The infrastructure of the network includes terminal stations located on the outermost planets of each route, represented as squares on the map. Key asteroid belts where the inner and outer routes intersect serve as transfer hubs, depicted as diamonds. Additionally, there are smaller waystations positioned between terminal stations and transfer hubs, marked as dots on the map. Spacecraft within this network operate exclusively on either the inner or outer route and never switch between the two. Travel times between adjacent waystations vary significantly between the inner and outer routes: the average travel time on the inner route is approximately 30 Earth hours, while on the outer route, it averages around 72 Earth hours. At transfer hubs, spacecraft typically spend about 6 Earth hours for operations such as cargo and passenger transfers, as well as refueling.
If a spacecraft departs from Earth with maximum fuel capacity and travels to Mars along the shortest possible route, how much fuel (in metric tons) will be consumed approximately?
- 2
- 3
- 4
- 5
The Correct Option is B
Solution and Explanation
Fuel consumption rate: 1 metric ton per 0.5 AU
Total distance: 1.5 AU
Total fuel consumption:
\(\frac{1 \text{ metric ton}}{0.5 \text{ AU}} = \frac{X \text{ metric tons}}{1.5 \text{ AU}}\)
\(X = \left( \frac{1 \text{ metric ton}}{0.5 \text{ AU}} \right) \times 1.5 \text{ AU} = 3 \text{ metric tons}\)
Considering the average distance between waystations and standard fuel consumption rates, calculate the maximum number of waystations a standard spacecraft can traverse on a single fuel tank on the outer route.
- 15
- 16
- 17
- 18
The Correct Option is B
Solution and Explanation
Total distance the spacecraft can travel with a full tank:
Fuel capacity: 100 metric tons
Consumption rate: 1 metric ton per 0.5 AU
\(\text{Total distance} = \text{Fuel capacity} \times \text{Distance per metric ton}\)
\(\text{Total distance} = 100 \text{ metric tons} \times \frac{0.5 \text{ AU}}{\text{metric ton}} = 50 \text{ AU}\)
Number of waystations:
Distance between waystations on the outer route: 3 AU
\(\text{Number of waystations} = \frac{\text{Total distance}}{\text{Distance between waystations}}\)
\(\text{Number of waystations} = \frac{50 \text{ AU}}{3 \text{ AU/waystation}} \approx 16.67 \text{ waystations}\)
Since a spacecraft can't traverse a fraction of a waystation, round down to the nearest whole number. So answer is 16
If a new type of spacecraft is introduced with a fuel efficiency of 1.2 times that of the standard spacecraft, and the distance between waystations on the inner route is reduced by 20%, calculate the percentage increase in the number of waystations a spacecraft can cover on a single fuel tank on the inner route compared to the standard spacecraft.
- 29.99%
- 76.78.%
- 25.81%
- 81.82%
The Correct Option is D
Solution and Explanation
New distance between waystations on the inner route.
Original distance: 1.5 AU
Reduction: 20%
New distance: \(1.5 \text{ AU} \times (1 - 0.2) = 1.2 \text{ AU}\)
New fuel consumption rate for the new spacecraft.
Standard fuel consumption rate: 1 metric ton per 0.5 AU
New spacecraft fuel efficiency: 1.2 times the standard
New fuel consumption rate: \(\frac{1 \text{ metric ton}}{0.5 \text{ AU} \times 1.2} = \frac{1}{0.6} \text{ metric ton per AU} = \frac{5}{3} \text{ metric tons per AU}\)
Number of waystations for both spacecraft.
Standard spacecraft:
Fuel capacity: 100 metric tons
Fuel consumption rate: 1 metric ton per 0.5 AU or 2 metric tons per AU
\(\text{Number of waystations} = \frac{\left(\frac{\text{Fuel capacity}}{\text{Fuel consumption rate}}\right)}{\text{Distance per waystation}}\)
\(\text{Number of waystations} = \frac{\left(\frac{100 \text{ metric tons}}{2 \text{ metric tons/AU}}\right)}{1.5 \text{ AU}}\)
\(\text{Number of waystations} = \frac{50}{1.5} = 33.33 \text{ waystations (rounded down to 33 waystations)}\)
New spacecraft:
Fuel capacity: 100 metric tons
\(\text{Fuel consumption rate} = \frac{5}{3} \text{ metric tons per AU}\)
\(\text{Number of waystations} = \frac{\left( \frac{\text{Fuel capacity}}{\text{Fuel consumption rate}} \right)}{\text{Distance per waystation}}\)
\(\text{Number of waystations} = \frac{\left( \frac{100 \text{ metric tons}}{\frac{5}{3} \text{ metric tons/AU}} \right)}{1.2 \text{ AU}} = 60 \text{ waystations}\)
Percentage increase.
\(\text{Percentage increase} = \left( \frac{\left( \text{New number of waystations} - \text{Old number of waystations} \right)}{\text{Old number of waystations}} \right) \times 100 = \left( \frac{60 - 33}{33} \right) \times 100 = \left( \frac{27}{33} \right) \times 100 \approx 81.82\%\)
Assuming a constant spacecraft speed of 50,000 km/h, calculate the approximate travel time (in Earth hours) for a spacecraft to journey from Earth to Neptune via the shortest route, considering the average distances between waystations and transfer hubs. (Note: 1 AU is approximately 150 million kilometers)
- Less than 10000 hours
- Between 10000 and 20000 hours
- Between 20000 and 30000 hours
- More than 30000 hours
The Correct Option is D
Solution and Explanation
Total distance in kilometers:
Distance in AU: 30 AU
Conversion factor: 1 AU = 150 million km
\(\text{Total distance} = 30 \text{ AU} \times 150 \text{ million km/AU} = 4.5 \text{ billion km}\)
Travel time:
Distance: 4.5 billion km
Speed: 50,000 km/h
\(\text{Time} = \frac{\text{Distance}}{\text{Speed}} = \frac{4.5 \text{ billion km}}{50,000 \text{ km/h}} = 90,000 \text{ hours}\)
If the price of fuel increases by 20%, and the distance between waystations on the outer route is increased by 10%, calculate the percentage increase in the total fuel cost for a spacecraft traveling from Jupiter to Neptune.
- Less than 30%
- Between 30% and 35%
- Between 35% and 40%
- More than 40%
The Correct Option is B
Solution and Explanation
Assuming
Original fuel price: P
Original distance between waystations: D
Original number of waystations between Jupiter and Neptune: N
Original total fuel cost:
Fuel consumption is directly proportional to distance.
\(\text{Total fuel cost} = P \times N \times D\)
New situation:
New fuel price: 1.2P (20% increase)
New distance between waystations: 1.1D (10% increase)
Number of waystations will decrease as the distance between them increases, but for simplicity, let's assume it remains N (this assumption might slightly overestimate the cost increase).
New total fuel cost:
\(\text{New total fuel cost} = 1.2P \times N \times 1.1D = 1.32 \times (P \times N \times D)\)
Percentage increase in total fuel cost:
=\(\left( \frac{\left( \text{New total fuel cost} - \text{Original total fuel cost} \right)}{\text{Original total fuel cost}} \right) \times 100\)
\(=\left( \frac{1.32 \times P \times N \times D - P \times N \times D}{P \times N \times D} \right) \times 100 \\= \left( \frac{(1.32 - 1) \times P \times N \times D}{P \times N \times D} \right) \times 100 = \left( \frac{0.32 \times P \times N \times D}{P \times N \times D} \right) \times 100 = 0.32 \times 100 = 32\%\)
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