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The Answers

The Answers. Senior. Challenge ‘11. Saturn the Bath.

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The Answers

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  1. The Answers Senior Challenge ‘11

  2. Saturn the Bath Saturn is so light in weight that it would float in a bath of water if one was big enough; its density is in fact 0.7kg/l. Given that Earth’s volume is 1,084,000 km³, and Saturn’s volume is 755 times larger than Earth’s, how much does Saturn weigh?

  3. Saturn the Bath 0.7kg/l = 700kg/m³ = 700,000,000,000kg/km³ = 700,000,000 tonne/km³ 1,084,000 × 755 × 700,000,000 = 572,894,000,000,000,000 Tonnes. = 5.72894×1020 Kg or = 5.72894×1017 Tonnes

  4. Space Times An old edition of ‘THE TIMES ATLAS of the WORLD’ is 16” × 10” and has 180 pages. The latest edition is 18” × 12½” and has 544 pages, including many details about Space! Clearly, it will need much more space on the bookshelf, but how much heavier would you expect it to be? Give reasons for your estimate.

  5. Space Times Height Scale factor of 1.25 Width Scale factor of 1.125 No. Pages Scale factor of 3.02 1.125 × 1.25 × 3.02 = 4.25

  6. At a base on Europa, a rectangular storage room, 32 yards × 48 yards, is divided by straight paths into six identical square areas. As a safety inspection, a robot has to traverse in turn each of the rectangular circuits possible in the room, but not starting the next circuit until the previous one is completed. The inspection starts by the Airlock at A, at one of the central crossings, and the finish is at the Bay door at B, at the middle of the shorter side farthest from A. How many distinct rectangular circuits are there?How many yards long is the inspection? Tour of Duty

  7. Tour of Duty

  8. Tour of Duty So that is: 6 single square circuits, 3 2x1 circuits, 4 1x2 circuits, 2 2x2 circuits, 2 1x3 circuits and 1 2x3 circuit, So 18 in total The length of the tour is: 6×64+3×96+4×96+2×128+2×128+160 =84+288+384+256+256+160 =1728 yards. A further 32 yards is required to move from A to B, so the total distance is 1760 yards = 1 mile.

  9. Work, Rest and Play A Mars day lasts for 24.6 Earth hours. A Mars year lasts for 668.6 Earth days. How many Mars days are in a Mars year? Devise a leap year system for Mars to use up the extra part days.

  10. Work, Rest and Play A Mars year lasts for 668.6 × 24=16046.4 hours This is 16046 ÷ 24.6 = 652.2927 Mars days.

  11. Work, Rest and Play 1 day every 4 years, and 1 day every 25 years, and 1 day every 400 years 2 days every 7 years 1 day after 3 years, then after 4 years, then 3...

  12. Fuel Moon A group of lunar buggies is stationed at a base at the Moon’s North Pole. The tank of each buggy holds enough fuel to take it exactly half way around the Moon. Any amount of fuel can be transferred from the tank of one buggy to the tank of another when they are together. The only source of fuel is at the base. Assume that the buggies all have the same speed and fuel consumption throughout a journey, and that they all return safely to the base. What is the smallest number of buggies required to allow one buggy to complete a full circular trip around the Moon, passing over the South Pole?

  13. Fuel Moon

  14. Fuel Moon • An alternative solution has the refuelling points at 45° and 90° away from the base. • ‘Never Leave a Man Behind’ • No marks were awarded for solutions which left buggies stranded around the moon.

  15. One Giant Leap for Man Luke Merseywalker is a high jump athlete with a Personal Best (PB) of 2.6 metres on Earth. He is selected to represent Earth at the 2112 Interplanetary Olympics on Titan.Based on his PB on Earth, what height would you expect him to achieve on Titan, given that the acceleration due to gravity there is 0.14g, where g its value on Earth? (The Games take place in a sealed Stadium Dome with an Earth-like atmosphere.)

  16. One Giant Leap for Man • The simple answer: 2.6 ÷ 0.14 ≈18.6m • However this models Luke as a point on the ground. Assuming his centre of gravity is 1m above the ground, so he raises it 1.6m on Earth, so: 1 + 1.6/0.14 = 12.4m

  17. Ways and Means In a certain Solar System, the orbit of each planet has a known mean distance, in Astronomical Units (AU), away from the Sun. The mean distances of the orbits, working away from the Sun, form a geometric progression, where the common ratio is an integer (1, 2, 4, 8, ... is an example of a geometric progression with common ratio 2). The sum of the mean distances of all of the orbits of the planets is 1640, and the mean distances of the first and fourth orbits from the sun are 0.5 and 13.5 AU respectively. What is the largest of the mean distances of the orbits in the system?

  18. Ways and Means • We now need to add planets to the system until we get a total of the orbital means of 1640

  19. Ways and Means 3 planets = 6.5AU 4 planets = 20AU 5 planets = 60.5AU 6 planets = 182AU 7 planets = 526.5AU 8 planets = 1640AU 8th planet orbits at an average of 1093.5AU

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