Set light year as length unit, year as time unit, kg as mass unit.

$(d s)^2=c^2 (d \tau)^2=(d \tau)^2=(1-\frac{r_s}{r}) (d t)^2-(1-\frac{r_s}{r})^{-1} (d r)^2-r^2 (d \Omega)^2$

With a 1-dimension pure radial movement $d \Omega=0$, let $r \equiv r_s +r_s x$ then

$(d s)^2=\frac{x}{1+x} (d t)^2 - \frac{1+x}{x} r_s^2 (d x)^2$

light movement

$0=-\frac{x}{1+x} (d t)^2+\frac{1+x}{x} r_s^2 (d x)^2$

aka

$d t = \pm \frac{1+x}{x} r_s d x$

a mass object movement

$\begin{aligned}&E=\left(1-\frac{r_s}{r}\right)\frac{d t}{d \tau}\\\\&\left(\frac{d r}{d \tau}\right)^2=E^2-\left(1-\frac{r_s}{r}\right)c^2=E^2-\left(1-\frac{r_s}{r}\right)\\\\&\text{accordingly,}\\\\&\left(1-\frac{r_s}{r}\right)^{-1}E^2 (d \tau)^2=\left(1-\frac{r_s}{r}\right) (d t)^2\\\\&\left(\left(1-\frac{r_s}{r}\right)^{-1}E^2-1\right)(d \tau)^2=\left(1-\frac{r_s}{r}\right)^{-1}(d r)^2\\\\&(d \tau)^2=\left(1-\frac{r_s}{r}\right) (d t)^2-\left(1-\frac{r_s}{r}\right)^{-1}(d r)^2\end{aligned}$

Note that Special Relativity is the case when $\frac{r}{r_s}$ is super large. Newton physics is further the case when $\frac{d r}{d \tau}$ is small and accordingly $E=1+\frac{1}{2} (\frac{d r}{d \tau})^2-\frac{r_s}{2r}$.

To calculate $\frac{d r}{d t}$ of an object moving toward the gravity center:

$\frac{d r}{d t}=\frac{d r}{d \tau}\frac{d \tau}{d t}=-\sqrt{E^2-(1-\frac{r_s}{r})}\frac{r-r_s}{r E}=-\sqrt{1-\frac{r-r_s}{r E^2}}\frac{r-r_s}{r}=-\sqrt{\frac{(E^2-1)r+r_s}{r E^2}}\frac{r-r_s}{r}$

Therefore,

$r_s d x =-\sqrt{\frac{1 +(1-\frac{1}{E^2})x}{1+x}} \frac{x}{1+x}d t$

$d t=-\frac{1+x}{x} \sqrt{\frac{1+x}{1 +(1-\frac{1}{E^2})x}} r_s d x$

Light movement is like a mass object movement with infinite energy per unit mass $E$. An almost idle object which has super large $r$ away from a gravity's center has $E=1$

rescue story

Let $0<h<A<B$. Alice who falls from $r_s+r_s A$ and caught at $r_s+r_s h$ by Bob who starts at $r_s+r_s B$ with high initial speed meaning Bob's super large $E$. Typically $E$ is a function of the $\tau$ so that the engine can change it according to $\frac{d E}{d \tau}=\frac{a}{c} \frac{d r}{d \tau}$ where $a$ is the acceleration, constant or not, with respect to proper time $\tau$ of the spaceship, but here the rescue spaceship is modeled as an initial high speed then free-fall.

Starting free-fall at zero speed, Alice's $E$ must be such that:

$0=E^2-(1-\frac{r_s}{r_s+r_s A})=E^2-\frac{A}{1+A}$

Therefore,

$k \equiv 1-\frac{1}{E^2}=1-\frac{1+A}{A}=-\frac{1}{A}$

Alice falling, formula $F_1$:

$\begin{aligned}&\int_h^A \frac{1+x}{x} \sqrt{\frac{1+x}{1-\frac{x}{A}}} d x \\&= \sqrt{A(A-h)(1+h)} +\frac{\sqrt{A}(3+A)}{2} \left(\frac{\pi}{2}+\arcsin \left( \frac{A-1-2h}{A+1}\right)\right)+\ln\left(\frac{2A+h(A-1)+2\sqrt{A(A-h)(1+h)}}{h (1+A)}\right)\end{aligned}$

Alice SOS signal to the space station, formula $F_0$:

$\int_A^B \frac{1+x}{x}d x=B+\ln(B)-A-\ln(A)$

Bob's rescue, formula $F_2$:

$\begin{aligned}&\int_h^B \frac{1+x}{x} \sqrt{\frac{1+x}{1+k x}}d x\\\\&=\left(B+\frac{1}{k}\right) \sqrt{\frac{1+B}{1+B k}}- \left( h+\frac{1}{k} \right) \sqrt{\frac{1+h}{1+h k}} + \ln(\frac{B}{h})-\ln \left(\frac{2+(1+k)B +2 \sqrt{(1+B)(1+B k)}}{2+(1+k)h +2 \sqrt{(1+h)(1+h k)}} \right)+ \frac{3k-1}{2k \sqrt{k}} \ln\left( \frac{1+(1+2B) k + 2\sqrt{k(1+B)(1+B k)}}{1+(1+2h) k + 2\sqrt{k(1+h)(1+h k)}} \right)\end{aligned}$

To make the rescue possible, say, a 10 years preparation means:

$r_s F_1 = r_s F_0 + 10 + r_s F_2$ aka $F_1-F_0-F_2=\frac{10}{r_s}$

Therefore $F_1 - F_0 - F_2$ must be positive:

$\begin{aligned}0&<\int_h^A \frac{1+x}{x} \sqrt{\frac{1+x}{1-\frac{x}{A}}} d x-\int_A^B \frac{1+x}{x}d x -\int_h^B \frac{1+x}{x} \sqrt{\frac{1+x}{1+k x}}d x\\&=\int_h^A \frac{1+x}{x} \left(\sqrt{\frac{1+x}{1-\frac{x}{A}}}- \sqrt{\frac{1+x}{1+k x}}\right) d x-\int_A^B \frac{1+x}{x} \left(1+\sqrt{\frac{1+x}{1+k x}}\right) d x\end{aligned}$

One can see the larger the $k$ and the smaller the $h$, the larger the $F_1 - F_0 - F_2$ difference. Hence, to check the rescue possibility, consider the case of $k=1$ meaning Bob at light speed and $h=0$ meaning as late rescues as possible. It is:

$0<2A+\frac{\sqrt{A}(3+A)}{2} \left(\frac{\pi}{2} + \arcsin \left( \frac{A-1}{A+1}\right)\right) +\ln\left( \frac{4A}{A+1}\right)+\ln(A)-2B-2\ln(B)$

Meaning, Bob spaceship station cannot be too far away from where Alice starts falling, formula $F_4$ whose two times of the left side minus the right side is the $F_1 - F_0 - F_2$:

$A+\frac{\sqrt{A}(3+A)}{4} \left(\frac{\pi}{2}+ \arcsin \left( \frac{A-1}{A+1}\right) \right) +\ln\left( \frac{2A}{\sqrt{A+1}}\right) >B+\ln(B)$

Alice will never really fall into the horizon and her headlamp is always seen by Bob although Bob must be long-lived to see the light. But, once $F_4$ is violated, any rescue information that Bob sends to Alice will not be seen by Alice. $A$ is typically large. For the Sun, $A=235829.5$ on its surface 695700 km away from the center and $90066127.7>B$ if all the mass is within its horizon.

plot

Alice and the space station is established beside the supermassive black hole at the center of our galaxy for the purpose of research. The space station is $700000r_s$ away from the black hole's horizon. Alice's spaceship loses engine power and starts free-falling at $10000 r_s$ away from the horizon. Alice has a beam-me-up machine which can encode the information of her body and her mind and transmit it back to the space station and then recreate the body and the mind but she needs a passing code to activate the task and the passing code is transmitted from the space station to Alice for the rescue. The black hole has horizon $2.60 \times 10^{10}$ meters. The left of $F_4$ is $790637.748$ aka $790624.167 >B$ aka $2.0556 \times 10^{16}$ meters away from the horizon. Luckily, location of the space station satisfies this condition.

