2.1 Tsiolkovsky’s Rocket Equation

Let us derive Tsiolkovsky’s Rocket Equation (see, for example, [2]). We begin with the fundamental principle of conservation of momentum, taking into account the rocket’s decreasing mass over time as it expels fuel. As the rocket ejects fuel backward, it gains an equal amount of forward momentum, generating thrust that propels the rocket forward. This process is consistent with Newton’s third law of motion.

The Tsiolkovsky Rocket Equation provides a deeper understanding by expressing a mathematical relationship among the rocket’s velocity change, the exhaust velocity, and the ratio of the rocket’s initial to final mass. The initial momentum of the rocket, \(p_i\), is expressed as \begin{equation}\tag{2} p_i = mv, \end{equation} where \(m=m(t)\) is the mass of the rocket and \(v=v(t)\) is the rocket's velocity at given time \(t\).

Figure 1: A rocket expels part of its mass to propel itself. The gas is expelled at velocity \(-u\) relative to the rocket, and at velocity \(v-u\) as observed from the laboratory frame.

The final momentum of the combined rocket–gas system can be expressed as, see Fig. 1, \begin{equation}\tag{3} p_f = (m - m_g)(v + dv) + m_g(v - u), \end{equation} where \(m_g\) is the mass of the gas expelled by the rocket, \(dv\) is the increase in the rocket’s velocity, and \(u\) is the velocity of the gas relative to the rocket.

Since momentum is conserved the initial and final momenta are set equal to each other: \begin{equation}\tag{4} mv = (m - m_g)(v + dv) + m_g(v - u), \end{equation} resulting in the following relation: \begin{equation}\tag{5} dv = -u \frac{dm}{m}, \end{equation} where we have replaced the mass of the expelled gas with the change in the rocket’s mass: \(m_g = -dm\).

One key insight from the equation above is that the change in velocity is directly proportional to the exhaust velocity and inversely proportional to the current mass. Moreover, integration plays a crucial role in deriving the total change in velocity, as this relationship must be integrated over the entire process of fuel consumption – from the initial to the final mass. \begin{equation}\tag{6} \int_{v_0}^{v} dv = -u \int_{m_0}^{m} \frac{dm}{m} \end{equation} Evaluating both integrals yields the following equation: \begin{equation}\tag{7} \Delta v = -u \ln \left( \frac{m}{m_0} \right). \end{equation}

Rearranging the equation to express \(m\) in terms of \(m_0\), \(\Delta v\), and \(u\), we obtain: \begin{equation}\tag{8} m = m_0 e^{-\frac{\Delta v}{u}}. \end{equation} This final form highlights how the rocket’s mass decreases exponentially as it gains velocity.

2.2 The Effects of Gravity on the Rocket Equation

Let us consider the motion of a rocket in a gravitational field. Incorporating gravity into the rocket equation alters the relationship between mass, thrust, and velocity. Since gravity continuously acts on the rocket, it introduces a force that must be countered by thrust. This results in time-dependent factors that significantly influence the rocket’s trajectory and fuel consumption. When operating in a gravitational field with acceleration \(g\), the rocket equation (7) becomes: \begin{equation}\tag{9} \Delta v = -u \ln \left( \frac{m}{m_0} \right) - gt. \end{equation} This equation illustrates how the rocket’s velocity is affected by both the reduction in mass due to fuel consumption and the constant downward force of gravity over time. Here, \(g\) is the acceleration due to gravity (approximately \(9.81\, \mathrm{m/s^2}\) on Earth), and \(t\) is the time elapsed since launch. The negative sign in the \(-gt\) term indicates that gravity opposes the rocket’s thrust, thereby reducing its net velocity over time.

Building upon the effects of mass loss and gravitational force on a rocket’s motion, it is important to consider the role of time in these dynamics. By expressing mass as a function of time, we have: \begin{equation}\tag{10} m(t) = m_0 - |\dot{m}|t, \end{equation} where \(\dot{m}\) is the rate at which the rocket burns fuel. Substituting this expression into equation (8) allows us to integrate over time and derive the velocity as a function of time: \begin{equation}\tag{11} v(t) = -u \ln\left(1 - \frac{|\dot{m}|}{m_0}t\right) - gt. \end{equation}

To extend these calculations, we introduce the concept of thrust, defined as \(f = |\dot{m}| u\), where \(f\) is the thrust force produced by the rocket’s engines. This expression shows how rockets generate the force needed to overcome gravity and achieve targeted acceleration.

