the mass of the sun is 2×10³⁰ kg, and the distance between neptune and the sun is 30 au. what is the orbital…

the mass of the sun is 2×10³⁰ kg, and the distance between neptune and the sun is 30 au. what is the orbital period of neptune in earth years?\n30 earth years\n164 earth years\n3.8×10¹¹ earth years\n2.3×10¹⁷ earth years

the mass of the sun is 2×10³⁰ kg, and the distance between neptune and the sun is 30 au. what is the orbital period of neptune in earth years?\n30 earth years\n164 earth years\n3.8×10¹¹ earth years\n2.3×10¹⁷ earth years

Answer

Explanation:

Step1: Convert distance to meters

1 AU = 1.496×10¹¹ m, so 30 AU = 30×1.496×10¹¹ m = 4.488×10¹² m.

Step2: Use Kepler's third - law in the form $T^{2}=\frac{4\pi^{2}r^{3}}{GM}$

where $G = 6.67×10^{- 11}\ Nm^{2}/kg^{2}$, $M = 2×10^{30}\ kg$, $r = 4.488×10^{12}\ m$. First, calculate $r^{3}=(4.488×10^{12})^{3}=8.99×10^{37}\ m^{3}$. Then, calculate $GM = 6.67×10^{-11}×2×10^{30}=1.334×10^{20}\ Nm^{2}/kg$. Next, $\frac{4\pi^{2}r^{3}}{GM}=\frac{4\times\pi^{2}\times8.99×10^{37}}{1.334×10^{20}}\ s^{2}$. $4\times\pi^{2}\approx39.48$, so $\frac{39.48\times8.99×10^{37}}{1.334×10^{20}}\ s^{2}=\frac{354.9252×10^{37}}{1.334×10^{20}}\ s^{2}\approx2.66×10^{18}\ s^{2}$. $T=\sqrt{2.66×10^{18}\ s^{2}}\approx1.63×10^{9}\ s$.

Step3: Convert seconds to years

1 year = 365×24×3600 s = 31536000 s. $T=\frac{1.63×10^{9}}{31536000}\ years\approx164\ years$.

Answer:

164 Earth years