertential energy}=\frac{1}{2} \dot{q}^2 +\frac{G M}{q}$ , the variation condition leads to Newton's law $\ddot{q}=-\frac{G M}{q^2}$

To avoid (1) ambiguity in variables and functions (2) pollution/crowdy of function names, let $F(\tau; x)=(F_0(\tau; x),\cdots,F_n(\tau; x))$ denote the functions for the point $q=(q_0,\cdots,q_n)$ with respect to the parametric named '$\tau$'. $L(\tau, q, \dot{q}) d \tau$ in the integral really means $L(x, F(\tau;x), F'(\tau;x)) d x$ and $x$ is just a place-holder in the integral. Therefore for $L(\tau,q_0, \cdots, q_n, \dot{q_0}, \cdots, \dot{q_n})$, the variation condition equations for $q_i(\tau)$ are

$\begin{aligned}i&=0 \cdots n\\\\F_i(\tau;\tau_0)&=q_{i,0}\\\\F_i(\tau;\tau_1)&=q_{i,1}\\\\\frac{d \left(\frac{\partial L}{\partial \dot{q_i}} \vert_{q=F(\tau;\tau),\dot{q}=F'(\tau;\tau)}\right)}{d \tau}&=\frac{\partial L}{\partial q_i} \vert_{q=F(\tau;\tau),\dot{q}=F'(\tau;\tau)}\end{aligned}$

The value of $L$ under a world line is uniquely determined. However, if the starting and end points $q_{i,0}, q_{i,1}$ are the only concern and the starting and end parameter $\tau$ are free to be change by re-parametric, like cases when $L$ has no $\tau$ in its argument, then a re-parametric from $\tau$ to another parametric $s$ starting from 0 could be arranged after a world line is given. The parameter value of starting and end points in the integral could change accordingly from $F_i(\tau;\tau_0)=q_{i,0}$ to $F_i(s;0)=q_{i,0}$ but formally the same variation condition:

$\frac{d \left(\frac{\partial L}{\partial \dot{q_i}} \vert_{q=F(s;s),\dot{q}=F'(s;s)}\right)}{d s}=\frac{\partial L}{\partial q_i} \vert_{q=F(s;s),\dot{q}=F'(s;s)}$

A simple corollary is that the variation condition claims $\frac{\partial L}{\partial \dot{q}} \vert_{q=F(\tau;\tau),\dot{q}=F'(\tau;\tau)}$ are constants if $L$ has no $q$ for its argument which is the law of "whatever" conservation in physics.

question 1

Find shortest $\int_{\tau_0}^{\tau_1} \sqrt{\dot{x}^2+\dot{y}^2}d \tau$ with (loosely wording) $x(\tau_0)=0,x(\tau_1)=10,y(\tau_0)=0,y(\tau_1)=10$

$\begin{aligned}&\text{constant A,B,C}\\\\&\frac{\dot{x}}{\sqrt{\dot{x}^2+\dot{y}^2}}=A\\\\&\frac{\dot{y}}{\sqrt{\dot{x}^2+\dot{y}^2}}=B\\\\&x(\tau)=\frac{A}{B}y(\tau)+C\\\\&0=\frac{A}{B} \cdot 0 +C\\\\&C=0\\\\&10=\frac{A}{B} \cdot 10\\\\&A=B\\\\&x(\tau)=y(\tau)\\\\&\int_{\tau_0}^{\tau_1} \sqrt{\dot{x}^2+\dot{y}^2}d \tau=\int_{\tau_0}^{\tau_1} \sqrt{2} \dot{x} d \tau=\sqrt{2} x \vert_{\tau_0}^{\tau_1}=10\sqrt{2}\end{aligned}$

Albian the starting/end points, any parametric $\tau$ satisfying $x(\tau)=y(\tau)$ is an optimal world line, i.e., a straight line from (0,0) to (10,10). For example, $x(\tau)=y(\tau)=\frac{\tau^2-\tau_0^2}{\tau_1^2-\tau_0^2} \cdot 10$

Given two parametric $\tau$ and $s$ so that $q=F(\tau; x)=F(s; y)$, it implies a change of parameter value $x=g(y)$ such that $F(\tau;g(y))=F(s;y)$; avoid the confusing wording like $q(x)=q(g(y))=q(y)$.

