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(a) Neglecting gravitational forces, what force would be re- quired to accelerate a 12001200-metric-ton spaceship from rest to one-tenth the speed of light in 33 days? In 22 months? (One metric ton =1000=1000 kg.) (b) Assuming that the engines are shut down when this speed is reached, what would be the time required to complete a 55-light-month journey for each of these two cases? (Use 11 month =30=30 days.)

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(a.)\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

F=mawhere: a=vt    F=mvt\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=3.00108 msc = 3.00 \cdot 10^8\mathrm{~\dfrac{m}{s}}, Then v=0.1cv=0.1c

Entering known values, we obtain FF as

In 3 days :

F=(1200 metric ton1000 kg1 metric ton)(0.1)(3.00108 ms)3 d86400 s1 d=1.4108 kgms21.4108 N\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*}

F=1.4108 N in 3 days\boxed{\therefore F = 1.4 \cdot 10^8\text{ N in 3 days} }

In 2 months :

F=(1200 metric ton1000 kg1 metric ton)(0.1)(3.00108 ms)60 d86400 s1 d=6.9106 kgms26.9106 N\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*}

F=6.9106 N in 2 months\boxed{\therefore F = 6.9 \cdot 10^6\text{ N in 2 months} }

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