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A Rocket's Distance from the center of the moon

  1. May 7, 2009 #1
    1. The problem statement, all variables and given/known data
    The Moon has a mass of M= 7.0E22 kg and a radius of R=1.75E6 m. A projectile with
    mass m=10 kg is shot straight up from the Moon’s surface with an initial speed of 500 m/s. What is the maximum distance from the center of the Moon that this projectile can reach?

    2. Relevant equations
    E_initial=E_final
    Kinetic_initial=Grav_potential
    .5mv^2=GMm/r


    3. The attempt at a solution
    Thinking through this I believe this is wrong, because essentially that is the equation for escape velocity. We don't know for sure that this rocket will escape... I am lost...
     
  2. jcsd
  3. May 7, 2009 #2

    rl.bhat

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    Thinking through this I believe this is wrong, because essentially that is the equation for escape velocity. We don't know for sure that this rocket will escape... I am lost...

    It is true for any velocity. Calculate r.
     
  4. May 7, 2009 #3
    I didn't get the right answer...
     
  5. May 7, 2009 #4

    rl.bhat

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    Show your calculations.
     
  6. May 7, 2009 #5
    r=(2GM)/v^2 is what i believe I worked it out to so

    r=(2*6.7E-11*7E20)/500^2

    r=3.752E5

    According to the answer key it is supposed to be 1.84E6
     
  7. May 7, 2009 #6
    So is there something I am missing I thought a simple energy conservation would cover it
     
  8. May 7, 2009 #7
    Use the equation :

    K+U = K_o+U_o

    where K could be 1/2mv^2
    and U = -GMm/r
     
  9. May 7, 2009 #8
    I am assuming that there is no kinetic in my final state because the rocket doesn't have enough velocity to escape according to sqrt(2GM/r) so then:

    (GMm/r_final)=.5mv^2+GMm/r_initial

    r=GMm/(.5mv^2+(GMm/r_initial))

    r= 6.7E-11*7E22*10/(.5*10*500^2+((6.7E-11*7E22*10)/1.75E6))

    r=1.672e6
    I still don't get the correct answer
     
  10. May 7, 2009 #9
    here :

    -GMm/ r =1/2m*v^2 - GMm/R_e

    r = (1/R_e - v^2 / (2*G*M_e ) ^-1

    r = 1.84E^6
     
  11. May 7, 2009 #10
    Could I get a short explanation on that? I don't follow your variables
     
  12. May 8, 2009 #11
    r=(-2*M_e*G*R_e)/(v^2*R_e-2*G*M_e)

    r=(-2*7e22*6.7e-11*1.75e6)?(500^2*1.75e6-2*6.7e-11*7e22)

    r=1.84e6

    Good luck with your test today, I've been up all night studying for it, hope this helps!
     
  13. May 8, 2009 #12
    Maybe my brain is physics fried haha But which Variables need to be negative and why? Is it both potentials? I thought the would have positive potential energy or is it the fact that it is in the negative direction? Oh boy I am going to fail haha
     
  14. May 8, 2009 #13
    Use exactly what tnutty has written here for positive and negative values.

    Dont worry about the final, I dont know if anyone is going to do well on it, its bad if Im aiming for a 50% on it.....
     
  15. May 8, 2009 #14

    LowlyPion

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    The gravitational potential expression is

    PE = -GM/r

    So for your problem they would both carry a negative sign.
     
  16. May 8, 2009 #15
    I also have my final today. Your's not with jones is it?
     
  17. May 8, 2009 #16
    Jones? I go to Purdue
     
  18. May 8, 2009 #17
    By the way I got it thanks guys... Those negative signs will get you every time haha
     
  19. May 8, 2009 #18
    NO problem
     
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