Butch Posted September 14, 2018 Posted September 14, 2018 Do gravitational force and escape velocity have a direct and constant relationship, specifically is the gravitational force at the schwarzild radius the same for any body? If so what is it? 1
Strange Posted September 14, 2018 Posted September 14, 2018 (edited) Escape velocity is given by [math]v_e = \sqrt{\frac{2GM}{r}}[/math] (https://en.wikipedia.org/wiki/Escape_velocity) Where M is the mass of the planet (or whatever) and r is its radius. If you substitute the Schwarzschild radius ([math]r_s = \frac{2 G M}{c^2}[/math]) for r then you will find the result comes to c. (https://en.wikipedia.org/wiki/Schwarzschild_radius) The equation for gravitational force is given by [math]F = G \frac{m M}{r^2}[/math]. As you can see, this decreases with r2 rather than square root of r for escape velocity. It also depends on the mass (m) of the object. (https://en.wikipedia.org/wiki/Newton's_law_of_universal_gravitation) So escape velocity and gravitational force are related but not in a direct way. The acceleration due to gravity (g) is independent of the mass of the object (because acceleration is force/mass): [math]g = G \frac{M}{r^2}[/math] If we substitute this into the equation for escape velocity, we end up with: [math]g = \frac{{v_e}^2}{2 r}[/math] p.s. I wouldn't be too surprised if someone comes along and points out that I have made a mistake there somewhere... Edited September 14, 2018 by Strange 1
Butch Posted September 15, 2018 Author Posted September 15, 2018 Thank you, you always make things pretty clear...
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