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sity P x We 79% History Bookmarks People Window Help ng.com/ibiscms/modlibis/view.php?id-3390903 Print Calculator Periodic Table 19 of 19 Sapling Learning The show a hypothetical at two times The spatial coordinates (x, y of the bodies are Units n the first picture, the velocity of the center of mass of the system is zero. Find the magnitude, ds. of the stars displacement. ms 1.8975 1000 kg mA, 2.4401 x102e kg AU me 6.4663 x 10 kg mc 84729 x 102 kg (0.4995, 1.3397) (0, 1.1779) (04815.0) (0.0) C (0,-0.1959) H0.7 177, -0. e Previous Check Answer e Next a Exit about us careers partners privacy policy l terms of use contact us help

Explanation / Answer

here,

ma = 2.4401 * 10^28 kg
mb = 6.4663 * 10^26 kg
mc = 8.4729 * 10^27 kg
ms = 1.8975 * 10^30 kg

The x- and y-coordinates of the CM of the system must be fixed, so
Xcm1 = ma1 * xa1 + mb1 * xb1 + mc1 * xc1 + ms1 * xs1
Xcm2 = ma2 * xa2 + mb2 * xb2 + mc2 * xc2 + ms2 * xs2
Xcm1 = Xcm2

2.4401 * 10^28 * 0.4815 + 6.4663 * 10^26 * 0.4995 + 8.4729 * 10^27 * 0 + 1.8975 * 10^30 * 0 = 2.4401 * 10^28 * 0 + 6.4663 * 10^26 * (-1.6629) + 8.4729 * 10^27 * (-0.7177) + 1.8975 * 10^30 * x

distance of planet s (x coordinate ) , x = 0.01013 Au

similarly for Y coordinate :

2.4401 * 10^28 * 0 + 6.4663 * 10^26 * 1.3397 + 8.4729 * 10^27 * 1.1779 + 1.8975 * 10^30 * 0 = 2.4401 * 10^28 * -0.1959 + 6.4663 * 10^26 * (0) + 8.4729 * 10^27 * (-0.7963) + 1.8975 * 10^30 * y

distance of planet s (y coordinate ) , y = 0.01179 Au

net distance, d = sqrt(x^2 + y^2)

net distance, d = sqrt(0.01013^2 + 0.01179^2)

net distance, d = 0.015544 Au

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