Typically the number $\int_{\tau_0}^{\tau_1} L(x,F(\tau;x),F'(\tau;x)) d x$ will not be the same for different parametric as different parametric change the integral bounds $\tau_0,\tau_1$ or even modify the argument of the parameter argument in $L$. Variation condition finds the optimal world line with fixed $\tau_0$ and $\tau_1$, but its value can change with another parametric even $F$ is already the optimal world line for $\int_{\tau_0}^{\tau_1} L(x,F(\tau;x),F'(\tau;x)) d x$. To make this value of a world lone intrinsic irrelevant to parametric, $L$ must have some special property.

Along a given world line $q=F(\tau;\tau)$, distance can be defined with a positive $M(q,\dot{q})$ who is multipliable in its dot parts and has no parameter argument. Meaning, for any function $\tau=g(s)$, the $M$ has such property:

$\begin{aligned}&M\left(F(\tau;g(s)),F'(\tau;g(s))\right) \cdot g'(s)=M\left(F(\tau;g(s)),F'(\tau;g(s)) \cdot g'(s)\right)\end{aligned}$

Then, a so-called distance metric parameter $s$ implies function $a,b$ of a change of parameter:

$\begin{aligned}&d s =M\left(F(\tau;\tau),F'(\tau;\tau)\right)d \tau\\\\&s \equiv a(\tau)=\int_{\tau_0}^\tau M\left(F(\tau;x),F'(\tau;x)\right)d x, \tau \equiv b(s)\\\\&a'(\tau)=M\left(F(\tau;\tau),F'(\tau;\tau)\right)\\\\&1=M\left(F(\tau;b(s)),F'(\tau;b(s))\right) \cdot b'(s)=M\left(F(\tau;b(s)),F'(\tau;b(s)) \cdot b'(s)\right)\\\\&=M\left(F(s;s),F'(s;s)\right)\\\\&\int_{\tau_0}^{\tau_1} M\left(F(\tau;x),F'(\tau;x)\right)d x=\int_{0}^{a(\tau_1)} M\left(F(s;y),F'(s;y)\right)d y\\\\&x=g(y),\int_{\tau_0}^{\tau_1} M\left(F(\tau;x),F'(\tau;x)\right)d x=\int_{g^{-1}(\tau_0)}^{g^{-1}(\tau_1)} M\left(F(\tau;g(y)),F'(\tau;g(y))\right) g'(y) d y \\\\&=\int_{g^{-1}(\tau_0)}^{g^{-1}(\tau_1)} M\left(F(\tau;g(y)),F'(\tau;g(y) \cdot g'(y))\right) d y=\int_{g^{-1}(\tau_0)}^{g^{-1}(\tau_1)} M\left(F(y;y),F'(y;y)\right) d y\end{aligned}$

And given a world line, value of $\int M(q,\dot{q}) d\tau$, so-called distance, remains intact regardless of whatever re-parametric about $\tau$.

question 2

re-parametric the solution path of question 1 by its $L$ which is also a distance metric.

The solution paths are the world line of $x=y=F(\tau;\tau)=\frac{\tau^2-\tau_0^2}{\tau_1^2-\tau_0^2} \cdot 10$, therefore,

$\begin{aligned}&d s=\sqrt{\dot{x}^2+\dot{y}^2}d \tau=\sqrt{2}\dot{x}d\tau\\\\&s=\sqrt{2}x=10\sqrt{2}\frac{\tau^2-\tau_0^2}{\tau_1^2-\tau_0^2} \equiv a(\tau)\\\\&\tau = b(s) \equiv \sqrt{\tau_0^2+\frac{(\tau_1^2-\tau_0^2)s}{10\sqrt{2}}}\\\\&y=x=F(\tau;\tau)=F(s;s)=F(\tau;b(s))=\frac{\tau^2-\tau_0^2}{\tau_1^2-\tau_0^2} \cdot 10 \vert_{\tau=b(s)}=\frac{s}{\sqrt{2}}\end{aligned}$

