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$$S_e = \sqrt{\frac{2GM}{R} } > c $$

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.

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closed as off-topic by Brandon Enright, Jim, Kyle Kanos, Kyle, rob Jun 26 at 22:42

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1  
You may want to read the Wikipedia article on the Schwarzschild radius. –  Kyle Kanos Jun 22 at 2:40
    
I just need to know where I'm going wrong...the post isn't off topic I'm just coming here for corrections bcuz I'm NOT a physics student so I don't have HOMEWORK or a textbook to consult...where am I going wrong in THIS equation I just seem to have a problem fully understanding G? –  user50517 Jun 27 at 15:10
    
M is the mass of the black hole, not the mass of the object trying to escape the black hole. –  Señor O Jun 27 at 15:34

2 Answers 2

Escape velocity = (√GM)/√r > C. G= universal gravitational constant (6.67384 × 10-11 m^3 kg^-1 s^-2) M= mass of black hole r=radius of black hole c=3.00x10^8 m/s
The escape velocity of a black hole is constant. The mass of the object "escaping" does not affect it.

Im not sure i understand your question but if you are saying the black holes mass is 2,030 tons (1,841,585.02 kg) just plug that in as M and determine the radius of that object.

If you mean that the object trying to escape has a mass of 2030 tons that doesnt matter. Determine the mass of the black hole.

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As @Drew says, it's not clear if you're talking about the mass of the thing trying to escape the black hole or the mass of the black hole. However, as you can see from your formula, it is independent of the mass of the object (usually denoted $m$ instead of the black hole mass denoted $M$).

Your question is contained in your forumala! A little rearrangement shows that if

$M>\frac{c^2R}{2G}$

(where $R$ is the radius of the "object" we're trying to escape) then this "object" is a black hole.

Take the Earth for example with $M=6 \times 10^{24}$ and $R=6.4 \times 10^3$, the equation above isn't satisfied, and hence the earth isn't a black hole. From here you can work out what radius the Earth must be contracted to for it be a black hole.

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@Drew smthg I'm missing is in the gravitational constant see in 6.67384 × 10-11 m^3 kg^-1 s^-2 is the m^3 the mass of the black hole or the object pulling into it? I thought it was mass of object since it was "m" and not "M"...which is it my friends? –  user50517 Jun 26 at 20:08
    
@user50517 if you're talking about the m in the units of the gravitational constant, that's meters, not mass. The mass of the object doesn't matter in this scenario, only the mass of the black hole –  user13223423 Jun 26 at 21:24

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