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In mission 1 we already agreed that the m3 of fuel left in the tank should be reserved for the re...

In mission 1 we already agreed that the m3 of fuel left in the tank should be reserved for the return home on the last trip (unless finally decided not to return). Establish connection with us NASA. Real images of the planet KA9701 have been obtained. They send us relevant information for the landing maneuver. The team must, in any case, study the possibility of returning home. Thus, the captain of the ship proposes to study the landing site before making the decision. The central screen tells us the situation of the planet KA9701. The elevation of the terrain is identified as well as the coordinates of some points to be able to develop the landing strategy as well as to determine the best zone for said maneuver. Below

are the images received on the central screen from NASA

Dr. Rovira together with Dr. Beneyto, space technicians specializing in space landings, inform us that it is essential to study the existing surface to verify that the spacecraft has enough space in the landing and takeoff. Your experience suggests landing in the area between 0 and 3 km.

By pressing the landing maneuver start button, the screen shows these images.

NASA reports that according to their calculations between CRA01, CRA02 and CRA03 and between CRA02, CRA03 and CRA05 two perfect equilateral triangles are defined. In addition, the indications from NASA are clear: - If you enter the expressions of the lines in the central computer, it offers geometric data. - The ideal surface for landing is shaded in red. - Its value is ... There is a communication failure and they do not receive the signal !! They have 20 minutes to determine whether or not they can land before they consume the right fuel to be able to return to Earth.

THE ACTIVITY3. The team must obtain the expressions of the lines drawn on the screen and enter them in the central computer. To introduce the lines you have to give your expression in three different forms as well as point, vector and angle that form with the horizonta line. Understanding the horizontal line of the image horizontally straight

mage NASAUSGS ESA/ DLR/ FU Berlin (G Neukum)CRA05 2 Image NASAJUSGS ESA DLR FU Berlin G Neukum).12 CRAO3 CRA02 (4000; 5000) CRA04 CRA01 (1000; 2000) Línea de unión L1

mage NASAUSGS ESA/ DLR/ FU Berlin (G Neukum)
CRA05 2 Image NASAJUSGS ESA DLR FU Berlin G Neukum)
12 CRAO3 CRA02 (4000; 5000) CRA04 CRA01 (1000; 2000) Línea de unión L1
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To define the equation of lines, lets take the horizontal line to be the line passing through CRA01, CRA04 and CRA02.

So line from CRA01 to CRA03 (green line) will start from CRA01 (1000; 2000) and makes an angle 60 degrees from horizontal (since CRA01, CRA02 and CRA03 makes an equilateral triangle). Thus vector representation has length = 3000V2 at angle 60 degrees.

Line from CRA04 to CRA03 (yellow line) makes an angle 90 degrees from horizontal. Start point is (2500, 3500) i.e. midpoint of CRA01 and CRA02. thus vector representation will have length = 3000 3/2 with angle 90 degrees

Line from CRA02 to CRA03 (red line) makes angle 120 degrees from horizontal with start point at CRA02 (4000, 5000). Thus in vector representation start point of (4000, 5000), length of 3000V2 and angle of 120 degrees.

line from CRA02 to CRA05 (black line) starts from (4000, 5000) and makes an angle 60 degrees. Length = length of equilateral triangle = 3000V2

and line from CRA03 to CRA05 is parallel to horizontal line. Thus angle with horizontal = 0 degrees. To find point location of CRA03, use y = (tan 60)(x-1000) + 2000

and 1500(x-1000) + 1500(y -2000) = (1/2)(3000*3000)

1500(x-1000)+1500(y-2000) = 1500*3000

(x-1000) + (tan60)(x-1000) = 3000

(1+tan60)x - (1+tan60)1000 = 3000

  (1 + V3):r _ (1 V3) 1000-3000

gives x = 2098.11

and thus y = (tan 60)(2098.11-1000) + 2000 = 3902

thus point CRA03 is (2098; 3902) and angle of line is 0 degrees to horizontal with length 3000V2

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