Question

Question: A rocket maintains a constant thrust F, giving it an acceleration of g

$\mathbf{(}\mathit{i}\mathbf{.}\mathit{e}\mathbf{.}\mathbf{,}\mathbf{9}\mathbf{.}\mathbf{8}\mathbf{}\mathit{m}\mathbf{/}{\mathit{s}}^{\mathbf{2}}\mathbf{)}$.

(a) If classical physics were valid, how long would it take for the rocket’s speed to reach $\mathbf{0}\mathbf{.}\mathbf{99}\mathit{c}\mathbf{?}$?

(b) Using the result of exercise 117(c), how long would it really take to reach $\mathbf{0}\mathbf{.}\mathbf{99}\mathit{c}\mathbf{?}$?

$\mathit{u}\mathbf{=}\frac{\mathbf{1}}{\sqrt{\mathbf{1}\mathbf{+}{\mathbf{(}\mathbf{F}\mathbf{t}\mathbf{/}\mathbf{m}\mathbf{c}\mathbf{)}}^{\mathbf{2}}}}\frac{\mathbf{F}}{\mathbf{T}}\mathit{t}$

Short answer

1. If classical approximations were valid for high speeds,it would take almost a year for the rocket to reach $\mathbf{0}\mathbf{.}\mathbf{99}\mathit{c}\mathbf{,}$
2. But in reality o.e., relativistic-ally,it would take around seven years.

Step-by-step solution

Determine the time required to reach 0.99 cclassically

(a) Using classical kinematic equations,

$v=u+at$

Let’s consider the rocket was initially at rest, and then due to constant thrust, it traveled at an acceleration of $g.$

$$\begin{gathered} v=gt \\ t=\frac{v}{g} \\ t=\frac{0.99\times 3\times {10}^8{\displaystyle \raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.}}{9.8{\displaystyle \raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s^2$}\right.}} \\ t=3.03\times {10}^{7}s \end{gathered}$$

Therefore, classically,the time required to reach the velocity of is0.96 years or around350 days.

Determine the time required to reach 0.99 crelativistically

(b) Now,relativistic-ally,

$$\begin{gathered} u=\frac{1}{\sqrt{1+{(Ft/mc)}^2}}\frac{F}{T} \\ t=\frac{mu}{F}\sqrt{1+{\left(\frac{Ft}{mc}\right)}^2} \\ {t}^2={\left(\frac{mu}{F}\right)}^2\left(1+{\left(\frac{Ft}{mc}\right)}^2\right) \end{gathered}$$

Here, $F=ma=mg$

$t^2={\left(\frac{u}{g}\right)}^2\left(1+{\left(\frac{gt}{c}\right)}^2\right)$

Rearranging further to get an equation for t ,

$$\begin{gathered} t=\frac{u}{g\sqrt{1-{\displaystyle \raisebox{1ex}{$u^2$}\!\left/ \!\raisebox{-1ex}{$c^2$}\right.}}} \\ =\frac{0.99c}{9.8{\displaystyle \raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s^2$}\right.}\left(\sqrt{1-{(0.99)}^2}\right)} \\ =2.143\times {10}^{8}s \\ =6.795yrs \end{gathered}$$

.

Hence, the time the rocket really takes to reach $0.99c$is$\mathbf{6}\mathbf{.}\mathbf{8}\mathbf{}\mathit{y}\mathit{e}\mathit{a}\mathit{r}\mathit{s}\mathbf{.}$