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Find a formula for the distance a ship falls under the influence of gravity

A spaceship is pulled toward the ground by gravity. Offsetting this force, the ship fires a rocket directly downward, consuming fuel at a rate of k pounds per second and the exhaust material has a constant speed of c feet per second relative to the rocket. Find a formula giving the distance the spaceship falls in time t if it is at rest at time t = 0 and has initial weight of w pounds.


The motion of the rocket is governed by the equation

    \[ m(t) r''(t) = m'(t) c(t) + F(t). \]

We are given

    \begin{align*}  m(t) &= \frac{w - kt}{g} & \implies && m'(t) = -\frac{k}{g} \\  c(t) &= -c \\  F(t) &= m(t) g = w - kt. \end{align*}

So,

    \begin{align*}  &&r''(t) &= \frac{kc}{w - kt} + g \\ \implies && r'(t) &= \int \left( \frac{kc}{w-kt} + g \right) \, dt \\  &&&= -c \log | w - kt| + gt + C_0. \end{align*}

Then, since r'(0) = 0 we have C_0 = c \log w. Therefore,

    \[ r'(t) = -c \log \left| \frac{w-kt}{w} \right| + gt. \]

And so,

    \begin{align*}  r(t) &= -\int \left( c \log \left| \frac{w-kt}{w} \right| + gt \right) \, dt \\  &= - \frac{c(w-kt)}{k} \log \left| \frac{w-kt}{w} \right| + \frac{c(w-kt)}{k} + \frac{1}{2} g t^2 + C_1. \end{align*}

Then since r(0) = 0 we have

    \[ C_1 = -\frac{cw}{k}. \]

Hence, the equation of motion is

    \[ r(t) = \frac{1}{2} gt^2 - ct + c\left( t- \frac{w}{k} \right) \log \left| 1- \frac{kt}{w} \right|. \]

2 comments

  1. tom says:

    The answer looks wrong in Apostol- the -ct element is supposed to be -cw/k. It would only be -ct if the position when the fuel runs out was requested, T=W/k.

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