Question

The figures show a hypothetical planetary system at two different times. The system has a star S and three planets, labeled A, B, and C. The table provides the mass of each body in the system, as well as their spatial coordinates (?,?) in their initial and final positions. The spatial coordinates of the bodies are given in Astronomical Units (AU). Body Mass (kg) Initital Position Final Position S ?S=2.0197×1030 (0,0) (?S,?S) A ?A=2.0735×1028 (0.3429,0) (0,−0.5335) B ?B=6.6485×1026 (0.5617,1.2553) (−1.5385,0) C ?C=8.4729×1027 (0,1.6711) (−0.8021,−0.6035) The initial velocity of the center of mass of the system is zero. Find the magnitude ?S of the star's displacement from the origin in its final position. An x y coordinate system. The star S is at the origin. Planet A lies on the positive x-axis. Planet B is in the first quadrant of the graph. Planet C lies on the positive y-axis. Initial Position An x y coordinate system. The star S is near the origin in the first quadrant. Planet A lies on the negative y axis. Planet B lies on the negative x axis. Planet C is in the third quadrant of the graph. Final Position ?S= AU

Answer 1

Given that velocity of the center of mass of system is zero, which during both configuration X and Y coordinates of center of mass of system will remain fixed.

Now Center of mass is given by:

In x-direction

Xcm = (mAx1 + mBx2 + mCx3 + mS*x4)/(mA + mB + mC + mS)

In y-direction

Ycm = (mAy1 + mBy2 + mCy3 + mS*y4)/(mA + mB + mC + mS)

(x1, y1) = (0.3429, 0) = coordinates of mA

(x2, y2) = (0.5617, 1.2553) = = coordinates of mB

(x3, y3) = (0, 1.6711) = = coordinates of mC

(x4, y4) = (0, 0) = = coordinates of mS

Now Since center of mass will remain fixed, So

Xcm of system 1 = Xcm of system 2

(mAx1 + mBx2 + mCx3 + mS*x4)/(mA + mB + mC + mS) = (mAx1' + mBx2' + mCx3' + mS*x4')/(mA + mB + mC + mS)

(mAx1 + mBx2 + mCx3 + mS*x4) = (mAx1' + mBx2' + mCx3' + mS*x4')

x4' = [mA*(x1 - x1') + mB*(x2 - x2') + mC*(x3 - x3') + mS*x4]/mS

x4' = [2.0735*10^28*(0.3429 - 0) + 6.6485*10^26*(0.5617 - (-1.5385)) + 8.4729*10^27*(0 - (-0.8021)) + 2.0197*10^30*0]/(2.0197*10^30)

x4' = 0.0075766 AU

Similarly

Ycm of system 1 = Ycm of system 2

(mAy1 + mBy2 + mCy3 + mS*y4)/(mA + mB + mC + mS) = (mAy1' + mBy2' + mCy3' + mS*y4')/(mA + mB + mC + mS)

(mAy1 + mBy2 + mCy3 + mS*y4) = (mAy1' + mBy2' + mCy3' + mS*y4')

y4' = [mA*(y1 - y1') + mB*(y2 - y2') + mC*(y3 - y3') + mS*y4]/mS

y4' = [2.0735*10^28*(0 - (-0.5335)) + 6.6485*10^26*(1.2553 - 0) + 8.4729*10^27*(1.6711 - (-0.6035)) + 2.0197*10^30*0]/(2.0197*10^30)

x4' = 0.015432 AU

Now initially Star was at (0, 0)

In System 2, Star is at (0.0075766, 0.015432)

So displacement of star will be

dS = sqrt [(0.0075766 - 0)^2 + (0.015432 - 0)^2]

dS = 0.01719 AU

Let me know if you've any query.