Skip to main content

Section 7.9 Mass Varying System

There are many systems of interest in which the mass of the system is not constant in time. For instance, a rocket loses fuel when it is in the accelerating phase, a squid propels itself forward by ejecting water from back, etc.

In these examples object, rocket or squid, accelerates even in the absence of an external force, violating \(m\vec a = \vec F\text{.}\) The reason is that \(m\vec a = \vec F\) is not correct if \(m\) changes with time. We need general form of second law of motion in terms of momentum, either in its differential form or integral form.

\begin{equation} \vec F = \frac{\Delta \vec p}{\Delta t} \rightarrow \dfrac{d\vec p}{dt}, \ \ \leftrightarrow\ \ \vec J = \int \vec F dt = \Delta \vec p .\tag{7.9.1} \end{equation}

Besides needing to work with \(\vec J = \Delta \vec p\text{,}\) there is also a question of what is included in the system and what is not. A good way to proceed is to look at how momentum changes in an infinitesimal interval of time \(\Delta t\text{.}\) We will work out examples to show this method below.

Subsection 7.9.1 Motion of a Rocket in Free Space

Consider a rocket far away from planets and stars so that we can focus exclusively on effects of changing mass. Let \(m(t)\) be the mass of the rocket and \(v_x(t)\) be its \(x\)-velocity at instant \(t\) with respect to a fixed frame as shown in Figure 7.9.1.

Figure 7.9.1.

During interval \(t\) to \(t + \Delta t\text{,}\) burnt fuel of mass \(\Delta m\) is ejected from back. Let us assume that the ejected gas has speed \(u\) with respect to the rocket at instant \(t\text{.}\) The remainder of the rocket has mass \(m(t)-\Delta m\) and has \(x\)-velocity \(v_x + \Delta v_x\) with respect to the fixed frame.

The interval between \(t\) to \(t+\Delta t\) is just like one-dimensional explosion problems you have studied. To apply conservation of momentum with respect to the fixed frame, we will need to express all velocities with respect to the fixed frame. The \(x\)-velocity of the ejected fueld with respect to the fixed frame will be \(v_x -u\text{.}\) The \(x\)-equation of conservation of momentum will be

\begin{equation*} m v_x = \Delta m ( v_x -u ) + ( m - \Delta m ) (v_x + \Delta v_x). \end{equation*}

Note that \(|\Delta m \Delta v_x| \ll |u\Delta m| \text{ or } | m\Delta v_x| \text{,}\) we drop \(\Delta m \Delta v_x\text{.}\) This gives the following relation.

\begin{equation*} m \Delta v_x = u \Delta m. \end{equation*}

Dividing by \(\Delta\) we obtain the replacement of \(F = ma\) for a rocket, which is sometimes called rocket equation.

\begin{equation} m \frac{\Delta v_x}{\Delta t} = u \frac{\Delta m}{\Delta t}.\label{eq-rocket-equation-non-calculus}\tag{7.9.2} \end{equation}

Left side is \(m a\) and right side is the force of thrust provided by ejecting burnt fuel. This equation can be solved by calculus to give the following expression for the final velocity \(v_f\text{,}\) when mass of the rocket has gone from \(m_i\) to \(m_f\text{.}\)

\begin{equation} v_{x,f} = v_{x,i} + u \ln\frac{m_i}{m_f}.\tag{7.9.3} \end{equation}

Subsection 7.9.2 (Calculus) Motion of a Rocket in Free Space

When we take infinitesimal time limit in Eq. (7.9.2), this equation will become a differential equation. But, we need to be careful when we convert \(\Delta m/\Delta t\) to \(dm/dt\) since \(m\) of the rocket is decreasing and \(dm/dt\) will be negative while \(\Delta m/\Delta t\) was positive. Hence, we will compensate by including a minus sign.

\begin{equation} m \frac{d v_x}{d t} =- u \frac{d m}{d t}.\label{eq-rocket-equation-non-calculus-to-calculus}\tag{7.9.4} \end{equation}

This can be written as

\begin{equation*} \dfrac{dv_x}{u} = -\dfrac{dm}{m}. \end{equation*}

Integrating this on the left from \(v_x = v_0 \) to \(v_x = v_x(t) \) and on the right from \(m = M \) to \(m = m(t) \text{,}\) we obtain the following for the final velocity of the rocket after rearranging terms.

\begin{equation*} v_x(t) - v_0 = u\, \ln \left( \dfrac{M}{m(t)} \right). \end{equation*}

Subsection 7.9.3 Example of Mass Accretion

As an example of momentum considerations when mass is accumulating, let us consider a spaceship moving in space where it is struck by a continuous stream of particles which stick to the ship.

