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## Q. 19.5

A spaceship is pointing towards a planet. The initial total mass of the ship is $M_{0}$ kg and the fuel is ejected at a rate of k kg $s^{-1}$ with a speed u $ms^{-1}$ relative to the spaceship. The ship starts at rest relative to the planet, and its speed, when all the fuel is exhausted, is 2u $ms^{-1}$. However, it is still so far from the planet that the gravitational force can be neglected.
i) From the equation of motion derive an expression for v in terms of the remaining mass m at any time.
Hence show that the total mass at the end is $\frac{M_{0}}{e^{2}}$ kg.
ii) Find the velocity of the ejected fuel relative to the planet just before the end of the period. Comment on this result.
iii) Show that the distance travelled during the period of the rocket firing is

$\frac{M_{0}u}{ke^{2}}(e^{2} – 3).$

## Verified Solution

i) After time t, the mass of the ship is $m = M_{0} – kt$. Since there are no external forces, the equation of motion is

$m\frac{dv}{dt} = ku$

$\frac{dv}{dt} = \frac{ku}{(m_{0} – ku)}.$

Integrating gives

$v = – u ln(M_{0} – kt) + c.$

When t = 0, v = 0, so c = u ln ($M_{0}$). This gives

$v = – u ln \left(\frac{M_{0}}{M_{0} – kt}\right) = u ln \left( \frac{M_{0}}{m}\right).$                ①

At the end of the burn, v = 2u, giving

$ln\left(\frac{M_{0}}{m}\right) = 2$

$\frac{M_{0}}{m} = e^{2}$

⇒         $m = \frac{M_{0}}{e^{2}}.$

ii) At the end of the burn period, the velocity of the fuel relative to the planet is 2u – u = u $ms^{-1}$. The expelled fuel is actually travelling towards the planet, although not as fast as the rocket! There is no paradox here; all velocities are relative.

iii) From ①,

$\frac{ds}{dt} = u ln\left(\frac{M_{0}}{m_{0} – kt}\right)$

where s is the displacement from the start. Integrating gives

$s = u \int{ln\left(\frac{M_{0}}{m_{0} – kt}\right) dt}$

$= u \int{ln M_{0} dt} – u \int{ln(M_{0} – kt)} dt.$          ②

The second integral is easier if you make the substitution $m = M_{0} – kt.$

$\begin{matrix} \int{ln(M_{0} – kt) dt} = – \frac{1}{k} \int{ln m dm} & \longleftarrow \boxed{\text{You can use integration by parts with u = ln x ,} \frac{dv}{dx} = 1 \text{ to show that } \int{ln x dx = x ln x – x.}} \end{matrix}$

$= – \frac{1}{k} (m ln m – m) + \text{constant.}$

Substituting this in ② gives

s = ut ln $M_{0} + \frac{mu}{k} (ln m – 1) + c.$

When s = 0, t = 0, m = $M_{0}$. So

$c = – \frac{M_{0}u}{k} (ln M_{0} – 1).$

Hence

s = ut ln $M_{0} + \frac{mu}{k} (ln m – 1) – \frac{M_{0}u}{k}(ln M_{0} – 1).$        ③

At burn out, $m = \frac{M_{0}}{e^{2}}$  (proved in part i) )

$⇒ M_{0} – kT = \frac{M_{0}}{e^{2}}$

where T is time of burn out

⇒           $T = \frac{M_{0}}{ke^{2}} (e^{2} – 1).$

Substituting $m = \frac{M_{0}}{e^{2}}$ and t = T = $\frac{M_{0}}{ke^{2}}(e^{2} – 1)$  in ③ gives

$s = \frac{uM_{0}}{ke^{2}} (e^{2} – 1) ln M_{0} + \frac{M_{0}u}{ke^{2}}(ln(\frac{M_{0}}{e^{2}}) – 1) – \frac{M_{0}u}{k}(ln M_{0} – 1)$

$= \frac{M_{0}u}{ke^{2}}[e^{2} ln M_{0} – ln M_{0}) + (ln M_{0} \underset{\underset{ln e^{2} = 2}{\uparrow}}{-2} – 1) – (e^{2} ln M_{0} – e^{2})]$

$= \frac{M_{0}u}{ke^{2}}(e^{2} – 3).$