(a) Neglecting gravitational forces, what force would be required 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 \(=1000 \mathrm{~kg} .)(b)\) 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.)

Convert the spaceship's mass from metric-ton to kg: \(1200 \times 1000 = 1200000 \mathrm{~kg}\). Convert the speed of light (\(3 \times 10^8 \mathrm{~m/s}\)) to one-tenth of its value: \(0.1 \times 3 \times 10^8 = 3 \times 10^7 \mathrm{~m/s}\). Convert time from days and months to seconds: \(3 \mathrm{~days}=3 \times 24 \times 60 \times 60=259200 \mathrm{~s}\) and \(2 \mathrm{~months}=2 \times 30 \times 24 \times 60 \times 60=5184000 \mathrm{~s}\).

Use the formula for acceleration: acceleration = final velocity/ time. Then use the force formula: force = mass x acceleration. Substituting the calculated acceleration, for 3 days, the force is \(1200000 \times \frac{3 \times 10^7}{259200} = 1.385 \times 10^{14} \mathrm{~N}\). And for 2 months, the force is \(1200000 \times \frac{3 \times 10^7}{5184000} = 6.9 \times 10^{12} \mathrm{~N}\).

Now using the velocity calculated in the first part and the given distance of 5 light-months, calculate the time required for each case. 1 light-month equals to the distance light travels in one month, which is \(3 \times 10^8 \times 30 \times 24 \times 60 \times 60 =2.592 \times 10^{15} \) m. So, 5 light-months is \(5 × 2.592 \times 10^{15} = 1.296 \times 10^{16} m\). Use time=distance/velocity formula. For 3 days case, time = \(1.296 \times 10^{16}/(3 \times 10^7) = 4.32 \times 10^8 \mathrm{~s} \approx 13.7 \mathrm{~years}\). And for 2 months case, time = \(1.296 \times 10^{16}/(3 \times 10^7) = 4.32 \times 10^8 \mathrm{~s} \approx 13.7 \mathrm{~years}\). So the time required to complete the journey is the same in both cases.