Chapter 10, Problem 83PQ

To determine

Show that for an open system such as a rocket, the second rocket equation is $M_R{\left({\overrightarrow{a}}_R\right)}_{\oplus }={{\displaystyle \sum \overrightarrow{F}}}_{\text{ext}}+{\overrightarrow{F}}_{\text{thrust}}$.

Answer to Problem 83PQ

It is shown that for an open system the second rocket equation is $M_R{\left({\overrightarrow{a}}_R\right)}_{\oplus }={{\displaystyle \sum \overrightarrow{F}}}_{\text{ext}}+{\overrightarrow{F}}_{\text{thrust}}$.

Explanation of Solution

Consider the closed rocket exhausted fuel system and imagine an external force acting on the system.

Write the expression for the Newton’s second law.

    ${\displaystyle \sum {\overrightarrow{F}}_{\text{ext}}=\frac{d{\overrightarrow{p}}_{\text{tot}}}{dt}}$                                                                                                        (I)

Here, ${\overrightarrow{F}}_{\text{ext}}$ is the external force, ${\overrightarrow{p}}_{\text{tot}}$ is the total momentum.

Write the expression for the total momentum.

    ${\overrightarrow{p}}_{\text{tot}}={\overrightarrow{p}}_{R,\oplus }+{\overrightarrow{p}}_{E,\oplus }$                                                                                                 (II)

Here, ${\overrightarrow{p}}_{R,\oplus }$ is the momentum of the rocket-exhaust system, and ${\overrightarrow{p}}_{E,\oplus }$ is the momentum of the Earth-exhaust system.

Use equation (II) in equation (I),

    ${\displaystyle \sum {\overrightarrow{F}}_{\text{ext}}=\frac{d{\overrightarrow{p}}_{R,\oplus }}{dt}}+\frac{d{\overrightarrow{p}}_{E,\oplus }}{dt}$                                                                                     (III)

The change in momentum of the exhaust is the same in both frame of reference of the Earth and the rocket so that,

    $d{\overrightarrow{p}}_{E,\oplus }=d{\overrightarrow{p}}_{E,R}$                                                                                                       (IV)

Use equation (IV) in equation (III),

    ${\displaystyle \sum {\overrightarrow{F}}_{\text{ext}}=\frac{d{\overrightarrow{p}}_{R,\oplus }}{dt}}+\frac{d{\overrightarrow{p}}_{E,R}}{dt}$                                                                                      (V)

Write the expression for the momentum of an moving object.

    $p=mv$                                                                                                                 (VI)

Here, $p$ is the momentum of the object, $m$ is the mass of the object, $v$ is the speed of the object.

When a small amount of mass $dm$ is ejected in the exhaust, the rockets velocity changes by a small amount $d{\overrightarrow{v}}_{R,\oplus }$, so that use equation (VI) in equation (V),

    ${\displaystyle \sum {\overrightarrow{F}}_{\text{ext}}=\frac{M_{R}d{\overrightarrow{v}}_{R,\oplus }}{dt}}+\frac{{\left({\overrightarrow{v}}_E\right)}_{R}dm}{dt}$                                                                      (VII)

The mass in the exhaust leaves the system so that the mass exhausted will equal the mass lost by the rocket so that,

  $dm=-d{M}_R$                                                                                                     (VIII)

Use equation (VIII) in equation (VII),

    ${\displaystyle \sum {\overrightarrow{F}}_{\text{ext}}=\frac{M_{R}d{\overrightarrow{v}}_{R,\oplus }}{dt}}-\frac{{\left({\overrightarrow{v}}_E\right)}_{R}d{M}_R}{dt}$                                                                     (IX)

Write the expression for the force on the rocket in the Earths frame of reference.

    $F_{R,\oplus }=M_R{\overrightarrow{a}}_{R,\oplus }$                                                                                                      (X)

The acceleration on the rocket in the Earths frame of reference is given by,

    ${\overrightarrow{a}}_{R,\oplus }=\frac{d{\overrightarrow{v}}_{R,\oplus }}{dt}$                                                                                                        (XI)

Use equation (XI) in (X),

    $F_{R,\oplus }=M_R\frac{d{\overrightarrow{v}}_{R,\oplus }}{dt}$                                                                                               (XII)

Write the expression for the force exerted by the fuel which is known as thrust,

    ${\overrightarrow{F}}_{\text{thrust}}={\left({\overrightarrow{v}}_E\right)}_R\frac{d{M}_R}{dt}$                                                                                         (XIII)

Use equation (XII) in equation (X), and use (XIII) in equation (IX)

    ${\displaystyle \sum F_{\text{ext}}=}{M}_R{\overrightarrow{a}}_{R,\oplus }-{\overrightarrow{F}}_{\text{thrust}}$

Rearrange the above equation,

    $m_R{\left({\overrightarrow{a}}_R\right)}_{\oplus }={\displaystyle \sum {\overrightarrow{F}}_{\text{ext}}+{\overrightarrow{F}}_{\text{thrust}}}$                                                                            (XIV)

Conclusion:

Therefore, it is shown from equation (XIV) that for an open system the second rocket equation is $M_R{\left({\overrightarrow{a}}_R\right)}_{\oplus }={{\displaystyle \sum \overrightarrow{F}}}_{\text{ext}}+{\overrightarrow{F}}_{\text{thrust}}$.