Thread: Kepler vector problem
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Old October 6th, 2008, 03:27 PM
icemanfan icemanfan is offline
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r(t) = <t^3, t^2, t^3>

r'(t) = <3t^2, 2t, 3t^2>

a(t) = r''(t) = <6t, 2, 6t>

The force required is F = ma:

F(t) = m(a(t)) = m<6t, 2, 6t> = <6mt, 2m, 6mt>

The actual magnitude of the force is:

\sqrt{(6mt)^2 + (2m)^2 + (6mt)^2}

= \sqrt{36m^2t^2 + 4m^2 + 36m^2t^2}

= \sqrt{72m^2t^2 + 4m^2}

= \sqrt{(4m^2)(18t^2 + 1)}

= 2m\sqrt{18t^2 + 1}
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