1 light-year is $9.4607 \times 10^{15}$ meter so the horizon is $r_s=\frac{2.60 \times 10^{10}}{9.4607 \times 10^{15}}=2.75 \times 10^{-6}$ light-year. The allowed time for rescue preparation is $r_s \times (790637.748 - 700000 - \ln(700000)) \times 2 =r_s \times (790637.748 - 700013.459) \times 2=0.498$ year beyond which Bob space station has no way to rescue Alice back.

When the beam-me-up method is not an option but a powerful spaceship with $E=2$ aka $k=0.75$ for the rescue, the max preparation time is $r_s \times ( F_1(h=0) - F_0 - F_2(h=0,k=0.75))=0.201$ year. If the preparation time is actually 0.1 year, less than 0.201 therefore feasible, then Alice would be caught up at $h=1440.30$ aka $1440.30 r_s$ away from the horizon by this powerful spaceship: $r_s \times ( F_1(h=1440.30) - F_0 - F_2(h=1440.30,k=0.75))=0.1$ year.

However, the required time back to the space station could be quite large due to the term $\ln(h)$ in $F_2$ if $h$ is near zero.

compared with classic Newton

$\begin{aligned}&\frac{1}{2} (\frac{d r}{d t})^2-\frac{G M}{r}=E\\\\&\frac{d r}{d t}=-\sqrt{\frac{2G M}{r}+2E}=-\sqrt{\frac{r_s c^2}{r}+2E}=-\sqrt{\frac{r_s}{r}+2E}\end{aligned}$

Alice's falling

no initial speed at $r_s+r_S A$ in both theories

$\begin{aligned}&E=-\frac{G M}{r_{initial}}=-\frac{r_s c^2}{2r_{initial}}=-\frac{r_s}{2r_{initial}}=-\frac{1}{2(1+A)}\\\\&r_s\frac{d x}{d t}=-\sqrt{\frac{1}{1+x}+2E}=-\sqrt{\frac{1}{1+x}-\frac{1}{1+A}}=-\sqrt{\frac{A-x}{(1+x)(1+A)}}=-\sqrt{\frac{1-\frac{1}{A}x}{(1+x)(1+\frac{1}{A})}}\end{aligned}$

$F_6$:

$\int_h^A \sqrt{\frac{(1+x)(1+\frac{1}{A})}{1-\frac{1}{A}x}} d x=\sqrt{(1+A)(A-h)(1+h)}+\frac{1}{2}(1+A)^{\frac{3}{2}}\left(\frac{\pi}{2}+\arcsin\left(\frac{A-1-2h}{A+1}\right)\right)$

Let $y \equiv \frac{1}{A}, f \equiv \frac{1+h}{1+A} <1, \frac{h}{A}=(1+\frac{1}{A})f-\frac{1}{A}=(1+y)f-y=f+(f-1)y$

Ratio of Alice's falling, $\frac{F_1}{F_6}$:

