Question

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Moon Pory om Earth Roe Sun For simplicity, assume the earth moves in a circular orbit around the sun, and the moon circles the earth, all in the horizontal plane. The center of the earth is a distance of 1.5*1011 m from the center of the sun, circling at an angular velocity of de = 2*10-7 rad/s (1 revolution per year). The center of the moon is a distance of 3.84*108 m from the center of the earth, circling at an angular velocity of Om = 2.7*10-6 rad/s (1 revolution in 27 days). Put your coordinate frame on the sun, with X pointing right and Y pointing up. The journey starts at perihelion on Jan 3, with the earth aligned directly to the right of the sun, and the moon directly to the right of the earth. 30 days later:

Answer 1

* Answers from the Given data : RSE QE = 1.5x 10"m 2 x 10 * rad's 3.84 x 108m On = 2.78106 Rme location of the moon: As moon completes in one revolution in 27 days So, after 27 days it will come back to it's original Position So, 30-97 The Net days moon 3 days t= 3x27 X 60X60 t = 9.592 x105 sec 5 x 2.599 X10 sec 2.7 X106 rad's # OM OM - 0. 69984 rad

in MM Ime cos om t&mesin Om 3. 84 x 108 cos (0.69984) 1 + 3.1 3.84x108 sin (0.69986) → 9.9378 X108 + 2.4933 9 x10m worm Magnitude of moon velocity (9.7x166) (3.84 X108) = 1036,8 mis Perpendicular vector to En is given by : sinom 1) EM dom (im cosomit im sinom. dom – im sinomi + rm cos oma .. Perpendicular unit vector is : = - sin om ? + cosom } = -sin (o. 69984) 1 + cos(0.6998u) î = -0.644095 î +0.764954 î .. velocity vector is 1036.8 (-0.644095 î+0.464953 668 T +793 m/s 3:"/