Given:
The orbital period of the Ceres, T=4.60 years
To find:
The semimajor axis of the orbit of the Ceres.
Explanation:
1 year is 3.154×10⁷ s
Thus the orbital period in seconds is given by,
[tex]\begin{gathered} T=4.60\times3.154\times10^7 \\ =1.451\times10^8\text{ s} \end{gathered}[/tex]From Kepler's third law, the orbital period is related to the semi-major axis as,
[tex]T^2=\frac{4\pi^2}{GM}a^3[/tex]Where G is the gravitational constant, M is the mass of the sun, and a is the semimajor axis.
On substituting the known values,
[tex]\begin{gathered} (1.451\times10^8)^2=\frac{4\pi^2}{6.67\times10^{-11}\times2\times10^{30}}\times a^3 \\ \Rightarrow a=\sqrt[3]{\frac{(1.451\times10^8)^2\times6.67\times10^{-11}\times2\times10^{30}}{4\pi^2}} \\ =4.14^{11}\text{ m} \end{gathered}[/tex]The semimajor axis of the mars is 2.28×10¹¹ m
The semimajor axis of the Jupiter is 7.78×10¹¹ m
Thus the Ceres is in between the mars and the Jupiter.
Final answer:
The semimajor axis of the Ceres is 4.41×10¹¹ m. Thus it is between Mars and Jupiter.