$\begin{aligned}&\frac{\sqrt{A(A-h)(1+h)} +\frac{\sqrt{A}(3+A)}{2} \left(\frac{\pi}{2}+\arcsin \left( \frac{A-1-2h}{A+1}\right)\right)+\ln\left(\frac{2A+h(A-1)+2\sqrt{A(A-h)(1+h)}}{h (1+A)}\right)}{\sqrt{(1+A)(A-h)(1+h)}+\frac{1}{2}(1+A)^{\frac{3}{2}}\left(\frac{\pi}{2}+\arcsin\left(\frac{A-1-2h}{A+1}\right)\right)}\\\\&=\frac{\sqrt{(1-\frac{h}{A})\frac{1+h}{A}} +\frac{\frac{3}{A}+1}{2} \left(\frac{\pi}{2}+\arcsin \left(\frac{1-\frac{1}{A}-\frac{2h}{A}}{1+\frac{1}{A}}\right)\right)+\frac{1}{A^{\frac{3}{2}}}\ln\left(\frac{\frac{2}{A}+\frac{h}{A}(1-\frac{1}{A})+2\sqrt{(1-\frac{h}{A})\frac{(1+h)}{A^2}}}{\frac{h}{A}(\frac{1}{A}+1)}\right)}{\sqrt{(\frac{1}{A}+1)(1-\frac{h}{A})\frac{1+h}{A}}+\frac{1}{2}(\frac{1}{A}+1)^{\frac{3}{2}}\left(\frac{\pi}{2}+\arcsin\left(\frac{1-\frac{1}{A}-\frac{2h}{A}}{1+\frac{1}{A}}\right)\right)}\\\\&=\frac{(1+y)\sqrt{(1-f)f}+\frac{3y+1}{2}\left(\frac{\pi}{2}+\arcsin(1-2f)\right)+y^{\frac{3}{2}}\ln\left(\frac{f-(f-1)y+2\sqrt{(1-f)fy}}{f+(f-1)y}\right)}{(1+y)^{\frac{3}{2}}\sqrt{(1-f)f}+\frac{1}{2}(y+1)^{\frac{3}{2}}\left(\frac{\pi}{2}+\arcsin(1-2f)\right)}\end{aligned}$

The General Relativity is the same as Newton is in the sense of constant $f$ and $A \rightarrow \infty$, then the ratio is:

$\frac{\sqrt{(1-f)f}+\frac{1}{2}\left(\frac{\pi}{2}+\arcsin(1-2f)\right)+0 \times \ln(\frac{f}{f})}{\sqrt{(1-f)f}+\frac{1}{2}\left(\frac{\pi}{2}+\arcsin(1-2f)\right)}=1$

Further, calculate $\frac{\frac{F_1}{F_6}-1}{y}=\frac{F_1-F_6}{y F_6}$ as $y \rightarrow 0$. By L'Hôpital rule,

$\begin{aligned}&\lim_{y \rightarrow 0}\frac{\left(1+y-(1+y)^{\frac{3}{2}}\right)\sqrt{(1-f)f}+\left(\frac{3y+1}{2}-\frac{1}{2}(y+1)^{\frac{3}{2}}\right)\left(\frac{\pi}{2}+\arcsin(1-2f)\right)+y^{\frac{3}{2}}\ln\left(\frac{f-(f-1)y+2\sqrt{(1-f)fy}}{f+(f-1)y}\right)}{y(1+y)^{\frac{3}{2}}\sqrt{(1-f)f}+\frac{1}{2}y(y+1)^{\frac{3}{2}}\left(\frac{\pi}{2}+\arcsin(1-2f)\right)}\\\\&=\frac{(1-\frac{3}{2})\sqrt{(1-f)f}+(\frac{3}{2}-\frac{3}{4})\left(\frac{\pi}{2}+\arcsin(1-2f)\right)+\frac{3}{2}\times 0^{\frac{1}{2}}}{(1+0)\sqrt{(1-f)f}+\frac{1}{2}(1+0)\left(\frac{\pi}{2}+\arcsin(1-2f)\right)}\\\\&=\frac{-\frac{1}{2}\sqrt{(1-f)f}+\frac{3}{4}\left(\frac{\pi}{2}+\arcsin(1-2f)\right)}{\sqrt{(1-f)f}+\frac{1}{2}\left(\frac{\pi}{2}+\arcsin(1-2f)\right)} \end{aligned}$ Therefore,

