Viva Rocketry! Part 1


In this post, we will first look at the main characteristics of rocket flight, and then examine the feasibility of launching a satellite as the payload of a rocket into an orbit above the earth.

A rocket accelerates itself by ejecting part of its mass with high velocity.

Screen Shot 2018-11-09 at 9.37.30 AM.png

Fig. 1

Fig. 1 shows a moving rocket. At time t+\Delta t, the mass \Delta m leaves the rocket in opposite direction. As a result, the rocket is being propelled away with an increased speed.


m(\square), m_{\square} – the mass of rocket at time {\square}

\vec{v}_{\square} – the velocity of rocket at time \square

v(\square), v_{\square} – the magnitude of \vec{v}_{\square}

\vec{v}^*_{t+\Delta t} – the velocity of ejected mass \Delta m at t + \Delta t

v^*_{t+\Delta t} – the magnitude of \vec{v}^*_{t+\Delta t}

u – the magnitude of \Delta m‘s velocity relative to the rocket when it is ejected. It is time invariant.

From Fig. 1, we have

\Delta m = m_t - m_{t + \Delta t},

\vec{v}_t = v_{t},

\vec{v}_{t + \Delta t} = v_{t + \Delta t}

and most notably, the relationship between v^*_{t+\Delta t}, v_{t+\Delta t} and u (see “A Thought Experiment on Velocities”):

 v^*_{t+\Delta t} = u - v_{t + \Delta t}.

It follows that

\vec{v}^*_{t+\Delta t} = -v^*_{t+\Delta t} = v_{t + \Delta t} - u,

momentum at time t: \vec{p}(t) = m_t \vec{v}_t = m_t v_t


momentum at time t+\Delta t\vec{p}(t+\Delta t) = m_{t+\Delta t}\vec{v}_{t+\Delta t} + {\Delta m} \vec{v}^*_{t+\Delta t}=m_{t+\Delta t}\vec{v}_{t+\Delta t} + (m_t - m_{t+\Delta t}) \vec{v}^*_{t+\Delta t}= m_{t+\Delta t}v_{t+\Delta t} + (m_t - m_{t+\Delta t})(v_{t+\Delta t}-u).

Consequently, change of momentum in \Delta t is \vec{p}(t+\Delta t)- \vec{p}(t) = m_t (v_{t + \Delta t} - v_t) + u (m_{t + \Delta t} - m_t).

Apply Newton’s second law of motion to the whole system,

\vec{F}= {d \over dt} \vec{p}(t)

= \lim\limits_{\Delta t \rightarrow 0} {{\vec{p}(t+\Delta t) - \vec{p}(t)} \over \Delta t}

= \lim\limits_{\Delta t \rightarrow 0} { {m_t (v_{t + \Delta t} - v_t) + u (m_{t + \Delta t} - m_t)} \over {\Delta t} }

= \lim\limits_{\Delta t \rightarrow 0} {m_t {{v_{t + \Delta t} - v_t} \over {\Delta t}} + {u {{m_{t + \Delta t} - m_t} \over {\Delta t}}}}

= m_t \lim\limits_{\Delta t \rightarrow 0}{(v_{t+\Delta t} - v_t) \over {\Delta t}} + u \lim\limits_{\Delta t \rightarrow 0}{(m_{t +\Delta t} - m_t) \over \Delta t}

That is,

\vec{F} = m(t) {d \over dt} v(t) + u {d \over dt} m(t)

where \vec{F} is the sum of external forces acting on the system.

To get an overall picture of the rocket flight, we will neglect all external forces.

Without any external force, \vec{F} = 0. Therefore

0 = m(t) {d \over dt} v(t) + u {d \over dt} m(t)


{d \over dt} v(t) = -{u \over m(t)} {d \over dt} m(t)\quad\quad\quad(1)

That fact that u, m(t) in (1) are positive quantities shows as the rocket loses mass ({d \over dt} m(t) < 0), its velocity increases ({d \over dt} v(t) > 0)

Integrate (1) with respect to t,

\int {d \over dt} v(t)\;dt = -u \int {1 \over m(t)} {d \over dt} m(t)\;dt


v(t) = -u \log(m(t)) + c

where c is the constant of integration.