Or, in this simple case, one sees:

$\begin{aligned}&s=\sqrt{2}x\\&y=x=F(s;s)=\frac{s}{\sqrt{2}}\end{aligned}$

Or, if $\tau$ happens to be distance metric, then $M(q,\dot{q}) \vert_{q=F(\tau;x),\dot{q}=F'(\tau;x)}=1$ because $d s=M(q,\dot{q})d s$. Often, this is the equation to re-parametric old parameter to distance metric, so-called "normalization". Going through variation condition again it comes $x=y$. Let $x=y=F(s;s)=f(s)$ by a function $f(s)$. Then by $\sqrt{f'(s)^2+f'(s)^2}=1$ , it follows $f'(s)=\frac{1}{\sqrt{2}}$. Combined with $f(0)=0$, it means $f(s)=\frac{s}{\sqrt{2}}$

some demonstration

The value for a path is irrelevant to different parametric:

$\begin{aligned}&\int_0^{10} \sqrt{\dot{x}^2+\dot{y}^2}dt\\&q_{,0}=(0,0),q_{,1}=(10,10)\\&x(t)=\frac{t^2}{10},y(t)=\frac{t^2}{10},\dot{x}=\dot{y}=\frac{t}{5},\int_0^{10} \frac{\sqrt{2}}{5} t d t=\frac{\sqrt{2}}{10} \cdot 100=10\sqrt{2}\\&w=t^2,\int_0^{100} \frac{\sqrt{2}}{10} d w=10\sqrt{2}\\&x(w)=y(w)=\frac{w}{10},\dot{x}=\dot{y}=\frac{1}{10},\int_0^{100}\frac{\sqrt{2}}{10} d w=10\sqrt{2}\end{aligned}$

The value for a path is relevant to the parametric, even the path is optimal:

$\begin{aligned}&\int_0^{10} (\dot{x}^2+\dot{y}^2) d t\\&q_{,0}=(0,0),q_{,1}= (10,10)\\\\&x(t)=\frac{t^2}{10},y(t)=\frac{t^2}{10},\dot{x}=\frac{t}{5},\dot{y}=\frac{t}{5},\int_0^{10} \frac{2}{25}t^2 dt=\frac{2000}{75}\\&x(r)=2 r, y(r)=2 r,\dot{x}=2,\dot{y}=2,\int_0^5 8 d r=40\\\\&\text{variation solving:}\\&x(t)=A t, y(t)=B t,A=B=1,x(t)=t,y(t)=t,\dot{x}=1,\dot{y}=1, \int_0^{10} 2 d t =20\\&d s =\sqrt{\dot{x}^2+\dot{y}^2}d t\\&d s = \sqrt{2}d t,s=\sqrt{2}t,t=\frac{s}{\sqrt{2}}\\\\&\text{re-parametric or variation:}\\&x(s)=A s, y(s)=B s, A=B=\frac{1}{\sqrt{2}},x(s)=\frac{s}{\sqrt{2}},y(s)=\frac{s}{\sqrt{2}},\int_0^{10\sqrt{2}} (\frac{1}{2} + \frac{1}{2}) d s=10 \sqrt{2}\\\\&\text{another distance metric:}\\&d s=\sqrt{\dot{x}\dot{y}}d t = d t,s=t,\int_0^{10} (1+1) d t=20\end{aligned}$

As one can see ironically the optimal problem of $\int_0^{10} (\dot{x}^2+\dot{y}^2) d \tau$ has higher value than that of $\int_0^{10\sqrt{2}} (\dot{x}^2+\dot{y}^2) d \tau$ although $10 < 10 \sqrt{2}$

general relativity

For general relativity, $L=\sum_{u,v} g_{u,v} \dot{q_u} \dot{q_v}$ where $g_{u,v}=g_{v,u}$ and $g_{u,v}$ has no $q_0$ and $\dot{q}$ and the parameter. $L$ is not a distance metric but $M=\sqrt{L}$ is a distance metric which is also the proper time.

