[Bored chemist]

What do you think "subjective" means?

[guest39538]

What do you think "subjective" means?

What do you want it to mean? 

[Kryptid]

That shows the approximation is a load of garbage and a broken calculation.

You can verify the calculation for yourself. The event horizon is simply the location where the black hole's escape velocity equals the speed of light. The equation used to calculate the escape velocity of an object is:

v_e = √((2GM)/r)

Where v_e is the escape velocity in meters per second, G is the gravitational constant (6.67 x 10^-11 m³•kg^-1•s^-2), M is the mass in kilograms and r is the radius from the center of the object in meters.

Plugging in the numbers for the Earth, we get:

v_e = √((2 x (6.67 x 10^-11 m³•kg^-1•s^-2) x (5.972 x 10²⁴ kg))/(6,371,000 meters))

v_e = 11,182 meters per second (11.182 kilometers per second)

Given how many times we've sent spacecraft into orbit, we've have plenty of occasion to thoroughly test the validity of this equation.

Now let's rearrange the equation so that what we are looking to find is not the escape velocity, but the radius at which the escape velocity takes on a particular value:

r = √((2GM)/(v_e²))

Put in the relevant data for the Earth and you can verify that this rearranged equation accurately predicts the radius of the Earth based on its mass and escape velocity.

Now we can enter the speed of light as the escape velocity (299,792,458 meters per second) in order to find the distance from the center of a black hole at which the event horizon must exist for a given mass. We’ll enter the measured mass of the Cygnus X-1 black hole of 14.8 solar masses:

r = (2 x (6.67 x 10^-11 m³•kg^-1•s^-2) x (2.943 x 10³¹ kg))/(299,792,458 meters per second)²

r = 43,682 meters (43.682 kilometers)

So there you have it, a step-by-step explanation on how to calculate the radius of a black hole’s event horizon based on an experimentally-verified equation.

Calculations aside, it should be pretty obvious that the orbiting star is outside of the black hole's event horizon because we can see it.

[guest39538]

That shows the approximation is a load of garbage and a broken calculation.

You can verify the calculation for yourself. The event horizon is simply the location where the black hole's escape velocity equals the speed of light. The equation used to calculate the escape velocity of an object is:

v_e = √((2GM)/r)

Where v_e is the escape velocity in meters per second, G is the gravitational constant (6.67 x 10^-11 m³•kg^-1•s^-2), M is the mass in kilograms and r is the radius from the center of the object in meters.

Plugging in the numbers for the Earth, we get:

v_e = √((2 x (6.67 x 10^-11 m³•kg^-1•s^-2) x (5.972 x 10²⁴ kg))/(6,371,000 meters))

v_e = 11,182 meters per second (11.182 kilometers per second)

Given how many times we've sent spacecraft into orbit, we've have plenty of occasion to thoroughly test the validity of this equation.

Now let's rearrange the equation so that what we are looking to find is not the escape velocity, but the radius at which the escape velocity takes on a particular value:

r = √((2GM)/(v_e²))

Put in the relevant data for the Earth and you can verify that this rearranged equation accurately predicts the radius of the Earth based on its mass and escape velocity.

Now we can enter the speed of light as the escape velocity (299,792,458 meters per second) in order to find the distance from the center of a black hole at which the event horizon must exist for a given mass. We’ll enter the measured mass of the Cygnus X-1 black hole of 14.8 solar masses:

r = (2 x (6.67 x 10^-11 m³•kg^-1•s^-2) x (2.943 x 10³¹ kg))/(299,792,458 meters per second)²

r = 43,682 meters (43.682 kilometers)

So there you have it, a step-by-step explanation on how to calculate the radius of a black hole’s event horizon based on an experimentally-verified equation.

Calculations aside, it should be pretty obvious that the orbiting star is outside of the black hole's event horizon because we can see it.

Interesting thanks, that will take me some time get my head around, in my conceptual version of a BH, light can escape the BH, but you can only see it if you are within range because of size and intensity.

[evan_au]

After the above post, this discussion thread collapsed under its own weight and become a black hole for human endeavours.
It is now locked - mod