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 of mass of the system is zero. Find the magnitude, ds, of the star's displacement. ms = 2.7529 x 1000 kg mA-2.8067 x1020 kg m-5.7375 x109 kg Incorrect You found the x coordinate of the star's location in the second picture. What about its y coordinate? Number ds0.00502AU mc-7.1397 x10kg (0, 1.6711) T (0.8105, 1.5507) (0.2967, 0) (1.4141,0 (0, 0) 0, -0.4491) (-0.5911, -0.5553)

Explanation / Answer

The x- and y-coordinates of the CM of the system must be fixed, so

X:

mS*0 + 2.8067e28*0.2967 + 5.7375e26*0.8105 + mC*0

= 2.7529e30*x + mA*0 – 5.7375e26*1.4141 – 7.1397e27*0.5911

x = 0.005021657 AU

Y:

mS*0 + mA*0 + 5.7375e26*1.5507 + 7.1397e27*1.6711

= 2.7529e30*y - 2.8067e28*0.4491 - mb*0 - 7.1397e27*0.5533

y = 0.010670984 AU

so d = ?(x² + y²) = 0.011793512 AU

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.