$d s=d \tau \sqrt{\sum_{u,v} g_{u,v} \dot{q_u} \dot{q_v}}$

For a typical gravity, $L(r,\phi,\theta,\dot{t},\dot{r},\dot{\phi},\dot{\theta})=g_0 \dot{t}^2-g_1 \dot{r}^2-g_2 \dot{\phi}^2-g_3 \dot{\theta}^2$ where $g_0, g_1, g_2, g_3$ are:

$\begin{aligned}g_0&=1-\frac{r_s}{r}\\g_1&=(1-\frac{r_s}{r})^{-1}\\g_2&=r^2\\g_3&=r^2 \sin^2(\phi)\end{aligned}$

The variation condition for $M$ and $f(M)$ are:

$\begin{aligned}&\frac{d \left(\frac{\partial M}{\partial \dot{q_i}} \vert_{q=F(s;s),\dot{q}=F'(s;s)}\right)}{d s}=\frac{\partial M}{\partial q_i} \vert_{q=F(s;s),\dot{q}=F'(s;s)}\\\\&f''(M)\frac{d \left(M(F(s;s),F'(s;s))\right)}{d s}\frac{\partial M}{\partial \dot{q_i}}+f'(M)\frac{d \left(\frac{\partial M}{\partial \dot{q_i}} \vert_{q=F(s;s),\dot{q}=F'(s;s)}\right)}{d s} \vert_{q=F(s;s),\dot{q}=F'(s;s)}=f'(M)\frac{\partial M}{\partial q_i} \vert_{q=F(s;s),\dot{q}=F'(s;s)}\end{aligned}$

But $M\left(F(s;s),F'(s;s)\right)=1$ , so the variation condition of $f(M)$ is again:

$\frac{d \left(\frac{\partial M}{\partial \dot{q_i}} \vert_{q=F(s;s),\dot{q}=F'(s;s)}\right)}{d s}=\frac{\partial M}{\partial q_i} \vert_{q=F(s;s),\dot{q}=F'(s;s)}$

With distance metric, the variation condition of $M$ and $L=f(M)=M^2$ are the same. Therefore technically one can state the variation condition for $L$ first which is cleaner on face of deduction, then impose the condition $M=1$ to re-parametric to the distance metric.

Then when $L$ optimal, $\frac{1}{2}\frac{\partial L}{\partial \dot{q_0}} \vert_{q=F(\tau;\tau)}$ is constant named law of energy conservation and total energy $E=g_0 \dot{t}$ which $\tau$ is re-parametric to match distance metric after variation imposed, i.e. $1=g_0 \dot{t}^2-g_1 \dot{r}^2-g_2 \dot{\phi}^2-g_3 \dot{\theta}^2$ . Therefore $E=g_0 \dot{t} = \sqrt{g_0 \left(1+g_1 \dot{r}^2+g_2 \dot{\phi}^2+g_3 \dot{\theta}^2\right)}$

Let $t=q_0(\tau), r=q_1(\tau),\phi=q_2(\tau),\theta=q_3(\tau)$ for a world line dictated by an engine thrust, then $1=g_0 \dot{q_0}^2-g_1 \dot{q_1}^2-g_2 \dot{q_2}^2-g_3 \dot{q_3}^2$ so that $\tau$ is itself distance metric $s$. There are 3 degree of freedom to control the engine thrust for the movement as $\dot{q_0}=\sqrt{g_0^{-1}\left(1+g_1 \dot{q_1}^2+g_2 \dot{q_2}^2+g_3 \dot{q_3}^2\right)}$ whose total energy, possibly not constant, is $E=g_0 \dot{q_0}=\sqrt{g_0\left(1+g_1 \dot{q_1}^2+g_2 \dot{1_2}^2+g_3 \dot{q_3}^2\right)}$. Not necessary optimal/geodesic as, say, being geodesic implies $\dot{q_0}$ of the form $C g_0^{-1}$ for some constant $C$. Optimal/geodesic world line is a global concept not a local concept. Locally, any short step is a distance. Summing over all the small steps connecting two points is not necessary the geodesic line.

