Maximal Velocity on a Circular Orbit Is Less than 43% of the Speed of Light
Since Kepler and Newton, we know classic trajectories of celestial bodies:
On a circular orbit, the speed of a celestial body is constant (only the direction of the velocity changes) because centripetal acceleration (caused by gravity) is balanced by centrifugal acceleration v²/R, where v is the celestial body’s speed and R is the radius of the orbit. Applying Newton’s formula for gravity, we get this balance as:
- G×M/R² = v²/R => G×M/R = v² => v = sqrt(G×M/R),
where G is the gravitational constant, M — mass of the attractor at the center of the circle, sqrt is square root. Here is the problem: for massive and dense celestial bodies at the center, G×M/R value can become greater than c², then v > c (where c is the speed of light ≈ 300,000 km/sec). That is impossible, because c is the universal speed limit. And even if the speed of the orbiting body drops, this celestial body will dive inside the circular orbit into an elliptical orbit, and its speed will start increasing (again). Is it a real problem, observed in the real world, or is it just a math mistake? No worries — it is just a math/physics mistake, and the solution lies in Dr. Vivian Robinson precise formula for gravity (we discussed in Real Gravity Does Not Produce Singularities — Black Holes Myth and Math Busted):
- G×M/[ R² × exp[ 2G×M/(R×c²) ] ], where exp(x)=eˣ and e≈2.72.
Thus, the balance between gravitational and centrifugal acceleration comes to:
- G×M/[ R² × exp[ 2G×M/(R×c²) ] ] = v²/R
- v² = G×M/R × exp[ — 2G×M/(R×c²)].
Let’s find max(v²) and at what radius R it is achieved. For that we have to find where the derivative of v² by R is 0:
- ( G×M/R × exp[ — 2G×M/(R×c²)] ) ' =
- G×M × ( -1/R² + 1/R×2G×M/(R²×c²) ) × exp[ — 2G×M/(R×c²)] =
- G×M/R² × ( -1 + 1/R×2G×M/c² ) × exp[ —2G×M/(R×c²)] = 0 =>
- -1 + 1/R×2G×M/c² = 0 => 1/R×2G×M/c² = 1 => R = 2G×M/c²
Thus, for R = 2G×M/c², v² reaches its maximum, which is:
- v² = G×M/R × exp[ —2G×M/(R×c²)] =
- G×M/(2G×M/c²) × exp[ — 2G×M/(2G×M/c² × c²)] = c²/(2e).
Therefore, max(v) = c/sqrt(2e), where e ≈ 2.72.
Thus, max(v) = c/sqrt(2e) ≈ 0.429×c.
We can see now that the speed on a circular orbit is far from the speed of light (it cannot even reach half c). How close is it to astronomical observations? — Check Gas bubble found going 30% speed of light around Milky Way’s supermassive black hole:
What about the escape velocity (the velocity at the perigee of a parabola)? Can it be close to c? The answer is in my next article:
