text: # What speed would a rocket below $1.5R_s$ require the least thrust to remain at constant radius?
  
  Take a rocket travelling in a circle at less than $$1.5R_s$$ away from a black hole. At that distance, no orbit is possible and the rocket must produce continuous force to stay outside the black hole. Travelling in a circle can reduce this force, but not eliminate it. In addition, if the rocket is travelling a near the speed of light, time dilation will cause the proper acceleration from the perspective of the rocket to be MUCH larger than from the perspective of a distant observer.
  
  At what speed will the required proper acceleration be minimised?
  
  blademan9999 asked: "At what speed will the required proper acceleration be minimised."
  
  In that example with a circular orbit in the $$\theta$$ direction, the coordinate acceleration is
  
  $$\rm \frac{d^2 x^{\mu}}{d \tau^2}=\{ \ddot{t}, \ \ddot{r}, \ \ddot{\theta}, \ \ddot{\phi}\}=\{0,\ 0, \ 0, \ 0\}$$
  
  and the 4-velocity
  
  $$\rm \frac{d x^{\mu}}{d \tau}=u^{\mu}=\{ \dot{t}, \ \dot{r}, \ \dot{\theta}, \ \dot{\phi}\}=\{ \gamma \surd g^{tt}, \ 0, \ v \gamma /r, \ 0\}$$
  
  so the force required for that is in natural units
  
  $$\rm F= \surd \ |\sum_{\mu, \nu} g_{\mu \nu} \ a^{\mu} a^{\nu}| = \surd \ |\frac{(r-2) \ v^2-1}{\sqrt{(r-2) \ r^3 \left(v^2-1\right)^2}}|$$
  
  where $$\rm v$$ is the local velocity relative to a stationary observer, and the proper acceleration
  
  $$\rm a^{\mu}= \frac{d^2 x^{\mu}}{d \tau^2}+\sum_{\alpha, \beta} \ \Gamma^{\mu}_{\alpha \beta} \ u^{\alpha}u^{\beta}$$
  
  so at $$\rm r=4$$ we get $$\rm F=0$$ with $$\rm v= \sqrt{1/(r-2)}=\sqrt{1/2}$$ as expected. At $$\rm r=1.5 \ r_s=3$$ the radial force $$\rm F=1/\sqrt{27}$$ (in units of $$\rm c^2 m/M/G$$) is independend of the angular velocity, and at $$\rm r<3$$ the required force to keep a constant $$\rm r$$ is smallest when $$\rm v=0$$.
  
  That is due to the $$-3 \dot{\theta}^2$$ term in the geodesic equation
  
  $$\rm \ddot{r}=-3 \dot{\theta}^2+ r \dot{\theta}^2 -1/r^2$$
  
  which differs from the Newtonian expression and makes the angular velocity contribute negatively below the critical radius $$\rm r=3$$, and cancel itself at that exact radius.
  
  This is assuming the Schwarzschild metric, for rotating black holes it would be a little bit more complicated. The solution might be shortened into a more elegant form, but since the coordinate acceleration is $$0$$ on a circular path the calculation is straightforward.
explanation: The document discusses the speed at which a rocket traveling in a circular orbit around a black hole would require the least thrust to maintain a constant radius, particularly when the radius is less than \(1.5R_s\). It includes a detailed mathematical derivation to find the required proper acceleration and the conditions under which it is minimized. The document contains a question and an answer to that question.
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  1. Briefly describe the document and mention if it contains a question or a question and answer.
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distilabel_metadata.role: system
distilabel_metadata.content: # What speed would a rocket below $1.5R_s$ require the least thrust to remain at constant radius?
  
  Take a rocket travelling in a circle at less than $$1.5R_s$$ away from a black hole. At that distance, no orbit is possible and the rocket must produce continuous force to stay outside the black hole. Travelling in a circle can reduce this force, but not eliminate it. In addition, if the rocket is travelling a near the speed of light, time dilation will cause the proper acceleration from the perspective of the rocket to be MUCH larger than from the perspective of a distant observer.
  
  At what speed will the required proper acceleration be minimised?
  
  blademan9999 asked: "At what speed will the required proper acceleration be minimised."
  
  In that example with a circular orbit in the $$\theta$$ direction, the coordinate acceleration is
  
  $$\rm \frac{d^2 x^{\mu}}{d \tau^2}=\{ \ddot{t}, \ \ddot{r}, \ \ddot{\theta}, \ \ddot{\phi}\}=\{0,\ 0, \ 0, \ 0\}$$
  
  and the 4-velocity
  
  $$\rm \frac{d x^{\mu}}{d \tau}=u^{\mu}=\{ \dot{t}, \ \dot{r}, \ \dot{\theta}, \ \dot{\phi}\}=\{ \gamma \surd g^{tt}, \ 0, \ v \gamma /r, \ 0\}$$
  
  so the force required for that is in natural units
  
  $$\rm F= \surd \ |\sum_{\mu, \nu} g_{\mu \nu} \ a^{\mu} a^{\nu}| = \surd \ |\frac{(r-2) \ v^2-1}{\sqrt{(r-2) \ r^3 \left(v^2-1\right)^2}}|$$
  
  where $$\rm v$$ is the local velocity relative to a stationary observer, and the proper acceleration
  
  $$\rm a^{\mu}= \frac{d^2 x^{\mu}}{d \tau^2}+\sum_{\alpha, \beta} \ \Gamma^{\mu}_{\alpha \beta} \ u^{\alpha}u^{\beta}$$
  
  so at $$\rm r=4$$ we get $$\rm F=0$$ with $$\rm v= \sqrt{1/(r-2)}=\sqrt{1/2}$$ as expected. At $$\rm r=1.5 \ r_s=3$$ the radial force $$\rm F=1/\sqrt{27}$$ (in units of $$\rm c^2 m/M/G$$) is independend of the angular velocity, and at $$\rm r<3$$ the required force to keep a constant $$\rm r$$ is smallest when $$\rm v=0$$.
  
  That is due to the $$-3 \dot{\theta}^2$$ term in the geodesic equation
  
  $$\rm \ddot{r}=-3 \dot{\theta}^2+ r \dot{\theta}^2 -1/r^2$$
  
  which differs from the Newtonian expression and makes the angular velocity contribute negatively below the critical radius $$\rm r=3$$, and cancel itself at that exact radius.
  
  This is assuming the Schwarzschild metric, for rotating black holes it would be a little bit more complicated. The solution might be shortened into a more elegant form, but since the coordinate acceleration is $$0$$ on a circular path the calculation is straightforward.
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document discusses the speed at which a rocket traveling in a circular orbit around a black hole would require the least thrust to maintain a constant radius, particularly when the radius is less than \(1.5R_s\). It includes a detailed mathematical derivation to find the required proper acceleration and the conditions under which it is minimized. The document contains a question and an answer to that question.
  
  Label: question-and-answer