Where Did I Go Wrong in Calculating the Kinetic Energy of an Asteroid Impact?

[newjerseyrunner]

I want to know how fast the space shuttle would have to hit the Earth to unleash the same kinetic energy as the asteroid that killed the dinosaurs. I've already determined that it's relativistic, by showing that Newtonian physics gives a v greater than c

Spoiler: Newtonian physics

4.2e25J = (2e6 * v ^ 2 ) / 2.
8.4e25J = 2e6 * v ^ 2
v = sqrt(8.4e25J / 2e6)
v = 6,480,740,698 m/s

But once I plug in the equations, I get a wrong result.

KE = (m * c ^ 2) / sqrt(1 - (v ^ 2 / c ^ 2))
KE / (m * c ^ 2) = sqrt(1 - (v ^ 2 / c ^ 2))
((KE / (m * c ^ 2)) ^ 2) - 1 = -(v ^ 2 / c ^ 2)
v = -sqrt((((KE / (m * c ^ 2)) ^ 2) - 1) * c ^ 2)

KE = 4.2e25J
c = 1
m = 2e6kg

v = -sqrt((((4.2e25 / (2e6)) ^ 2) - 1) * 1)
v = -2.1e+19

I'm expecting an answer between 0 and 1, where did I go wrong?

 

[BvU]

In the first step.

 

[fresh_42]

I want to know how fast the space shuttle would have to hit the Earth to unleash the same kinetic energy as the asteroid that killed the dinosaurs. I've already determined that it's relativistic, by showing that Newtonian physics gives a v greater than c

Spoiler: Newtonian physics

4.2e25J = (2e6 * v ^ 2 ) / 2.
8.4e25J = 2e6 * v ^ 2
v = sqrt(8.4e25J / 2e6)
v = 6,480,740,698 m/s

But once I plug in the equations, I get a wrong result.

KE = (m * c ^ 2) / sqrt(1 - (v ^ 2 / c ^ 2))
KE / (m * c ^ 2) = sqrt(1 - (v ^ 2 / c ^ 2))
((KE / (m * c ^ 2)) ^ 2) - 1 = -(v ^ 2 / c ^ 2)
v = -sqrt((((KE / (m * c ^ 2)) ^ 2) - 1) * c ^ 2)

KE = 4.2e25J
c = 1
m = 2e6kg

v = -sqrt((((4.2e25 / (2e6)) ^ 2) - 1) * 1)
v = -2.1e+19

I'm expecting an answer between 0 and 1, where did I go wrong?

You mixed nominator and denominator of the quotient on the RHS of your second line.

 

[BvU]

and with setting c = 1

 

[newjerseyrunner]

You mixed nominator and denominator of the quotient on the RHS of your second line.

Where? I divided both sides of the equation by the term m * c ^ 2? It was in the numerator of the RHS in the 1st line and in the denominator of the LHS on the second line. I'm sure this is going to be one of those things that's stupidly obvious once I see it, but I'm still missing it.

and with setting c = 1

I did? Isn't that the value I'm supposed to use? If I want v as a percentage of c, shouldn't c be 1 exactly?

 

[BvU]

You have $a = b/c$ and proceed with $a/b = c$

Isn't that the value I'm supposed to use?

No: c is 3 x 10⁸ in your system of units.

v/c is usually called $\beta$. Your numerical value for KE/m has a factor c² that you can't simply set to 1.

 

[fresh_42]

... instead of $a/b = 1/c$

 

[newjerseyrunner]

You have $a = b/c$ and proceed with $a/b = c$

No: c is 3 x 10⁸ in your system of units.

v/c is usually called $\beta$. Your numerical value for KE/m has a factor c² that you can't simply set to 1.

Oh, see, there's the stupid mistake. I need caffeine. I still would have gotten the wrong answer though so thanks pointing out my unit mistake too. :)

 

[BvU]

There's a few other problems to deal with in this scenario: they had big problems with re-entry even at very low velocities. Like most meteorites, your speeding shuttle would evaporate in the first few km of atmosphere -- fortunately.

 

[BvU]

I get $\beta = {465\over 466}$ and worry about fuel prices (but only for a short time).

[edit] Missed ! see #14 below

 

[newjerseyrunner]

I'm still getting something very wrong.

