# How to calculate the gravitational force of a black hole outside event horizon at a given distance $r$? [duplicate]

Let $$g_{bh}$$ $$[kg·\frac{m}{s^2}]$$ be the force caused by the gravity of a black hole.

How do we calculate the force $$g_{bh}(r)$$ outside event horizon $$R_{Sch} = \frac{2GM}{c^2}$$ onto a satellite with mass $$m$$ at a given distance $$r$$ from that black hole?

And with that formula, is it possible to also derive the speed $$v_s$$ of this satellite around that black hole (also outside event horizon $$R_{Sch}$$)? If yes, how?

I will answer the second question first as I find it easier one on the computations and since the calculation for the first question is similar, I will only comment on it and leave the details for you.

First of all, there are no stable circular orbits under $$r=3r_s$$ and no circular orbits under $$r=3r_s/2$$ for Schwarzschild black hole.

Now, the worldline of satellite on circular orbit is $$x^\mu(t)=(t,r,\pi/2,\omega t)$$, where $$\omega$$ is angular velocity in Schwarschild coordinates. We can thus compute 4-velocity:

$$v^\mu=\frac{dx^\mu}{d\tau}=(1,0,0,\omega)\frac{dt}{d\tau},$$ where $$\tau$$ is proper time along the curve, i.e.: $$d\tau^2=g_{tt}dt^2+g_{\phi\phi}\omega^2dt^2=dt^2\left(g_{tt}+g_{\phi\phi}\omega^2\right).$$ Thus the 4-velocity is: $$v^\mu=\frac{dx^\mu}{d\tau}=(1,0,0,\omega)\frac{1}{\sqrt{g_{tt}+g_{\phi\phi}\omega^2}}.$$

Note the 4-velocity in these coordinates is constant. Therefore the 4-acceleration is simply given by Christoffel symbols: $$a^\lambda=\Gamma^\lambda_{\nu\mu}v^\mu v^\nu.$$

We are interested in nonzero components, so only in $$\Gamma^\lambda_{tt}$$, $$\Gamma^\lambda_{t\phi}$$, $$\Gamma^\lambda_{\phi t}$$, $$\Gamma^\lambda_{\phi\phi}.$$You can google or compute that out of these the only nonzero are

$$\Gamma^r_{tt}=-\frac{r_s}{2r^2}g_{tt}$$ $$\Gamma^r_{\phi\phi}=r g_{tt}$$

Of course circular orbit is geodesic, so there is no acceleration. Therefore we demand:

$$0=\Gamma^r_{tt}v^t v^t+\Gamma^r_{\phi\phi}v^\phi v^\phi$$ and use this to compute $$\omega:$$ $$0=-\frac{r_s}{2r^2}+\omega^2r \Rightarrow \omega=\sqrt{\frac{r_s}{2r^3}}.$$ Now the velocity of orbiting depends on the observer. If we take observer at rest in schwarzshild coordinates, this one will measure velocity to be: $$v=\frac{\sqrt{g_{\phi\phi}}d\phi}{\sqrt{-g_{tt}}dt}=\omega\sqrt{\frac{g_{\phi\phi}}{-g_{tt}}}=\sqrt{\frac{r_s}{2r^3}\frac{r^2}{1-\frac{r_s}{r}}}=\sqrt{\frac{r_s}{2(r-r_s)}}.$$

As you can see, this will give you speed of light $$v=1$$ for $$r=3r_s/2$$. So this is the closest circular orbit that can exist.

Now you can do the same computation to get gravitational "force" at some distance $$r$$. But first realize, there is no gravitational force in GR. But there is an acceleration. So you can compute 4-acceleration of some object the same way I just did and than look what kind of force is producing this acceleration. If this object is at rest w.r.t. black hole (i.e. follows the worldline $$x^\mu(t)=(t,0,0,0)$$ in Schwarzschild coordinates) this force needed to keep the object at place will represent how strong of a pull the gravitation exerts on the object, even though conceptually this is not what is really happening. I leave the computation to you.

• Looking good, thx! Commented Oct 7, 2020 at 16:36
• There is another question of mine about similar topic here: physics.stackexchange.com/q/584596 (I know the answer is probably $r = 3r_s$, but I'd like to see how this is proved...) Commented Oct 7, 2020 at 17:57