At t = 0, v(0)=0, m(0) = m_1 + P where m_1 is the initial rocket mass (liquid or solid fuel + casing and instruments, exclude payload) and P the payload.

It means c = u \log(m_1+P).

As a result,

v(t) = -u \log(m(t)) + u \log(m_1+P)

= -u (\log(m(t) - \log(m_1+P))

= -u \log({m(t) \over m_1+P})


v(t) = -u \log({m(t) \over m_1+P})\quad\quad\quad(2)

Since m_1 is divided into two parts, the initial fuel mass \epsilon m_1 (0 < \epsilon < 1), and the casing and instruments of mass (1-\epsilon)m_1, m(0) can be written as

m(0) = \epsilon m_1 + ( 1 - \epsilon) m_1 + P

When all the fuel has burnt out at t_1,

m(t_1) = (1 - \epsilon)m_1 + P

By (2), the rocket’s final speed at t_1

v(t_1) = -u \log({m(t_1) \over {m_1+P}})

= -u \log({(1-\epsilon)m_1+P  \over {m_1 + P}})

= -u \log({{m_1 + P -\epsilon m_1} \over {m_1+P}})

= -u \log(1-{{\epsilon m_1} \over {m_1+P}})

= -u \log(1-{\epsilon \over {1 + {P \over m_1}}})

= -u \log(1-{\epsilon \over {1 + \beta}})

where \beta = {P \over m_1}.

In other words,

v(t_1) =-u \log(1-{\epsilon \over {1 + \beta}})\quad\quad\quad(3)

Hence, the final speed depends on three parameters

u, \epsilon and \beta


u = 3.0\;km\;s^{-1}, \epsilon = 0.8 and \beta = 1/100.

Using these values, (3) gives

Screen Shot 2018-11-05 at 9.21.14 PM.png

v_1 = 4.7\;km\;s^{-1}\quad\quad\quad(4)

This is an upper estimate to the typical final speed a single stage rocket can give to its payload. Neglected external forces such as gravity and air resistance would have reduced this speed.

With (4) in mind, let’s find out whether a satellite can be put into earth’s orbit as the payload of a single stage rocket.

We need to determine the speed that a satellite needs to have in order to stay in a circular orbit of height h above the earth, as illustrated in Fig. 2.

Screen Shot 2018-11-08 at 3.47.09 PM.png

Fig. 2

By Newton’s inverse square law of attraction, The gravitational pull on satellite with mass m_{s} is

{\gamma \; m_{s} M_{\oplus} \over (R_{\oplus} + h)^2}

where universal gravitational constant \gamma = 6.67 \times 10^{-11}, the earth’s mass M_{\oplus} = 5.9722 \times 10^{24}\; kg, and the earth’s radius R_{\oplus} = 6371\;km.

For a satellite to circle around the earth with a velocity of magnitude v, it must be true that

{\gamma \; m_{s} M_{\oplus} \over (R_{\oplus} + h)^2} = {m_{s} v^2 \over (R_{\oplus}+h) }


v = \sqrt{\gamma \; M_{\oplus} \over (R_{\oplus}+h)}

On a typical orbit, h = 100\;km above earth’s surface,

Screen Shot 2018-11-02 at 10.04.41 AM.png

v = 7.8\;km\cdot s^{-1}

This is far in excess of (4), the value obtained from a single stage rocket.

The implication is that a typical single stage rocket cannot serve as the launching vehicle of satellite orbiting around earth.

We will turn to multi-stage rocket in “Viva Rocketry! Part 2”.



3 thoughts on “Viva Rocketry! Part 1

  1. Pingback: Viva Rocketry! Part 1.1 | Vroom

  2. Pingback: Viva Rocketry! Part 1.2 | Vroom

  3. Pingback: An alternative derivation of ideal rocket’s flight equation (Viva Rocketry! Part 1.3) | Vroom

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