Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

B Where is the bug in my math?

  1. Jan 12, 2017 #1
    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. jcsd
  3. Jan 12, 2017 #2

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    In the first step.
     
  4. Jan 12, 2017 #3

    fresh_42

    Staff: Mentor

    You mixed nominator and denominator of the quotient on the RHS of your second line.
     
  5. Jan 12, 2017 #4

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    and with setting c = 1
     
  6. Jan 12, 2017 #5
    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?
     
  7. Jan 12, 2017 #6

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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.
     
  8. Jan 12, 2017 #7

    fresh_42

    Staff: Mentor

    ... instead of ##a/b = 1/c##
     
  9. Jan 12, 2017 #8
    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. :)
     
  10. Jan 12, 2017 #9

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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.
     
  11. Jan 12, 2017 #10

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

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

    [edit] Missed ! see #14 below
     
    Last edited: Jan 12, 2017
  12. Jan 12, 2017 #11
    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
     
  13. Jan 12, 2017 #12
    Agreed, this is purely theoretical assuming perfect conditions. No atmosphere.
     
  14. Jan 12, 2017 #13

    fresh_42

    Staff: Mentor

    I think you still have a sign error which gives you the imaginary solution. ##\sqrt{-v^2} = vi \neq -v##.
     
  15. Jan 12, 2017 #14

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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):$$
    1-\beta ={\textstyle {1\over 2} } {1\over 233^2} \ \Rightarrow \ \beta = 1 - {1\over 108889} = {108888\over 108889} \ \Rightarrow v = 0.999991\ {\rm c} $$
     
  16. Jan 12, 2017 #15

    TeethWhitener

    User Avatar
    Science Advisor
    Gold Member

    ##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.
     
  17. Jan 12, 2017 #16
    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.
     
  18. Jan 12, 2017 #17
    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.
     
  19. Jan 12, 2017 #18

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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 :smile:
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Where is the bug in my math?
  1. Math is my religion (Replies: 4)

  2. My math question (Replies: 8)

Loading...