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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

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

4 comments

  1. Artem says:

    I think the setup of the problem is wrong, as well as Apostol answer. When the rocket is landing, the direction of the exhaust is the same as the direction of the speed of the rocket. Thus, the equation setup should be c(t) = c \Rightarrow r''(t) = g - kc\frac{1}{w - kt}, and there is one term in Apostol answer that is wrong in the end: has k instead of w inside the first parenthesis before log

    • Anonymous says:

      Greetings! I agree with you. Definitely something fishy. Testing with some intuition reveals that according to the current signs the force produced by the fuel being hurled away, would propel the rocket to the ground, which is the exact opposite of the desired effect! And also contradicts the problem statement. Where it is stated that the force produced by the rocket is offsetting the force of gravity, but given the equation for r''(x) they would in fact act in unison!

      If you look at r(t) and evaluate the integral, you clearly get -g\frac{t^2}{2} not g\frac{t^2}{2} as OP stated. The rest of the expression has also the wrong sign.

  2. 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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