Rocket Speed

Relevant wiki: Newton's Laws of Motion

$Force$ = $Mass \times Acceleration$

As the rocket ejects fuel, its total $Mass$ (rocket mass + fuel mass) decreases. If the $Force$ (thrust) is to remain constant, the $Acceleration$ (rate of speed increase) must increase.

Acceleration = (rocket Thrust)/(rocket Mass) or A=T/M

T is constant but M is constantly decreasing as fuel is burned off at a constant rate. The value of T/M then increases over time and so A will increase in a hyperbolic fashion (y= a/x) until all fuel is spent at which point A drops to zero and the rocket reaches terminal velocity.

We let $m(t)=m_0 - \alpha t$ , $0\leq t < \frac{m_0}{\alpha}$ , $m_0$ is the initial mass of the rocket and $\alpha$ is some positive number.

Force remains constant, so let $F=F_0$

$F=ma$

$F_0 = a(m_0 - \alpha t)$

$a(t)= \frac{F_0}{m_0 - \alpha t}$

when $t$ get bigger , $a$ always be positive, so velocity increasing.

Everyone is quoting that F = ma, but we can't use that in this problem because the mass is changing. Instead let's use the more general version of Newton's second law that says that force is the time derivative of momentum.

$F = \frac{dP}{dt}$

Momentum is mass times velocity. To find what the time derivative of momentum is, consider the rocket at an arbitrary time t and a small amount of time later: t+∆t. At the end we will take the limit as ∆t goes to zero.

Since we know that the rocket is losing mass at a constant rate, we can write the mass of the rocket as:

$m(t) = m_0 - at$

where $m_0$ is the time of the rocket when t=0 and a is the rate at which the rocket is losing mass.

Let $v(t)$ be the velocity of the rocket at time t. This is the function we are trying to find. Thus, at time t, the momentum of the rocket will be:

$P(t) = v(t)m(t) = v(t)(m_0-at)$

At time t+∆t, the speed of the rocket will have changed by some amount ∆v, which may be positive or negative. Thus:

$P(t+\Delta t) = (v(t)+\Delta v)m(t+\Delta t) = (v(t)+\Delta v)(m_0-at-a\Delta t) = v(t)m_0 + m_0\Delta v - at\Delta v - v(t)at -av(t)\Delta t-a\Delta t\Delta v$

Now consider the value of P(t+∆t)-P(t).

$P(t+\Delta t)-P(t) = v(t)m_0 + m_0\Delta v - at\Delta v - v(t)at -av(t)\Delta t-a\Delta t\Delta v - v(t)m_0 + av(t)t = m_0\Delta v - at\Delta v -av(t)\Delta t-a\Delta t\Delta v$

Divide both sides of the equation by ∆t and we get:

$\frac{P(t+\Delta t)-P(t)}{\Delta t} = m_0\frac{\Delta v}{\Delta t} - at\frac{\Delta v}{\Delta t} - av(t) - a\Delta v = (m_0-at)\frac{\Delta v}{\Delta t} - av(t) - a\Delta v$

Now we take the limit as ∆t (and subsequently ∆v) go to zero. From the definition of the derivative, $\frac{P(t+\Delta t)-P(t)}{\Delta t}$ becomes the derivative of P with respect to time and likewise, $\frac{\Delta v}{\Delta t}$ becomes the derivative of velocity with respect to time. However, the a∆v term simply goes to zero (by direct substitution). Hence, we get:

$\frac{dP}{dt} = (m_0-at)\frac{dv}{dt} - av(t)$

By Newton's second law, however, we know that the time derivative of momentum is the force applied, which is a constant thrust, which we may call $F_0$ . We thus have a differential equation:

$F_0 = (m_0-at)\frac{dv}{dt} - av(t)$

with initial condition $v(0) = 0$ (i.e. it starts at rest). Using separation of variables we get:

$\frac{F_0}{a} + v(t) = (\frac{m_0}{a}-t)\frac{dv}{dt}$

$\frac{dv}{\frac{F_0}{a} + v(t)} = \frac{dt}{\frac{m_0}{a}-t}$

$\int_{0}^{v} \frac{dv}{\frac{F_0}{a} + v(t)} = \int_{0}^{t} \frac{dt}{\frac{m_0}{a}-t}$

$ln(\frac{F_0}{a}+v) - ln(\frac{F_0}{a}) = -ln(\frac{m_0}{a}-t) + ln(\frac{m_0}{a})$

$ln(\frac{F_0+av}{F_0}) = ln(\frac{m_0}{m_0-at})$

$\frac{F_0+av}{F_0} = \frac{m_0}{m_0-at}$

$1 + \frac{a}{F_0}v = \frac{m_0}{m_0-at}$

$\frac{a}{F_0}v = \frac{m_0}{m_0-at}- 1$

$v(t) = \frac{F_0}{a}( \frac{m_0}{m_0-at}-1)$

We can check that when t=0, $\frac{m_0}{m_0-at}-1 = \frac{m_0}{m_0} - 1 = 1 -1 = 0$ , which is what we expected. Notice also that when $t = \frac{m_0}{a}$ , we reach a vertical asymptote. Thus as t grow from 0 to $\frac{m_0}{a}$ , $\frac{m_0}{m_0-at}$ grows from 1 to infinity and so $v(t)$ grows from 0 to infinity. Therefore, it's clear that the velocity is increasing. Furthermore, the rate at which the velocity increases must also be increasing because otherwise, the graph would be concave down and it would never be able to reach a vertical asymptote approaching positive infinity.

Therefore: the speed is increasing at an increasing rate.

Note: the velocity will not actually reach infinity because the fuel will run out before $t=\frac{m_0}{a}$ . If it didn't, then that would mean that the rocket was made out of fuel and you not only burnt your fuel but your rocket as well. After the fuel runs out, the rocket would just move at the velocity it was going at when the fuel ran out. This is because with no fuel there is no thrust and thus, by Newton's second law, $\frac{dP}{dt} = 0$ , which implies (since the mass is no longer changing) that $\frac{dv}{dt} = 0$ (i.e. the acceleration is zero and the velocity is constant).

Fun problem.

Assuming we stay well below relativistic speeds. =)

Logically if we say more force is required for more massive body to have same acceleration , or if you know simple laws of motion i.e F=ma (m=mass, a=acceleration) so as rocket starts it has mass say M( which is max because it is sum of mass of rocket + the fuel for thrust) at any time it has mass less than M, And the engine is supplying constant force but the mass is decreasing so less massive body gets more accelerated by the same force so the increasing speed has a increasing nature(increasing acceleration)