JEROME - Black holes do not exist

Now you see in the movie that the King (us) does not cut off the Black-Knight's head, that would be futile.

Paul

:) :) :)
 
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Let me try my hand at Jerome translation:

Objects are made of matter. Maybe he is trying to say that he doesn't understand how black holes affect light since light isn't an "object".

Or something.
 
You're wasting your time. Jerome is nothing but a passive-aggressive troll. He doesn't want knowledge, he just want to yank your chains.

Fol de rol de rol.
 
JEROME DA GNOME said:
Please, how does one measure the gravity force of an orange?
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You measure it's mass.

Assumptions
I don't happen to have an orange handy to measure, so for argument's sake and ease of mathematics, let's say it has a mass of 100 grams or 0.1 kg. Let's also assume that an orange is spherical and has roughly uniform density and that we're dealing with Newtonian distances.

It is entirely likely that none of these assumptions will be true for any specific orange, but this is a first-approximation calculation. If you want to get more specific, we can make better assumptions later.

Calculation
The Newtonian equation for the force of gravity is

[latex]$$ F = G\frac{m_1 \times m_2}{r^2} $$[/latex]

Where,
  • F = Force measured in Newtons
  • G = gravitational constant = 6.67 × 10−11 N m2 kg-2
  • m1 = mass of the orange in kg = 0.1 kg
  • m2 = mass of whatever the gravity of the orange is acting upon in kg.
  • r = the distance from the center of orange to m2
Plugging these values in:
[latex]$$ F = (6.67 \times 10^{-11}\frac{N m^2}{kg^2}) \frac{{0.1 kg} \times m_2}{r^2} $$[/latex]
simplifying
[latex]$$ F = (6.67 \times 10^{-12}\frac{N m^2}{kg}) \frac{m_2}{r^2} $$[/latex]

So, the attractive force between an orange of mass 0.1 kg and another object is directly proportional to the mass of the second object and inversely proportional to the square of the distance between them. Ignoring the distance element for a second, we can calculate the strength of the orange's gravitational field by using an m2 of 1 kg and then multiplying the result by the actual mass of the second object when it is known. In other words:

[latex]$$ F = (6.67 \times 10^{-12}\frac{N m^2}{kg}) \frac{1 kg}{r^2} $$[/latex]
simplifying
[latex]$$ F = (6.67 \times 10^{-12}N m^2) \frac{1}{r^2} $$[/latex]


Answer
From this, we can now plot the strength of the gravitational force radially outward from the orange as a function of distance
r = 1 meter: F = 6.67 x 10-12N
r = 2 meter: F = 1.67 x 10-12N
r = 3 meter: F = 7.41 x 10-13N
etc.

Again, the strength of that field is multiplied by the mass of the object (in kg) the field is acting upon.

That's how you measure the gravitational force, in Newtons, of an orange. You measure it's mass and calculate the strength of its gravitational field radially outward.

If you'd like the calculation to be more specific, I need more details.
 
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OK, for those in the know.

The escape velocity of the Earth is 11.186 km/s

The escape velocity of the Moon is 2.38 km/s

We can work out the Moon’s escape velocity once knowing the Earth’s escape velocity, size and weight, and knowing the Moon’s size and weight.

So the moon is about .27265 is size of the earth, worked out above.

The mass of the moon is about 1/81 that of the Earth’s mass.

So if the earth was shrunk to the size of the moon, the escape velocity would be

.27265 ^ .5 = .5221, then we divide .5221 into 1 and get 1.91153

1.91153 time 11.186 km/s equals 21.425 km/s for the new size of the earth.

But the moon is 1/81 of the earth mass, so…….

81 ^ .5 = 9, then we divide 9 into 1 and get .11111

.11111 times 21.425 km/s we get 2.38 km/s for the moon’s escape velocity.

Now lets play with the shrinkage of the earth some more, let’s shrink is down to a radius of .82296 cm.

.82296 cm = 1.39755 ^ -9 of the Earth’s radius,

(1.39755 ^ -9) ^ .5 = 3.738 ^ -5

3.738 ^ -5 divide into 1 = 26,749.5

26,749.5 times Earth’s escape velocity of 11.186 km/s equals 299,200 km/s.

Does anybody but one person on this thread know what the last speed is about equal too.

Paul

:) :) :)
 
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JEROME DA GNOME said:
That was measuring the effects of a specific object in relation to other objects.
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Jerome, you are quite correct. However, the "effect" it was measuring is gravitational force.

You seem to be working under the impression that force is independent from the objects between which it interacts. Is that your understanding? If so, why do you think that?
 
Paulhoff said:
Does anybody but one person on this thread know what the last speed is about equal too.
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Pffft, you can't measure the speed of light. You can only measure the effects of light in relation to other objects. (insert non-sequitur picture here)
 
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MattusMaximus said:
To speak of a single, isolated "force" in terms of only one object makes no sense; how can something in isolation interact with something when it's the only thing? I really think that you've got your definitions all mixed up - perhaps you are talking about a field?
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D'oh. You beat me to it. I essentially calculated the gravitational field (classical version) for a hypothetical orange above.
 
JEROME DA GNOME said:
I do not deny that gravity exists, that was a straw-man someone threw out there.

The point is the force is neither measurable nor controllable.
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And yet we know how strong it is on the surface of the Earth.

Already addressed. That is the relationship between two objects that is being measured, not a measure of the force of gravity.
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What is a force is not a relationship between two objects, JEROME ?
 
Oh, now I it, we're not talking about "The Force" from Star Wars.

Paul

:) :) :)
 
Belz... said:
And yet we know how strong it is on the surface of the Earth.



What is a force is not a relationship between two objects, JEROME ?
Click to expand...


How can the force of gravity of the Earth be constant in relation to various objects that have different masses thus different gravity forces of their own?
 
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