Go spaceship go
Show that even a very powerful spaceship would eventually run out
of overtaking power
Age
16 to 18
Challenge level
A spaceship driven by an infinite source of power accelerates with constant power through deep space. We watch it fly away from Earth and record its speed and progress.
Show that after a certain time of travelling it will take longer than one day for the velocity of the spaceship to increase by $1 \mathrm{\ m\ s^{-1}}$.
Find this time of travelling (in years) if the spaceship starts travelling from rest and has a power to weight ratio of $500 \mathrm{\ W\ kg^{-1}}$. How fast would the spaceship be travelling at this time?
NOTES AND BACKGROUND
Of course, this problem uses entirely classical, Newtonian physics. A proper analysis would need to take into account special relativity, and it is worth considering how important this might be given the answer. It is also worth noting that the spaceship is actually quite powerful, given that a typical sportscar has a power to weight ratio of around $200 \mathrm{\ W\ kg^{-1}}$ .
Recall that the Power is the rate at which energy is produced, and in this problem the energy is entirely kinetic.
From stationary, we can say that
$$Pt = \frac{1}{2}mv^2 \quad \therefore \quad v = \sqrt{\frac{2Pt}{m}} \quad \therefore \quad \frac{\mathrm{d}v}{\mathrm{d}t} = \sqrt{\frac{P}{2mt}}$$
In general, $Pt = \frac{1}{2}m\left( v_2^2-v_1^2 \right)$.
If $v_2 = v_1 + 1$, then
$$Pt = \frac{1}{2}m\left( (v_1+1)^2-v_1^2 \right) = \frac{1}{2}m(2v_1 + 1)$$
If $t = 1 \textrm{ day } = 86400 \mathrm{\ s}$
$$v_1 = \frac{Pt}{m} - \frac{1}{2} = \frac{86400P}{m} - \frac{1}{2}$$
$$\frac{86400P}{m} - \frac{1}{2} = \sqrt{\frac{2Pt}{m}} \Rightarrow \frac{2Pt}{m} = \left( \frac{86400P}{m} - \frac{1}{2} \right) ^2$$
$$\therefore t = \frac{m}{2P} \left( \frac{86400P}{m} - \frac{1}{2} \right) ^2$$
after which it will take one day to accelerate by $1 \mathrm{\ m\ s^{-1}}$.
We are told that $\frac{P}{m} = 500 \mathrm{\ W\ kg^{-1}}$
$$\therefore t = \frac{1}{1000} \left( 43200000 - \frac{1}{2} \right) ^2 = 1.86624 \times 10^{12} \mathrm{\ s} = 21.6 \textrm{ million days}$$
$v = 4.32 \times 10^7 \mathrm{\ m\ s^{-1}}$, which is around $15\%$ of the speed of light.