Using this, \(f\) can be substituted into equation (10) to yield: \begin{equation}\tag{12} v(t) = -u \ln\left(1 - \frac{f t}{m_0 u}\right) - gt. \end{equation}

This modified equation balances the upward acceleration generated by propulsion with the downward pull of gravity, offering a clearer picture of the rocket’s changing velocity over time as it burns fuel: \begin{equation}\tag{13} \frac{v(\tau)}{u} = -\ln\left(1 - \frac{f}{m_0 g} \cdot \frac{g t}{u} \right) - \frac{g t}{u} = -\ln\left(1 - n \tau \right) - \tau, \end{equation} where the thrust-to-weight ratio (\(n = {f}/{m_0 g}\)) measures the rocket’s thrust relative to its weight, determining its ability to overcome gravity; the dimensionless time ratio (\(\tau = {g t}/{u}\)) scales time by gravitational acceleration and exhaust velocity, enabling a simplified analysis of rocket motion. Figure 2 visually represents this equation, showing the rocket’s velocity as a function of time in a constant gravitational field. Remarkably, at \(\tau = 1/n\), the rocket exhausts all its fuel and, according to the idealized model, reaches infinite velocity.

Figure 2: Velocity of a rocket as a function of time in a constant gravitational field \(g\), under constant thrust characterized by the thrust-to-weight ratio \(n\), for values n = 2, 3, 4, and 5. The vertical dashed lines indicate the moments when the rocket exhausts its fuel, corresponding to the condition \(m = m_0 - |\dot{m}|t\). Higher values of \(n\) lead to faster acceleration and earlier fuel depletion.

2.3 Derivation of Rocket Altitude as a Function of Time

When a rocket launches, its height over time is influenced by the engine’s thrust propelling it upward, Earth’s gravity pulling it downward, and the continuous loss of fuel, which reduces the rocket’s mass as it ascends. Because the rocket’s mass changes during flight, calculating its exact height at any given moment is not straightforward. Engineers apply the Tsiolkovsky Rocket Equation to determine how the rocket’s velocity evolves as it loses mass. By analyzing this velocity as a function of time, we can estimate the rocket’s altitude. For simplified scenarios, basic physics provides a rough approximation, but real-world missions require more detailed modeling.

From Tsiolkovsky’s Rocket Equation, the velocity at any time \(t\) is given by: \begin{equation}\tag{14} v(t) = -u \ln\left(1 - n \frac{g t}{u}\right) - g t. \end{equation}

To determine height as a function of time, we integrate the velocity: \begin{equation}\tag{15} h(t) = \int v(t)\, dt. \end{equation}

To integrate the logarithmic part of Eq. (14), we use the substitution \(w = 1 - n \tau\), which gives \(dw = -n\, d\tau = -{n g}/{u}\, dt\). This transforms the height integral into: \begin{equation}\tag{16} h(t) = \frac{u^2}{n g} \int \ln(w)\, dw - \frac{g t^2}{2}. \end{equation} Now we integrate Eq. (16) by parts: \begin{equation}\tag{17} h(t) = \frac{u^2}{n g} \left[ w \ln(w) - w \right] \Big|_1^{1 - n \tau} - \frac{g t^2}{2}. \end{equation} Evaluating the integrals yields an expression for \(h(t)\) in terms of \(n\), \(u\), and \(\tau = {g t}/{u}\): \begin{equation}\tag{18} h(t) = \frac{u^2}{n g} \left[ (1 - n \tau) \ln(1 - n \tau) - (1 - n \tau) + 1 \right] - \frac{g t^2}{2}, \end{equation} or equivalently, \begin{equation}\tag{19} h(t) = \frac{u^2}{n g} \left[ (1 - n \tau) \ln(1 - n \tau) + n \tau \right] - \frac{g t^2}{2}. \end{equation}

Expressing height as a function of the dimensionless time \(\tau\), we get: \[ h(\tau) = \frac{u^2}{n g} \left[ (1 - n \tau) \ln(1 - n \tau) + n \tau \right] - \frac{u^2 \tau^2}{2 g} = \] \[ = \frac{u^2}{2 g} \left[ \frac{2}{n} \left[ (1 - n \tau) \ln(1 - n \tau) + n \tau \right] - \tau^2 \right] = \] \begin{equation}\tag{20} = h_0 \left[ \frac{2}{n} \left[ (1 - n \tau) \ln(1 - n \tau) + n \tau \right] - \tau^2 \right], \end{equation} where \(h_0 = {u^2}/{2 g}\) is the characteristic height scale.

Figure 3 illustrates the altitude \(h/h_0\) of a rocket as a function of dimensionless time \(\tau = g t / u\) in a gravitational field. This formulation enables us to calculate a rocket’s altitude while accounting for fuel consumption (mass loss) and the downward pull of gravity. By carefully integrating the effects of thrust, gravity, and changing mass, we obtain a realistic model that describes the rocket’s trajectory. This result enhances our understanding of how these forces interact to shape the rocket’s path during ascent.

Figure 3: Altitude of a rocket $h$ as a function of dimensionless time \(\tau = g t / u\). The curves show the normalized altitude \(h / h_0\) for rockets with different thrust-to-weight ratios n = 2, 3, 4, and 5. The vertical lines indicate the moments when each rocket depletes its fuel. Rockets with higher \(n\) values gain altitude more rapidly but consume their fuel in a shorter amount of time.