$$S_e = \sqrt{\frac{2GM}{R} } > c $$
Once an object hits the event horizon it should be pulled at the speed of light. Or because of the mass of the object, while light has zero mass, it may have to get closer than just the speed of light.
The equation above shows how much speed is needed to escape the bh's pull beyond the event horizon but, my question is, how do I figure it for an object with the mass of 2,030 tons (1,841,585.02 kg). Also the bh is hypothetical so it can be any size needed to pull the mass. Honestly, as I'm completely new to physics, don't know where to start in the equation.
Now again I apologize if any of my question comes off as unintelligent or uninformed but I'm just beginning to learn physics so plz take it easy on my guys and gals.
Ok, sorry, the bh has a mass equal to the object...we're just going to say it's 1.8*10^6 so here's my dilemma...
2GM/R
√((2×(6.67×10^(−11))×(1.8×10^(6)))÷(14×10^(−4))) =0.4141428325
What's wrong with this?
-edit for Kyle- With the Rs I got smthg like 1 nanometer which I don't think is right but it could be Rs=2GM/c^2 where M is 1.8*10^6 n c is 300,000,000 meters So... (2×(6.67×10^(−11))×(1.8×10^(6)))÷((3×10^(8))^(2)) =2.66800000E−21 meters=1.12227082 nanometers
So a black hole with the mass of a space shuttle would only be a nanometer? Just doesn't seem right.