## Related questions with answers

(a) Neglecting gravitational forces, what force would be re- quired to accelerate a $1200$-metric-ton spaceship from rest to one-tenth the speed of light in $3$ days? In $2$ months? (One metric ton $=1000$ kg.) (b) Assuming that the engines are shut down when this speed is reached, what would be the time required to complete a $5$-light-month journey for each of these two cases? (Use $1$ month $=30$ days.)

Solution

Verified$\text{\textcolor{#c34632}{(a.)}}$ Calculate the force applied to accelerate a spaceship from rest to one-tenth the speed of light. The net force formula is given by

$\begin{align*} F & =ma \hspace{10mm} \text{where: } a = \dfrac{v}{t}\\ \implies F & =\dfrac{m\cdot v}{t} \end{align*}$

From Appendix B, speed of light $c = 3.00 \cdot 10^8\mathrm{~\dfrac{m}{s}}$, Then $v=0.1c$

Entering known values, we obtain $F$ as

In 3 days :

$\begin{align*} F & = \dfrac{\left(1200\text{ metric ton}\cdot \dfrac{1000\text{ kg}}{1\text{ metric ton}} \right)\left(0.1 \right)\left( 3.00 \cdot 10^8\mathrm{~\dfrac{m}{s}} \right)}{3\text{ d} \cdot \dfrac{86400\text{ s}}{1\text{ d}}}\\ & = 1.4 \cdot 10^8\mathrm{~\dfrac{kg \cdot m}{s^2}} \approx 1.4 \cdot 10^8\text{ N} \end{align*}$

$\boxed{\therefore F = 1.4 \cdot 10^8\text{ N in 3 days} }$

In 2 months :

$\begin{align*} F & = \dfrac{\left(1200\text{ metric ton}\cdot \dfrac{1000\text{ kg}}{1\text{ metric ton}} \right)\left(0.1 \right)\left( 3.00 \cdot 10^8\mathrm{~\dfrac{m}{s}} \right)}{60\text{ d} \cdot \dfrac{86400\text{ s}}{1\text{ d}}}\\ & = 6.9 \cdot 10^6\mathrm{~\dfrac{kg \cdot m}{s^2}} \approx 6.9 \cdot 10^6\text{ N} \end{align*}$

$\boxed{\therefore F = 6.9 \cdot 10^6\text{ N in 2 months} }$

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