The variation conditions of $L$ are:

$\begin{aligned}&\sum_u g_{u,0} F'_u(\tau;\tau) \vert_{q=F(\tau;\tau)}=E\\\\&i=1 \cdots 3\\\\&\frac{d}{d \tau}\left(\sum_u g_{u,i} F'_u(\tau;\tau) \vert_{q=F(\tau;\tau)} \right)=\frac{1}{2}\sum_{u,v}\frac{\partial g_{u,v}}{\partial q_i} \vert_{q=F(\tau;\tau)} F'_u(\tau;\tau) F'_v(\tau;\tau)\end{aligned}$

Solving, then an optimal world line $q=F(\tau; x)$ obtained. A change of parameter from $\tau$ to the distance metric $d s=M(F(\tau;x),F'(\tau;x)) d x,s=a(\tau)=\int_{\tau_0}^{\tau} M(F(\tau;x),F'(\tau;x)) d x,q=F(\tau;x)=F(s;y)$ with the help of the equation $M=1$. Then the variation problem becomes to optimize $\int_0^{a(\tau_1)} L(q, \dot{q})d s$ and the optimal world line is $q=F(s;y)$

As a solving process demo, to optimize $\int_0^1 L d\tau$ for a math artificial parameter $\tau$ where 0 is the staring point and 1 is the end point, apply the optimal conditions, two constants $E,K$:

$\begin{aligned}&L=(1-\frac{r_s}{r})\dot{t}^2-(1-\frac{r_s}{r})^{-1}\dot{r}^2-r^2\dot{\phi}^2-r^2\sin^2(\phi)\dot{\theta}^2\\\\&(1-\frac{r_s}{r})\dot{t}=E\\\\&-2r_s(r-r_s)^{-2}\dot{r}^2-2(1-\frac{r_s}{r})^{-1}\ddot{r}=\frac{r_s}{r^2}\dot{t}^2-r_s(r-r_s)^{-2}\dot{r}^2-2r\dot{\phi}^2-2r\sin^2(\phi)\dot{\theta}^2\\\\&-4r\dot{r}\dot{\phi}-2r^2\ddot{\phi}=-2r^2\sin(\phi)\cos(\phi)\dot{\theta}^2\\\\&r^2\sin^2(\phi)\dot{\theta}=K\end{aligned}$

By $0=\frac{r_s}{2r} \dot{t}^2-r^2\dot{\phi}^2-r^2\sin^2(\phi)\dot{\theta}^2$ seen above or for simplicity $\dot{t}$ is independent of how one puts the sphere coordinate for this circle orbit, put $\phi=\frac{\pi}{2}$, then it is:

$\begin{aligned}&(1-\frac{r_s}{r})\dot{t}=E\\\\&0=\frac{r_s \dot{t}^2}{r^2}-2r\dot{\theta}^2\\\\&r^2\dot{\theta}=K\end{aligned}$

Simplify, it is

$\begin{aligned}&0=r_s r (1-\frac{r_s}{r})^{-2} E^2-2K^2\\\\&\sqrt{r_s r}(1-\frac{r_s}{r})^{-1}E=\sqrt{2}K\end{aligned}$

re-parametric to distance metric,

$\begin{aligned}&1=(1-\frac{r_s}{r})\dot{t}^2-r^2 \dot{\theta}^2=(1-\frac{r_s}{r})^{-1}E^2-\frac{K^2}{r^2}=\left((1-\frac{r_s}{r})^{-1}-\frac{r_s}{2r}(1-\frac{r_s}{r})^{-2}\right)E^2\\\\&E=\frac{1-\frac{r_s}{r}}{\sqrt{1-\frac{r_s}{r}-\frac{r_s}{2r}}}=\frac{1-\frac{r_s}{r}}{\sqrt{1-\frac{3r_s}{2r}}}\\\\&K=\sqrt{\frac{r_s r}{2-\frac{3r_s}{r}}}\end{aligned}$