Figure 7.9.2.

Let \(m(t)\) and \(\vec v(t)\) be the mass and velocity of the spaceship at instant \(t\text{.}\) Let particles move with velocity \(\vec u\) and add a mass \(\Delta m\) to the spaceship in duration from \(t\) to \(t + \Delta t\text{.}\) Let \(\vec v' = \vec v + \Delta \vec v\) be the velocity of the spacecraft at \(t+\Delta t\text{.}\) Then, change in momentum in this duration will be

\begin{equation*} \Delta \vec p = (m+\Delta m)\, \vec v' - (m\, \vec v + \Delta m\, \vec u). \end{equation*}

Simplifying we have

\begin{equation*} \Delta \vec p = m\, \Delta \vec v + \Delta m\, ( \vec v - \vec u ) + \Delta m \Delta \vec v. \end{equation*}

Dividing by \(\Delta t\) we get

\begin{equation*} \frac{\Delta \vec p}{\Delta t} = m\, \frac{\Delta \vec v}{\Delta t} + \frac{\Delta m}{\Delta t}\, ( \vec v - \vec u ) + \Delta m \frac{\Delta \vec v}{\Delta t}. \end{equation*}

(Calculus Part) Taking \(\Delta t \rightarrow 0\) limit will mean \(\Delta m \rightarrow 0\) and \(\Delta v\rightarrow 0\) as well. In this limit we see that the last term drops out and the other terms turn into derivatives.

\begin{equation*} \frac{d\vec p}{dt} = m\, \frac{d \vec v}{dt} + \frac{d m}{d t}\, ( \vec v - \vec u ). \end{equation*}

Let \(\vec F\) be the net external force on the spaceship. By, \(\vec F = d\vec p/dt\text{,}\) we must have

\begin{equation} \vec F = m\, \frac{d \vec v}{d t} + \frac{d m}{d t}\, ( \vec v - \vec u ).\label{eq-spaceship-accretion-calculus}\tag{7.9.5} \end{equation}

In the absence of any force \(\vec F=0\text{,}\) the velocity of the spaceship will change according to

\begin{equation*} m\, \frac{d \vec v}{d t} = - \frac{d m}{d t}\, ( \vec v - \vec u ), \end{equation*}

which is usullay analyzed in component form, e.g., for \(x\)-component we will get the following (and similarly for other components).

\begin{equation} \frac{dv_x}{ v_x - u_x} = - \frac{dm}{m}.\tag{7.9.6} \end{equation}

Equation (7.9.5) also says that if you wanted the velocity of the spaceship steady at \(\vec v\text{,}\) you will need a force, e.g., by rockets on the ship, of the magnitude and direction given by

\begin{equation*} \vec F = \frac{d m}{d t}\, ( \vec v - \vec u ). \end{equation*}

Of course, if the striking particle had the same velocity as the velocity of the ship, i.e., if \(\vec u = \vec v\text{,}\) you do not need any force to maintain steady velocity. Of course, you will need \(|\vec u|\) to be a little greater than \(|\vec v|\) for the particles to at least catch up to the ship.

Subsection 7.9.4 Momentum Transport by Massless Radiation

Light particles, called photons, do not have mass but carry momentum and energy. The formula for the magnitude of momentum for a photon in light of wavelength \(\lambda\) is given by

\begin{equation} p = \frac{h}{\lambda},\tag{7.9.7} \end{equation}

where \(h\) is called Planck's constant and has the following value.

\begin{equation*} h = 6.626 \times 10^{-34}\, \text{m}^2\text{kg/s}. \end{equation*}

An example of application of momentum from photons, consider a payload that uses large solar sail to collect momentum from photons from the star for its navigation. Suppose area of the sail is \(A\text{.}\) Let intensity of radiation from the star be such that \(N\) photons of wavelength \(\lambda\) per unit volume are traveling in the direction of the payload.