$F_1=F_6 \cdot \left(1+\frac{-\frac{1}{2}\sqrt{(1-f)f}+\frac{3}{4}\left(\frac{\pi}{2}+\arcsin(1-2f)\right)}{\sqrt{(1-f)f}+\frac{1}{2}\left(\frac{\pi}{2}+\arcsin(1-2f)\right)} \cdot y+O(y^2)\right)$

Alice SOS

$F_5$ : $\int_A^B d x=B-A$

Let $y \equiv \frac{1}{B}, f \equiv \frac{1+A}{1+B} <1, \frac{A}{B}=(1+\frac{1}{B})f-\frac{1}{B}=(1+y)f-y=f+(f-1)y$

$\frac{F_0}{F_5}$:

$\begin{aligned}&\frac{B+\ln(B)-A-\ln(A)}{B-A}=\frac{1-\frac{A}{B}-\frac{1}{B}\ln(\frac{A}{B})}{1-\frac{A}{B}}=\frac{1-f-y\ln(f)}{1-f}=1-\frac{\ln(f)}{1-f} \cdot y\\\\&F_0=F_5 \cdot \left(1-\frac{\ln(f)}{1-f} \cdot y\right)\end{aligned}$

Bob's rescue

Both with unit mass energy $E>1$ at $r_s+r_S B$ in both theories. But Newton does not count mass as energy, so plus 1 to Newton's $E$ to be compatible with General Relativity.

$\begin{aligned}&k \equiv \frac{2E-2}{2E-1}<1,E=\frac{2-k}{2-2k},2E-1=\frac{1}{1-k}\\\\&r_s\frac{d x}{d t}=-\sqrt{\frac{1}{1+x}+2(E-1)}=-\sqrt{\frac{2E-1+(2E-2) x}{1+x}}=-\sqrt{\frac{(2E-1)(1+\frac{2E-2}{2E-1}x)}{1+x}}=\sqrt{\frac{\frac{1}{1-k}(1+k x)}{1+x}}\end{aligned}$

$F_7$: $\begin{aligned}&\int_h^B \sqrt{\frac{(1+x)(1-k)}{1+k x}} d x\\&=\left(B+\frac{1}{k}\right)\sqrt{\frac{(1+B)(1-k)}{1+B k}}-\left(h+\frac{1}{k}\right)\sqrt{\frac{(1+h)(1-k)}{1+h k}}-\frac{1}{2}(\frac{1}{k}-1)^{\frac{3}{2}}\ln\left(\frac{1+k+2B k+2\sqrt{(1+B)k(1+B k)}}{1+k+2h k+2\sqrt{(1+h)k(1+h k)}}\right)\end{aligned}$

Let $y \equiv \frac{1}{B}, f \equiv \frac{1+h}{1+B} <1,\frac{h}{B}=f+(f-1)y$

Ratio of Bob's rescue, $\frac{F_2}{F_7}$:

