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Sagot :
Given:
The radius of the orbit of the asteroid, R=4 AU
To find:
The orbital speed of the asteroid.
Explanation:
1 AU=1.5×10¹¹ m
Thus the radius of the orbit in meters is given by,
[tex]\begin{gathered} R=4\times1.5\times10^{11} \\ =6\times10^{11}\text{ m} \end{gathered}[/tex]The orbital period is given by,
[tex]T=2\pi\sqrt{\frac{R^3}{GM}}[/tex]Where G is the gravitational constant and M is the mass of the sun.
[tex]\begin{gathered} G=6.67\times10^{-11}\text{ m}^3\text{kg}^{-1}\text{s}^{-2}\text{ } \\ M=2\times10^{30}\text{ kg} \end{gathered}[/tex]On substituting the known values in the equation of the orbital period,
[tex]\begin{gathered} T=2\pi\sqrt{\frac{(6\times10^{11})^3}{6.67\times10^{-11}\times2\times10^{30}}} \\ =2.52\times10^8\text{ s} \end{gathered}[/tex]The circumference of the orbit of the asteroid is given by,
[tex]c=2\pi R[/tex]On substituting the known values,
[tex]\begin{gathered} c=2\pi\times6\times10^{11} \\ =3.77\times10^{12}\text{ m} \end{gathered}[/tex]Thus the orbital speed of the asteroid is given by,
[tex]v=\frac{c}{T}[/tex]On substituting the known values,
[tex]\begin{gathered} v=\frac{3.77\times10^{12}}{2.52\times10^8} \\ =14.96\times10^3\text{ m/s} \end{gathered}[/tex]Final answer:
The orbital speed of the asteroid is 14.96×10³ m/s
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