To conclude, with a given $E$, $r$ is obtained, $K,\dot{t},\dot{\theta}$ are derived. Some corollary:

Light orbit

Light is with infinite $E$ so its circle orbit is at $r=1.5r_s$ as $\lim_{E \rightarrow \infty} r=1.5 r_s$

GPS adjustment

The satellite is "free-fall" so its $\dot{t_{sat}}=g_0^{-1} E=\frac{1}{\sqrt{1-\frac{3 r_s}{2r_{sat}}}}$

The ground station on the Earth surface is with some "engine thrust" to fight against gravity and the rotation. Supposed it is at latitude $\phi$ whose distance to Earth center is $r_{\phi}$ due to Earth is not a perfect ball. Also note that it is 86164 instead of 86400 seconds per day after considering the spin of Earth:

$\begin{aligned}1&=\left(1-\frac{r_s}{r_{\phi}}\right) \dot{t_{\phi}}^2-r_{\phi}^2 \cos^2(\phi) \dot{\theta}^2=\left(1-\frac{r_s}{r_{\phi}}\right)\dot{t_{\phi}}^2- r_{\phi}^2\cos^2(\phi)\left(\frac{2\pi}{86164}\right)^2\\\\\dot{t_{\phi}}&=\sqrt{\frac{1+r_{\phi}^2\cos^2(\phi)\left(\frac{2\pi}{86164}\right)^2}{1-\frac{r_s}{r_{\phi}}}}\end{aligned}$

Therefore,

$\begin{aligned}\frac{d \tau_{sat}}{d \tau_{\phi}}&=\frac{\dot{t_{\phi}}}{\dot{t_{sat}}}=\sqrt{\frac{\left(1-\frac{3 r_s}{2r_{sat}}\right)\left(1 + r_{\phi}^2\cos^2(\phi)\left(\frac{2\pi}{86164}\right)^2\right)}{1-\frac{r_s}{r_{\phi}}}}\\\\&=\sqrt{1+\frac{\frac{r_s}{r_{\phi}}+r_{\phi}^2 \cos^2(\phi)\left(\frac{2\pi}{86164}\right)^2-\frac{3r_s}{2r_{sat}}\left(1+r_{\phi}^2 \cos^2(\phi)\left(\frac{2\pi}{86164}\right)^2\right)}{1-\frac{r_s}{r_{\phi}}}}\\\\&\approx 1+\frac{\frac{r_s}{r_{\phi}}+r_{\phi}^2 \cos^2(\phi)\left(\frac{2\pi}{86164}\right)^2-\frac{3r_s}{2r_{sat}}\left(1+r_{\phi}^2 \cos^2(\phi)\left(\frac{2\pi}{86164}\right)^2\right)}{2\left(1-\frac{r_s}{r_{\phi}}\right)}\end{aligned}$

Assume ground station at latitude 45 degree. Earth's $r_s$ is $8.8698 \times 10^{-6}$ km. Earth's radius is 6378 km. GPS satellite is of high 20200 km aka $r_{sat}=26578$ km and accordingly this number is higher than 1, meaning, satellite clock is faster than that of ground stations. Put all numbers in, it is $3.840 \times 10^{-5}$ second per day. However, Starlink satellite is of height 550 km and this number is lower than 1 and its clock is slower than that of ground stations. As $r_{\phi}^2 \cos^2(\phi)\left(\frac{2\pi}{86164}\right)^2=1.203 \times 10^{-12}$ , the break-even satellite orbit is roughly 3180 km above the sea level:

$\frac{r_{sat}}{r_{\phi}}=\frac{3}{2} \cdot \frac{1+r_{\phi}^2 \cos^2(\phi)\left(\frac{2\pi}{86164}\right)^2}{1+\frac{r_{\phi}}{r_s} \cdot r_{\phi}^2 \cos^2(\phi)\left(\frac{2\pi}{86164}\right)^2} \approx 1.5$