Figure 7.9.3.

Since speed of light in space will be equal to \(c=3\times 10^{8}\,\text{m/s}\text{,}\) the number of photons striking the collector in time \(\Delta t\) will be

\begin{equation*} \Delta n = N A c. \end{equation*}

For the sake of simplicity, let us suppose that momentum of photon flips in direction upon reflection. That will mean that each photon will increase the momentum of the payload by twice its own momentum.

\begin{equation*} \Delta p_\text{payload} = 2 N A c h /\lambda \Delta t. \end{equation*}

Now, let us look at some numbers to get a sense of the size of the effect. Consider visible light photon of wavelength \(0.5\,\mu\text{m}\text{,}\) area \(A=50\text{ m}^2\text{,}\) number density \(N=10^{12}\text{ m}^{-3}\text{.}\) This will impose follwing force on the payload.

\begin{align*} F \amp = \Delta p_\text{payload}/\Delta t\\ \amp = 2\times 10^{12}\times 50\times3\times 10^8\times 6.626\times10^{-34}/(0.5\times 10^{-6})\\ \amp = 4.0\times 10^{-5}\text{ N}. \end{align*}

This is not too small to ignore.

A spacecraft is moving at a constant speed of \(100\text{ m/s}\) in gravity-free space along a straight path. The crew decided to accelerate the spacecraft. When the pilot turns on one of the thrusters, it ejects burnt fuel in the back at the rate of \(10.0 \text{ kg/s}\) at a speed of \(1,000\text{ m/s}\) relative to the craft. At the moment the thruster was turned on, the mass of the spacecraft with crew plus the unburnt fuel was \(5.0\times 10^{5}\text{ kg}\text{.}\)

(a) What is the thrust on the spacecraft?

(b) Express the acceleration of the craft as a function of time, \(t \text{,}\) and evaluate the acceleration at the following values \(t \) : (i) \(0 \text{,}\) (ii) \(10\text{ s} \text{,}\) (iii) \(100\text{ s}\text{,}\) where \(t=0 \) is the instant when the thruster was turned on:

(c) Find the speed of the spacecraft as a function of time, and evaluate the speed at the following values \(t \) : (i) \(0 \text{,}\) (ii) \(100\text{ s} \text{,}\) (iii) \(10000\text{ s}\text{.}\)


Use the formulas.


(a) \(10,000\text{ N} \text{,}\) (b) \(0.02\text{ m/s}^2\text{,}\) \(0.02004\text{ m/s}^2\text{,}\) \(0.025\text{ m/s}^2\text{,}\) (c) Not provided.

Solution 1 (a)

(a) The thrust \(T \) on the spacecraft comes from the rate at which mass is lost and the velocity of that ejected mass.

\begin{equation*} T = \alpha u = 10.0 \text{ kg/s} \times 1,000\text{ m/s} = 10,000\text{ N}. \end{equation*}

The direction of the thrust is towards forward of the spacecraft since the burnt fuel is being ejected in the backward direction.

Solution 2 (b)

(b) Using the full Newton's equation equation with no external force:

\begin{equation*} m\vec a = \vec T, \end{equation*}

where mass \(m \) is decreasing with time with the following value at instant \(t \text{.}\)

\begin{equation*} m = m_0 - \alpha\, t, \end{equation*}


\begin{equation*} m_0 = 5.0\times 10^{5}\text{ kg} ,\ \ \alpha = 10.0 \text{ kg/s}. \end{equation*}

Let \(x \) axis be pointed in the forward direction. Then, we have the following for the \(x\) component of acceleration at instant \(t\text{.}\)

\begin{equation*} a_x = \dfrac{T_x}{m} = \dfrac{T}{m_0 - \alpha\, t}. \end{equation*}

