# B Where is the bug in my math?

1. Jan 12, 2017

### 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
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?

2. Jan 12, 2017

### BvU

In the first step.

3. Jan 12, 2017

### Staff: Mentor

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

4. Jan 12, 2017

### BvU

and with setting c = 1

5. Jan 12, 2017

### newjerseyrunner

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.
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?

6. Jan 12, 2017

### BvU

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.

7. Jan 12, 2017

### Staff: Mentor

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

8. Jan 12, 2017

### newjerseyrunner

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. :)

9. Jan 12, 2017

### 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.

10. Jan 12, 2017

### BvU

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

 Missed ! see #14 below

Last edited: Jan 12, 2017
11. Jan 12, 2017

### 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

12. Jan 12, 2017

### newjerseyrunner

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

13. Jan 12, 2017

### Staff: Mentor

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

14. Jan 12, 2017

### 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):$$1-\beta ={\textstyle {1\over 2} } {1\over 233^2} \ \Rightarrow \ \beta = 1 - {1\over 108889} = {108888\over 108889} \ \Rightarrow v = 0.999991\ {\rm c}$$

15. Jan 12, 2017

### TeethWhitener

$v \approx c$, 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.

16. Jan 12, 2017

### newjerseyrunner

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.

17. Jan 12, 2017

### newjerseyrunner

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.

18. Jan 12, 2017

### BvU

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

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