$\begin{aligned}&\lim_{B \rightarrow \infty} \frac{\left(B+\frac{1}{k}\right) \sqrt{\frac{1+B}{1+B k}}- \left( h+\frac{1}{k} \right) \sqrt{\frac{1+h}{1+h k}} + \ln(\frac{B}{h})-\ln \left(\frac{2+(1+k)B +2 \sqrt{(1+B)(1+B k)}}{2+(1+k)h +2 \sqrt{(1+h)(1+h k)}} \right)+ \frac{3k-1}{2k \sqrt{k}} \ln\left( \frac{1+(1+2B) k + 2\sqrt{k(1+B)(1+B k)}}{1+(1+2h) k + 2\sqrt{k(1+h)(1+h k)}} \right)}{\left(B+\frac{1}{k}\right)\sqrt{\frac{(1+B)(1-k)}{1+B k}}-\left(h+\frac{1}{k}\right)\sqrt{\frac{(1+h)(1-k)}{1+h k}}-\frac{1}{2}(\frac{1}{k}-1)^{\frac{3}{2}}\ln\left(\frac{1+k+2B k+2\sqrt{(1+B)k(1+B k)}}{1+k+2h k+2\sqrt{(1+h)k(1+h k)}}\right)}\\\\&=\frac{\frac{1}{\sqrt{k}}-\frac{f}{\sqrt{k}}-0 \times \ln(f)+0 \times \ln\left(f\right)-0 \times \frac{3k-1}{2k\sqrt{k}} \ln\left(4f\right)}{\sqrt{\frac{1-k}{k}}-f\sqrt{\frac{1-k}{ k}}-0 \times \frac{1}{2}\left(\frac{1}{k}-1\right)^{\frac{3}{2}}\ln\left(\frac{1}{f}\right)}\\\\&=\frac{1}{\sqrt{1-k}} \approx 1 +\frac{1}{2}k\\\\&\lim_{k \rightarrow 0} \frac{\left(B+\frac{1}{k}\right) \sqrt{\frac{1+B}{1+B k}}- \left( h+\frac{1}{k} \right) \sqrt{\frac{1+h}{1+h k}} + \ln(\frac{B}{h})-\ln \left(\frac{2+(1+k)B +2 \sqrt{(1+B)(1+B k)}}{2+(1+k)h +2 \sqrt{(1+h)(1+h k)}} \right)+ \frac{3k-1}{2k \sqrt{k}} \ln\left( \frac{1+(1+2B) k + 2\sqrt{k(1+B)(1+B k)}}{1+(1+2h) k + 2\sqrt{k(1+h)(1+h k)}} \right)}{\left(B+\frac{1}{k}\right)\sqrt{\frac{(1+B)(1-k)}{1+B k}}-\left(h+\frac{1}{k}\right)\sqrt{\frac{(1+h)(1-k)}{1+h k}}-\frac{1}{2}(\frac{1}{k}-1)^{\frac{3}{2}}\ln\left(\frac{1+k+2B k+2\sqrt{(1+B)k(1+B k)}}{1+k+2h k+2\sqrt{(1+h)k(1+h k)}}\right)}\\\\&=\lim_{k \rightarrow 0}\frac{\left(k B+1\right) \sqrt{\frac{1+B}{1+B k}}- \left(k h+1 \right) \sqrt{\frac{1+h}{1+h k}} + k\ln(\frac{B}{h})-k\ln \left(\frac{2+(1+k)B +2 \sqrt{(1+B)(1+B k)}}{2+(1+k)h +2 \sqrt{(1+h)(1+h k)}} \right)+ \frac{3k-1}{2} \frac{\ln\left( \frac{1+(1+2B) k + 2\sqrt{k(1+B)(1+B k)}}{1+(1+2h) k + 2\sqrt{k(1+h)(1+h k)}} \right)}{\sqrt{k}}}{\left(k B+1\right)\sqrt{\frac{(1+B)(1-k)}{1+B k}}-\left(k h+1\right)\sqrt{\frac{(1+h)(1-k)}{1+h k}}-\frac{1}{2}(1-k)\sqrt{1-k}\frac{\ln\left(\frac{1+k+2B k+2\sqrt{(1+B)k(1+B k)}}{1+k+2h k+2\sqrt{(1+h)k(1+h k)}}\right)}{\sqrt{k}}}\\\\&=\frac{\sqrt{1+B}-\sqrt{1+h}+0 \times \ln\left(\frac{B}{h}\right)-0 \times \ln\left(\frac{2+B+2\sqrt{1+B}}{2+h+2\sqrt{1+h}}\right)+\frac{-1}{2} \times 0}{\sqrt{1+B}-\sqrt{1+h}-\frac{1}{2} \times 0}=1\end{aligned}$

Therefore,

$F_2=F_7 \cdot \left(1+\frac{1}{2}k+O(k^2,k y, y^2)\right)$