Therefore, magnitude \(a \) of acceleration is

\begin{equation*} a(t) = \dfrac{10,000}{5\times 10^5 - 10\, t}. \end{equation*}


\begin{equation*} a(0) = \dfrac{10,000}{5\times 10^5 } = 0.02\text{ m/s}^2 \end{equation*}


\begin{equation*} a(100\text{ s}) = \dfrac{10,000}{5\times 10^5 - 1\times 10^3} = 0.02004\text{ m/s}^2 \end{equation*}


\begin{equation*} a(10^4\text{ s}) = \dfrac{10,000}{5\times 10^5 - 1\times 10^5} = 0.025\text{ m/s}^2 \end{equation*}
Solution 3 (c)

(c) Use the following equation to find the values of speed.

\begin{equation*} v = v_0 + u \ln\left( \dfrac{m_0}{m_0-\alpha t}\right). \end{equation*}

Rocket motion near the surface of the Earth is subject to an approximately constant gravitational force of the Earth.

Consider a rocket of initial total mass \(M_0 \) with \(M_r \) the mass of the rocket without the fuel. The rocket is rising straight up near Earth with the fuel being ejected at a constant speed \(u\) with respect to the rocket and at a constant rate \(\alpha \text{ kg/s}\text{.}\)

Prove that the vertical component of the velocity at time \(t \) will be

\begin{equation*} v = v_0 - gt + u\ln\left(\frac{M_0}{M_0-\alpha t}\right) \end{equation*}

Set up equation similar to the example with no external force.


Already given in the statement.


We will work with positive \(y\) axis pointed up. Let \(v_y \) denote \(y \) component of the velocity of the rocket at instant \(t \) and let \(\Delta v_y \) be the change in it during an interval between \(t \) and \(t + \Delta t \text{.}\) Notice also that velocity of the ejected fuel with respect to the inertial frame will be \(v_y - u\text{.}\)

Therefore, the change in momentum of the rocket plus burnt fuel combined system along the \(y \) axis is

\begin{align*} \Delta p_y \amp = \left[ \Delta m ( v_y - u ) + (m - \Delta m) (v_y + \Delta v_y ) \right] - m v_y,\\ \amp \rightarrow - u \Delta m + m \Delta v_y \text{ (as } \Delta t \rightarrow 0\text{.)} \end{align*}

The impulse of the external force of gravity on the rocket during the intreval is

\begin{equation*} J_y = - m(t) g \Delta t. \end{equation*}

Using the momentum-impulse relation we get

\begin{equation*} m \Delta v_y - u \Delta m = - m g \Delta t. \end{equation*}

Divide both sides by \(m\text{.}\) Now, taking the infinitesimal limit, the change \(\Delta v_x \) is replaced by \(dv_x \) and the change in mass \(\Delta m \) is replaced by \(-dm \) since \(dm \) is negative. We also write \(dm/m \) as \(d\ln m\text{.}\)

\begin{equation*} d v_y + u\, d \ln m = - g\, dt. \end{equation*}

Integrating from \(t = 0\) to \(t = t\) we find

\begin{equation*} v_y(t) - v_y(0) + u\,\ln\left(\frac{M_0-\alpha t}{M_0}\right) = - gt. \end{equation*}

In this equation we can write \(v_y(t) \) as \(v\) and \(v_y(0) \) as \(v_0\text{,}\) and rerarrange to get the desired result.

\begin{equation*} v = v_0 - gt + u\,\ln\left(\frac{M_0}{M_0-\alpha t}\right). \end{equation*}

A cart of mass \(M\) is moving at a constant velocity \(v\) as shown. When the cart enters the space below a large funnel, grain starts to fill the cart at constant mass rate, \(\Delta m/\Delta t =\alpha\text{.}\)

Figure 7.9.7.

In order to keep the track moving at constant velocity, you need to constantly apply a force \(F\text{.}\) Find an expression of this force.


Use \(\vec F\Delta t = \Delta \vec p\) in a duration \(\Delta t\text{.}\)


\(\alpha v\text{.}\)


Consider a interval from \(t\) to \(t+\Delta t\) when the mass of grain and cart goes from \(m\) to \(m + \alpha \Delta t\text{.}\) Let \(x\)-axis be pointed in the direction of constant velocity. Using \(J_x = \Delta p_x\text{,}\) we will get

\begin{equation*} F\Delta t = (m + \alpha \Delta t)\, v - m v= \alpha \Delta t)\, v. \end{equation*}

Therefore, \(F = \alpha v\text{.}\)