The figures show a hypothetical planetary system at two different times. The spa

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 mass of the system is zero. Find the magnitude, ds, of the star's displacement.

Map Sapling Learning The figures show a hypothetical planetary system at two different times. The spatial coordinates (x, y) of the bodies are given in Astronomical Units (AL) 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 ms 22641 100 kg m.-2.5623x 102 kg mB=6.8307×104kg Mc 8.4729 x 1027 kg Number d,-11.012799 | AU| (0. 1.2601) 0 8727, 1.3397) (0 5277. 0) (-1 9117. 0) (0. 0) (0-0.4913) 0.7177, -0.9891) d) Previous Give Up & View Solution #Try Agan eNext aExit

Explanation / Answer

Fisrt case:

mS = 2.2641 * 1030 kg at (x, y) = (0, 0)

mA = 2.5623 * 1028 kg at (x, y) = (0.5277 AU, 0)

mB = 6.8307 * 1026 kg at (x, y) = (0.8727 AU, 1.3397 AU)

mC = 8.4729* 1027 kg at (x, y) = (0, 1.2601 AU)

Thus, the center of mass (CM) of the system is,

M = mS + mA + mB + mC

M = [ 22641 + 256.23 +  6.8307 + 84.729 ] * 1026 kg

M = 22988.7897 * 1026 kg = 2.298 * 1030 kg

Position of the CM is,

X1 = [22641*0 + 256.23*0.5277 +  6.8307*0.8727 + 84.729*0] / 22988.78

X1 = 0.0061 AU

Y1 = [22641*0 + 256.23*0 +  6.8307*1.3397 + 84.729*1.2601] / 22988.78

Y1 = 0.0050 AU

Second Case:

mS = 2.2641 * 1030 kg at (x, y) = (xS, yS)

mA = 2.5623 * 1028 kg at (x, y) = ( 0 , -0.4913 AU)

mB = 6.8307 * 1026 kg at (x, y) = (-1.9117 AU, 0 AU)

mC = 8.4729* 1027 kg at (x, y) = (-0.7177, -0.9891 AU)

Position of the CM is,

X2 = [22641*xS + 256.23*0 +  6.8307*-1.9117 + 84.729*-0.7177] / 22988.78

X2 = 0.9848 * xS -0.0031 AU  

Y2 = [22641*yS + 256.23*-0.4913 +  6.8307*0 + 84.729*-0.9891] / 22988.78

Y2 = 0.9848 * yS - 0.0091 AU

Since the velocity of the CM is zero, (X2 - X1)/(t2 - t1) = (Y2 - Y1)/(t2 - t1) = 0

Since (t2 - t1) is not zero, (X2 - X1) = (Y2 - Y1) = 0

That is,

(X2 - X1) = 0.9848 * xS - 0.0031 AU - 0.0061 AU = 0

0.9848 * xS - 0.0092 AU = 0

xS = 0.0092 / 0.9848 = 0.0093 AU

(Y2 - Y1) = 0.9848 * yS - 0.0091 AU - 0.0050 AU = 0

0.9848 * yS - 0.0141 AU = 0

yS = 0.0141 / 0.9848 = 0.0142 AU

Hence, the magnitude of the displacement dS of the star is,

dS = [ (0.0093-0)2 + (0.0142-0)2 ]1/2

dS = 0.01697439 AU = 0.0169 AU

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