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(II) Our Sun rotates about the center of our Galaxy$\left(m_{\mathrm{G}} \approx 4 \times 10^{41} \mathrm{kg}\right)$ at a distance of about $3 \times 10^{4}$ light-years $\left[11 \mathrm{y}=\left(3.00 \times 10^{8} \mathrm{m} / \mathrm{s}\right) \cdot\left(3.16 \times 10^{7} \mathrm{yr}\right) \cdot(1.00 \mathrm{yr})\right] .$ What is theperiod of the Sun's orbital motion about the center of the Galaxy?
Step 1
We know that 1 light year is equal to $(3.00 \times 10^{8} \mathrm{m} / \mathrm{s}) \cdot\left(3.16 \times 10^{7} \mathrm{yr}\right) \cdot(1.00 \mathrm{yr})$. Therefore, the distance from the Sun to the center of the galaxy in meters is $3 \times 10^{4}$ Show more…
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(II) Our Sun rotates about the center of the Galaxy $\left(M_{G} \approx 4 \times 10^{41} \mathrm{kg}\right)$ at a distance of about $3 \times 10^{4}$ light- years $\left(1 \mathrm{ly}=3 \times 10^{8} \mathrm{m} / \mathrm{s} \times 3.16 \times 10^{7} \mathrm{s} / \mathrm{y} \times 1 \mathrm{y}\right) .$ What is the period of our orbital motion about the center of the Galaxy?
(II) Our Sun revolves about the center of our Galaxy ($m_G \approx 4 \times 10^{41} kg$) at a distance of about 3 $\times$ 10$^4$ light years [$1 ly = (3.00 \times 10^8 m/s) \cdot (3.16 \times 10^7 s/yr) \cdot (1.00 yr)$ D.What is the period of the Sun's orbital motion about the center of the Galaxy?
CIRCULAR MOTION; GRAVITATION
Planets, Kepler's Laws, and Newton's Synthesis
'(II) Our Sun revolves about the center of our Galaxy (mG 4 X 1041 kg) at a distance of about 3 X 104 light- years [1ly = (3.00 X 108m/s) . (3.16 X 107 s/yr) . (1.00 yr)]: What is the period of the Sun s orbital motion about the center of the Galaxy?'
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