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

Homework Help: Satellite Slingshot manoeuvre question

  1. Nov 20, 2008 #1
    I've already attempted to work everything out by the way but my answers seem a little off.

    1. The problem statement, all variables and given/known data
    A satellite performs a sling shot manoeuvre around a planet. The mass of the planet is [itex]6.00 * 10^{24} kg[/itex] and the mass of the satellite is [itex]7.00 * 10^{2} kg[/itex]. The satellite approaches the planet from a great distance with relative speed [itex]2 * 10^{3} ms^{-1}[/itex] and impact parameter [itex]6.0 * 10^{7} m[/itex]. Determine the closest approach distance, the maximum speed of the satellite relative to the planet and the angle through which it is deflected.

    2. Relevant equations
    [itex] 1/r = \alpha \sqrt {2c/l^{2} + \alpha^{2}} \hspace {5 mm}sin (\theta + \theta_{o})[/itex]

    where

    [itex] \alpha = {G(M_{1}+M_{2})}/l^{2}[/itex]

    [itex] l = V_{0}b[/itex]

    [itex] c = 1/2 V_{0}^{2}[/itex]


    3. The attempt at a solution

    I've attempted to solve the whole question, but I think my values for the closest approach distance and maximum speed are quite wrong.

    [itex]M_{planet} = 6.00 * 10^{24} kg[/itex]
    [itex]M_{satellite} = 7.00 * 10^{2} kg[/itex]
    [itex]V_{0} = 2*10^{3} ms^{-1}[/itex]
    [itex]Impact Parameter = b = 6.00 * 10^{7}m[/itex]

    Deflection Angle
    [itex]\sqrt {1 + 2c/l^{2}\alpha^{2}} \hspace {5 mm} sin \theta = -1[/itex]

    [itex] \alpha = {G(M_{1}+M_{2})}/l^{2}[/itex]

    [itex] l = V_{0}b[/itex]

    [itex] c = 1/2 V_{0}^{2}[/itex]

    [itex] c = (2 * 10^{3})^{2}/2 = 2*10^{6}[/itex]

    [itex] l = (2 * 10^{3}) * (6.0*10^{7}) = 1.2*10^{11}[/itex]

    [itex] \alpha = G(6.00*10^{24} + 7*10^2)/l^{2} = G(6*10^{24})/(1.44*10^{22}) = 4.002*10^{14}/1.44*10^{22} = 2.78*10^{-8}[/itex]

    [itex]\sqrt {1 + 2c/l^{2}\alpha^{2}} = \sqrt {1 + \dfrac{2*(2*10^{6})} {(1.2*10^{11})^{2} (2.78*10^{-8})^{2}}[/itex]

    [itex] \sqrt {1 + 0.36} = \sqrt {1.36}[/itex]

    [itex] sin \theta = {\dfrac{-1} {\sqrt 1.36}}[/itex]

    [itex] \theta = -59.04[/itex]

    Angle of Deflection = 118.08

    Unless anyone can point out a problem here, this seems fine.

    Closest Approach Distance

    [itex]\dfrac {1}{r_{min}} = \alpha \sqrt {\dfrac {2c}{l^{2}} + \alpha^{2}}[/itex]

    [itex]\dfrac {1}{r_{min}} = 2.78*10^{-8} * \sqrt {\dfrac{4*10^{6}}{(1.2*10^{11})^{2}} + (2.78*10^{-8})^{2}}[/itex]

    [itex]\dfrac {1}{r_{min}} = 2.78*10^{-8} \sqrt {1.051*10^{-15}}[/itex]

    [itex]\dfrac {1}{r_{min}} = 9.01*10^{-16}[/itex]

    [itex]r_{min} = 1.11*10^{15}m[/itex]

    This seems quite large, and it means that my speed at closest approach is very small.

    Speed at closest approach

    [itex]\dfrac{v_{0}}{b} = \dfrac {1.2*10^{11}}{1.11*10^{15}} = 1.08*10^{-4}ms^{-1}[/itex]

    This seems very small which is why i'm wondering if someone could tell me where I went wrong.

    This is also the first time i've tried extensively using latex so pardon me if it's hard to read.

    Thanks guys.
     
  2. jcsd
  3. Nov 20, 2008 #2
    an unfortunate bump :(
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook