Chapter 6, Problem 58P

A spaceship at rest relative to a nearby star in interplanetary space has a total mass of 2.50 × 10⁴ kg. Its engines fire at t = 0, steadily burning fuel at. 76.7 kg/s with an exhaust speed of 4.25 × 10³ m/s. Calculate the spaceship’s (a) acceleration at t = 0, (b) mass at t = 125 s, (c) acceleration at t = 125 s, and (d) speed at t = 125 s, relative to the same nearby star.

To determine

The magnitude of acceleration of the spaceship at time equal to zero seconds

Answer to Problem 58P

The acceleration at time zero seconds is $13.04\text{m}\cdot {\text{s}}^{-2}$.

Explanation of Solution

Explanation

Given Info:

Mass of the spaceship is $2.50\times {10}^4\text{kg}$ , rate of fuel burning is $76.7\text{kg}\cdot {\text{s}}^{-1}$ , the exhaust speed is $4.25\times {10}^3\text{m}\cdot {\text{s}}^{-1}$.

Formula to calculate acceleration at time zero seconds is,

$a_0=\frac{T}{M}$ (I)

• T is the thrust on the spaceship
• M is the total mass of the spaceship

Formula to calculate thrust on the rocket is,

$T=\left|v_{\text{e}}\left(\frac{\Delta M}{\Delta t}\right)\right|$ (II)

• $v_e$ is the exhaust velocity
• $\Delta M/\Delta t$ is the rate of change of mass of spaceship or fuel

Rewrite t the equation (I) for acceleration using the expression $\left|v_{\text{e}}\left(\Delta M/\Delta t\right)\right|$ for T

$a_0=\frac{\left|v_{\text{e}}\left(\frac{\Delta M}{\Delta t}\right)\right|}{M}$

Substitute $2.50\times {10}^4\text{kg}$ for M, $76.7\text{kg}\cdot {\text{s}}^{-1}$ for $\Delta M/\Delta t$ and $4.25\times {10}^3\text{m}\cdot {\text{s}}^{-1}$ for $v_e$ in the above equation to calculate $a_o$ ,

$\begin{array}{c}{a}_0=\frac{\left|\left(4.25\times {10}^3\text{m}\cdot {\text{s}}^{-1}\right)\left(76.7\text{kg}\cdot {\text{s}}^{-1}\right)\right|}{2.50\times {10}^4\text{kg}}\\ =\frac{130.4}{10}\\ =13.0\text{m}\cdot {\text{s}}^{-2}\end{array}$

Conclusion:

Therefore the acceleration at time zero seconds is $13.0\text{m}\cdot {\text{s}}^{-2}$

To determine

The mass of the spaceship at time 125 seconds

Answer to Problem 58P

Answer The mass of the spaceship at time 125 seconds is $1.54\times {10}^4\text{kg}$

Explanation of Solution

Explanation

Given Info:

Formula to calculate mass of the space ship at a particular time is,

$M_{\text{f}}=M-\left(\frac{\Delta M}{\Delta t}\right)t$

• $M_{\text{f}}$ is the mass of the space ship at time t
• t is the time

Substitute $2.50\times {10}^4\text{kg}$ for M, $76.7\text{kg}\cdot {\text{s}}^{-1}$ for $\Delta M/\Delta t$ and $125\text{s}$ for t in the above equation to calculate the mass at $125\text{s}$

$\begin{array}{c}{M}_{\text{f}}=\left(2.50\times {10}^4\text{kg}\right)-\left(76.7\text{kg}\cdot {\text{s}}^{-1}\right)\left(125\text{s}\right)\\ =1.54\times {10}^4\text{kg}\end{array}$

Conclusion:

Therefore the mass of the spaceship at time 125 seconds is $1.54\times {10}^4\text{kg}$

To determine

The magnitude of acceleration at 125 seconds

Answer to Problem 58P

Answer The acceleration at time 125 seconds is $21.17\text{m}\cdot {\text{s}}^{-2}$

Explanation of Solution

Explanation

Given Info:

Formula to calculate acceleration of the spaceship is,

$a_{\text{f}}=\frac{\left|v_{\text{e}}\left(\frac{\Delta M}{\Delta t}\right)\right|}{M_{\text{f}}}$

• $a_{\text{f}}$ is the acceleration of the spaceship at time 125 seconds
• $M_{\text{f}}$ is the mass of the space ship at time 125 seconds
• $\Delta M/\Delta t$ is the rate of change of mass of spaceship or fuel

Substitute $76.7\text{kg}\cdot {\text{s}}^{-1}$ for $\Delta M/\Delta t$ , $4.25\times {10}^3\text{m}\cdot {\text{s}}^{-1}$ for $v_e$ and $1.54\times {10}^4\text{kg}$ for $M_{\text{f}}$ in the above equation to calculate acceleration at 125 seconds

$\begin{array}{c}{a}_{\text{f}}=\frac{\left|\left(4.25\times {10}^3\text{m}\cdot {\text{s}}^{-1}\right)\left(76.7\text{kg}\cdot {\text{s}}^{-1}\right)\right|}{\left(1.54\times {10}^4\text{kg}\right)}\\ =\frac{211.7}{10}\\ =21.17\text{m}\cdot {\text{s}}^{-2}\\ \approx 21.2\text{m}\cdot {\text{s}}^{-2}\end{array}$

Conclusion:

Therefore the acceleration at time 125 seconds is $21.2\text{m}\cdot {\text{s}}^{-2}$

To determine

The magnitude speed of the space ship at time 125 seconds

Answer to Problem 58P

The speed of the space ship at 125 seconds is $\text{2}\text{.14}\times {\text{10}}^3\text{m}\cdot {\text{s}}^{-1}$

Explanation of Solution

Given Info:

Acceleration at zero seconds is $13.04\text{m}\cdot {\text{s}}^{-2}$, acceleration at 125 seconds is $21.17\text{m}\cdot {\text{s}}^{-2}$

Formula to calculate average acceleration is,

$a_{av}=\frac{a_o+a_{\text{f}}}{2}$ (III)

• $a_o$ is the acceleration of the spaceship at time 0 seconds
• $a_{\text{f}}$  is the acceleration of the spaceship at time 125 seconds

Formula to calculate average acceleration in terms of speed is,

$a_{\text{av}}=\frac{v_{\text{f}}-v_{\text{i}}}{t}$

But $v_i$ is the initial speed which is zero ,

Rewrite the above equation using the value of $v_i$ to calculate $a_{av}$

$a_{\text{av}}=\frac{v_{\text{f}}}{t}$ (IV)

Equate (III) and (IV) to calculate the speed of rocket at time 125 seconds.

$\frac{a_o+a_{\text{f}}}{2}=\frac{v_{\text{f}}}{t}$

$v_f=\left(\frac{a_o+a_{\text{f}}}{2}\right)t$

Substitute $13.04\text{m}\cdot {\text{s}}^{-2}$ for $a_0$, $21.17\text{m}\cdot {\text{s}}^{-2}$ for $a_{\text{f}}$ and $125\text{s}$ for t in the above equation for speed to calculate $v_{\text{f}}$

$\begin{array}{c}{v}_f=\left(\frac{\left(13.04\text{m}\cdot {\text{s}}^{-2}\right)+\left(21.17\text{m}\cdot {\text{s}}^{-2}\right)}{2}\right)125\text{s}\\ \text{=2}\text{.14}\times {\text{10}}^3\text{m}\cdot {\text{s}}^{-1}\end{array}$

Conclusion:

Therefore the speed of the space ship at 125 seconds is $\text{2}\text{.14}\times {\text{10}}^3\text{m}\cdot {\text{s}}^{-1}$