Relativistic Kinetic Energy
KE = (m * c ^ 2) / sqrt(1 - (v ^ 2 / c ^ 2))

Divide both sides by m * c ^ 2
(m * c ^ 2) / KE = sqrt(1 - (v ^ 2 / c ^ 2))

Square both sides
((m * c ^ 2) / KE)^2 = 1 - (v ^ 2 / c ^ 2)

Replace the subtraction by addition of a negative
((m * c ^ 2) / KE)^2 = 1 + -(v ^ 2 / c ^ 2)

Subtract 1 from both sides
((m * c ^ 2) / KE)^2 - 1 = -(v ^ 2 / c ^ 2)

Simplify the negative
((m * c ^ 2) / KE)^2 - 1 = -v ^ 2 / c ^ 2

Multiply both sides by c ^ 2
(((m * c ^ 2) / KE)^2 - 1) * c ^ 2 = -v ^ 2

Take the square root of both sides
sqrt((((m * c ^ 2) / KE)^2 - 1) * c ^ 2) = -v

Multiple both sides by -1
v = -sqrt((((m * c ^ 2) / KE)^2 - 1) * c ^ 2)

Input known values
v = -sqrt((((2e6 * 3e5 ^ 2) / 4.2e25)^2 - 1) * 3e5 ^ 2)

Solve
v = -300000 i

 

[newjerseyrunner]

There's a few other problems to deal with in this scenario: they had big problems with re-entry even at very low velocities. Like most meteorites, your speeding shuttle would evaporate in the first few km of atmosphere -- fortunately.

Agreed, this is purely theoretical assuming perfect conditions. No atmosphere.

 

[fresh_42]

I think you still have a sign error which gives you the imaginary solution. $\sqrt{-v^2} = vi \neq -v$.

 

[BvU]

Yes, and this is a not-so-smart way to find something close to 300000.

You have $\displaystyle {\rm{E} \over \ mc^2} = 233 = \gamma = {1\over \sqrt{1-\beta^2}} \$, so clearly $\beta\approx 1$ and you can write

$1-\beta^2 = (1+\beta) (1-\beta) = 2(1-\beta)$

and now I have to correct meself (post #10):$$
1-\beta ={\textstyle {1\over 2} } {1\over 233^2} \ \Rightarrow \ \beta = 1 - {1\over 108889} = {108888\over 108889} \ \Rightarrow v = 0.999991\ {\rm c} $$

 

[TeethWhitener]

There's a few other problems to deal with in this scenario: they had big problems with re-entry even at very low velocities. Like most meteorites, your speeding shuttle would evaporate in the first few km of atmosphere -- fortunately.

Agreed, this is purely theoretical assuming perfect conditions. No atmosphere.

$v \approx c$, and $c \approx 3 \times 10^8 m/s$, so if the atmosphere is 100km thick, the shuttle will take approximately 0.3 milliseconds to traverse it. Will the spaceship evaporate in that period? Does it matter? You're still depositing a mole of joules into the system.

Edit: Of some relevance.

 

[newjerseyrunner]

Yes, and this is a not-so-smart way to find something close to 300000.

You have $\displaystyle {\rm{E} \over \ mc^2} = 233 = \gamma = {1\over \sqrt{1-\beta^2}} \$, so clearly $\beta\approx 1$ and you can write

$1-\beta^2 = (1+\beta) (1-\beta) = 2(1-\beta)$

and now I have to correct meself (post #10):$$
1-\beta ={\textstyle {1\over 2} } {1\over 233^2} \ \Rightarrow \ \beta = 1 - {1\over 108889} = {108888\over 108889} \ \Rightarrow v = 0.999991\ {\rm c} $$

Awesome, that's the exact same answer I got when I put in 299792458 instead of 3e5, just with an i in it. I'm sure I'd find the mistake if I looked, but I don't really care if my mistake only changes the imaginary factor. Thanks for the simplification too.

 

[newjerseyrunner]

$v \approx c$, and $c \approx 3 \times 10^8 m/s$, so if the atmosphere is 100km thick, the shuttle will take approximately 0.3 milliseconds to traverse it. Will the spaceship evaporate in that period? Does it matter? You're still depositing a mole of joules into the system.

Edit: Of some relevance.

I was thinking that too. I doesn't really make a difference, you're still adding the amount of energy that killed the dinosaurs into the system. It'd go off like a bomb and obliterate the biosphere no matter where something with that much power exploded.

 

[BvU]

$v \approx c$, and $c \approx 3 \times 10^8 m/s$, so if the atmosphere is 100km thick, the shuttle will take approximately 0.3 milliseconds to traverse it. Will the spaceship evaporate in that period? Does it matter? You're still depositing a mole of joules into the system.

Granted. I got stuck in the immediate vicinity of the actual speeds of descent and forget to 'think' relativistically.
Does it matter ? No. There's no way you can get this kind of energy in a 2000 metric ton vehicle. Think of it: the energy equivalent of 470 kilotonne of matter ...

But it sure is fun as a thought experiment... and to bicker about