Yoda could throw Hulk into space with such an acceleration that, by the time Hulk realized what was happening, the distance between him and the planet they were on would already be equal to several planetary radii.
Lets do some maths, based on Yoda's mountain-lifting feat. Assume the gravitational acceleration on that planet was 1 g, the mass of the mountain 2E14 kg, and the mass of Hulk 600 kg.
CALCULATION: (you can skip this, and go directly to RESULTS below)
The force F (actual physical force, not the Force) required to lift the mountain equals 1.96E15 newtons. Such a force when applied on Hulk will accelerate him to relativistic speeds in a fraction of a second, so a special-relativistic analysis is required. In relativity, the relation between (proper) force F and proper acceleration α is given by (where m = Hulk's mass) :
F = mα (eqn 1)
This resembles the famous Newton's second law. The proper acceleration is related to coordinate acceleration "a" by
α = γ^3 a
where γ is the Lorentz factor. Substituting this into (eqn 1) and integrating yields the speed v (of Hulk) at time t:
v = c / sqrt[1 + (mc/Ft)^2] (eqn 2)
Here, c is the speed of light. Integration of (eqn 2) with respect to t yields the distance d traveled (by Hulk) in time t.
RESULTS:
In the table below, t is the time from the start of Yoda's force-push on Hulk, v is Hulk's speed at time t due to the force-push, and d is the distance traveled by Hulk at time t.
[CODE]
t (seconds) v (in units where c = 1.00) d (in units where Earth's diameter = 1.00)
-----------------------------------------------------------------------------------------
0 0 0
0.0001 0.737 0.001
0.001 0.996 0.022
0.01 0.99996 0.23
0.1 ~1 2.4
1 ~1 24
[/CODE]
Within one second, Hulk would be 24 Earth's diameters away from the planet they were fighting on. Now that's BFR!
Originally posted by Magnon
Yoda could throw Hulk into space with such an acceleration that, by the time Hulk realized what was happening, the distance between him and the planet they were on would already be equal to several planetary radii.Lets do some maths, based on Yoda's mountain-lifting feat. Assume the gravitational acceleration on that planet was 1 g, the mass of the mountain 2E14 kg, and the mass of Hulk 600 kg.
[B]CALCULATION:
(you can skip this, and go directly to RESULTS below)
The force F (actual physical force, not the Force) required to lift the mountain equals 1.96E15 newtons. Such a force when applied on Hulk will accelerate him to relativistic speeds in a fraction of a second, so a special-relativistic analysis is required. In relativity, the relation between (proper) force F and proper acceleration α is given by (where m = Hulk's mass) :F = mα (eqn 1)
This resembles the famous Newton's second law. The proper acceleration is related to coordinate acceleration "a" by
α = γ^3 a
where γ is the Lorentz factor. Substituting this into (eqn 1) and integrating yields the speed v (of Hulk) at time t:
v = c / sqrt[1 + (mc/Ft)^2] (eqn 2)
Here, c is the speed of light. Integration of (eqn 2) with respect to t yields the distance d traveled (by Hulk) in time t.
RESULTS:
In the table below, t is the time from the start of Yoda's force-push on Hulk, v is Hulk's speed at time t due to the force-push, and d is the distance traveled by Hulk at time t.
[CODE]
t (seconds) v (in units where c = 1.00) d (in units where Earth's diameter = 1.00)
-----------------------------------------------------------------------------------------
0 0 0
0.0001 0.737 0.001
0.001 0.996 0.022
0.01 0.99996 0.23
0.1 ~1 2.4
1 ~1 24
[/CODE]
Within one second, Hulk would be 24 Earth's diameters away from the planet they were fighting on. Now that's BFR! [/B]
I love it
Originally posted by Magnon
Yoda could throw Hulk into space with such an acceleration that, by the time Hulk realized what was happening, the distance between him and the planet they were on would already be equal to several planetary radii.Lets do some maths, based on Yoda's mountain-lifting feat. Assume the gravitational acceleration on that planet was 1 g, the mass of the mountain 2E14 kg, and the mass of Hulk 600 kg.
[B]CALCULATION:
(you can skip this, and go directly to RESULTS below)
The force F (actual physical force, not the Force) required to lift the mountain equals 1.96E15 newtons. Such a force when applied on Hulk will accelerate him to relativistic speeds in a fraction of a second, so a special-relativistic analysis is required. In relativity, the relation between (proper) force F and proper acceleration α is given by (where m = Hulk's mass) :F = mα (eqn 1)
This resembles the famous Newton's second law. The proper acceleration is related to coordinate acceleration "a" by
α = γ^3 a
where γ is the Lorentz factor. Substituting this into (eqn 1) and integrating yields the speed v (of Hulk) at time t:
v = c / sqrt[1 + (mc/Ft)^2] (eqn 2)
Here, c is the speed of light. Integration of (eqn 2) with respect to t yields the distance d traveled (by Hulk) in time t.
RESULTS:
In the table below, t is the time from the start of Yoda's force-push on Hulk, v is Hulk's speed at time t due to the force-push, and d is the distance traveled by Hulk at time t.
[CODE]
t (seconds) v (in units where c = 1.00) d (in units where Earth's diameter = 1.00)
-----------------------------------------------------------------------------------------
0 0 0
0.0001 0.737 0.001
0.001 0.996 0.022
0.01 0.99996 0.23
0.1 ~1 2.4
1 ~1 24
[/CODE]
Within one second, Hulk would be 24 Earth's diameters away from the planet they were fighting on. Now that's BFR! [/B]
The Force is strooooong in you!
Originally posted by LordGodTK-wise, Vader can likely approach Yoda/Palpatine-level, but they're still superior.
Is Yoda really that far beyond Darth vader?
...There's also the fact that Vader doesn't have a TK feat on par with hurling a mountain-sized creature to the ground, while also holding back an entire group of force-sensitives.
Originally posted by StiltmanFTWAnd rightfully so. That kind of BS would ruin the franchise for me.
Star Wars characters were never meant to be taking on threats such as the Hulk.