The figures show a hypothetical planetary system at two different times. The spa
ID: 2240595 • Letter: T
Question
The figures show a hypothetical planetary system at two different times. The spatial coordinates (x, y) of the bodies are given in Astronomical Units (AU). In the first picture, the velocity of the center of mass of the system is zero. Find the magnitude, dS, of the star's displacement.
The figures show a hypothetical planetary system at two different times. The spatial coordinates (x, y) of the bodies are given in Astronomical Units (AU). In the first picture, the velocity of the center of mass of the system is zero. Find the magnitude, ds, of the star's displacement.Explanation / Answer
x direction:
ms xsi + mA xAi + mB xBi + mC xCi = ms xsf + mA xAf + mB xBf + mC xCf
2.7529e30 * 0 + 2.9289e28 * 0.3891 + 6.2841e26 * 0.9349 + 7.5841e27 * 0 = 2.7529e30 * xsf + 2.9289e28 * 0 + 6.2841e26 * (-1.8495) + 7.5841e27 * (-0.8443)
==> xsf = 0.007101366 AU
y direction:
ms ysi + mA yAi + mB yBi + mC yCi = ms ysf + mA yAf + mB yBf + mC yCf
2.7529e30 * 0 + 2.9289e28 * 0 + 6.2841e26 * 1.5929 + 7.5841e27 * 1.6711 = 2.7529e30 * ysf + 2.9289e28 * (-0.1959) + 6.2841e26 * (0) + 7.5841e27 * (-0.8445)
==> ysf = 0.0093782089 AU
====> ds = srt(xsf^2 + ysf^2) = sqrt(0.007101366y2 + 0.0093782089y2) = 0.0117635 = 0.011764 AU
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.