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Gravity - maximum height

  1. Oct 31, 2008 #1
    Gravity -- maximum height

    An object is projected upward from the surface of the Earth with an initial speed of 3.7 km/s. Find the maximum height it reaches.
    m

    Ive used the Energy conservation
    Kf + Uf = Ki + Ui
    1/2 mvf^2 - G*M_e*m/rf = 1/2 mvi^2 - G*M_e*m/ri
    = 0 - G*M_e*m/rf = 1/2 mvi^2 - G*M_e*m/R_e

    R_e = radius of earth = 6.37 x 10^6m
    M_e = mass of earth = 5.98 x 10^24kg
    G = 6.67 x 10^-11 N*m^2/kg^2

    1/radius_f = -Vi^2/2*G*M_e + 1/R_e
    1/radius_f = (-3700 m/s)^2 / 2(6.67 x 10^-11)(5.98 x 10^24) + 1/6.37 x 10^6
    =13690000 / 7.97732 x 10^13 + 1/6.37 x 10^6
    1.716115187^-7 + 1.569858713^-7 = 3.2859739^-7
    1/3.2859739^-7 = 3043237.806
    then I do
    3043237.806 - 6.37 x 10^6 but that gives me a negative height.

    Can someone help me? Ive been going at this problem for hours.
     
  2. jcsd
  3. Oct 31, 2008 #2
    Re: Gravity -- maximum height

    Sorry, but this is really hard to follow. I find it much easier to write out the whole problem algebraically before subbing in any numbers, and also using consistant significant figures.

    If you could do this first it would help a lot.
     
  4. Oct 31, 2008 #3
    Re: Gravity -- maximum height

    This is the final equation:
    1/radius_f = -Vi^2/2*G*M_e + 1/R_e

    then I solve for radius_f then I divide it by 1

    then I subtract it from the earth's radius to get the height but it gives me a negative height.
     
  5. Oct 31, 2008 #4
    Re: Gravity -- maximum height

    I'm not entirely sure why that is, but it could have something to do with order of operations (don't quote me on that). Using your formula, I would take a reciprocal of it first: [tex]R + r = \frac{2GM_e}{-v^2} + R [/tex]

    From this we can see that the R terms cancel and we're left with r that we want to find. I have not solved this mathematically yet though.
     
    Last edited: Oct 31, 2008
  6. Oct 31, 2008 #5
    Re: Gravity -- maximum height

    well the textbook has the same problem explained but different velocity and it uses that equations and same steps I posted. But I don't know what I did wrong mathematically to give me a negative height. I was wondering if someone can work it out and see if they get the same answer or diff.
     
  7. Oct 31, 2008 #6
    Re: Gravity -- maximum height

    you go wrong right here: you forgot a minus sign in the first term on the rhs going from the 1st to 2nd line
     
  8. Nov 1, 2008 #7

    LowlyPion

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    Homework Helper

    Re: Gravity -- maximum height

    Gravitational Potential is given by -GMm/r
    So let u = GM so that

    1/2mV2 - um/R = - um/(R + h)

    V2/(2*u) = - 1/(R + h) + 1/R = h/(R2 +R*h)

    You might recognize then in the near field with R >> h :

    V2 = 2 * (u/(R2 +R*h)) * h ≈ 2*(g)*h

    Rearranging:
    2*u*h = V2R2 + V2*R*h

    Solving for h then I get
    h = V2R2/(2*u - V2*R)

    Using km and u in km as ≈ 4*105 and R as 6400 km V as 3.7 km/s then ...
     
  9. Nov 1, 2008 #8
    Re: Gravity -- maximum height

    h = V^2R^2/(2*u - V^2*R)

    h = 13.69 * 40960000 / (800000 - 87616)
    h = 560742400 / 712384
    h = 787.1350283 km
    h = .7871350283 m

    Do I add this to the Radius of the earth?
     
  10. Nov 1, 2008 #9

    LowlyPion

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    Homework Helper

    Re: Gravity -- maximum height

    The way it was originally constructed h was height above the radius.
    1/2mV2 - um/R = - um/(R + h)

    Better check your conversion however. The bolded line looks OK. The one below - not.

    And significant digits wise you should round likely to the nearest km.

    The value of u for Earth for instance I already rounded.

    See: http://en.wikipedia.org/wiki/Gravitational_constant#The_GM_product
     
    Last edited: Nov 1, 2008
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