Question

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?

• A. 2
• B. 3
• C. 4
• D. 5
• A. 15
• B. 16
• C. 17
• D. 18
• A. 29.99%
• B. 76.78.%
• C. 25.81%
• D. 81.82%
• A. Less than 10000 hours
• B. Between 10000 and 20000 hours
• C. Between 20000 and 30000 hours
• D. More than 30000 hours
• A. Less than 30%
• B. Between 30% and 35%
• C. Between 35% and 40%
• D. More than 40%

Answer 1

The Correct Option is B

Fuel consumption rate: 1 metric ton per 0.5 AU

Total distance: 1.5 AU

Total fuel consumption:

\(\frac{1 \text{ metric ton}}{0.5 \text{ AU}} = \frac{X \text{ metric tons}}{1.5 \text{ AU}}\)

\(X = \left( \frac{1 \text{ metric ton}}{0.5 \text{ AU}} \right) \times 1.5 \text{ AU} = 3 \text{ metric tons}\)

Answer 2

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

Answer 3

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.

\(\text{Percentage increase} = \left( \frac{\left( \text{New number of waystations} - \text{Old number of waystations} \right)}{\text{Old number of waystations}} \right) \times 100 = \left( \frac{60 - 33}{33} \right) \times 100 = \left( \frac{27}{33} \right) \times 100 \approx 81.82\%\)

Answer 4

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

\(\text{Time} = \frac{\text{Distance}}{\text{Speed}} = \frac{4.5 \text{ billion km}}{50,000 \text{ km/h}} = 90,000 \text{ hours}\)

Answer 5

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:
=\(\left( \frac{\left( \text{New total fuel cost} - \text{Original total fuel cost} \right)}{\text{Original total fuel cost}} \right) \times 100\)

 \(=\left( \frac{1.32 \times P \times N \times D - P \times N \times D}{P \times N \times D} \right) \times 100 \\= \left( \frac{(1.32 - 1) \times P \times N \times D}{P \times N \times D} \right) \times 100 = \left( \frac{0.32 \times P \times N \times D}{P \times N \times D} \right) \times 100 = 0.32 \times 100 = 32\%\)