Respuesta :
Given,
The height at which the satellite was orbiting the moon, h=122 km=122×10³ m
The radius of the moon, R=1840×10³ m
The mass of the moon, m=7.3×10²² kg
a.
The orbital velocity of the Apollo 11 is given by,
[tex]v=\sqrt[]{\frac{GM}{R+h}}[/tex]Where G=6.67×10⁻¹¹ m³ kg⁻¹ s⁻² is the gravitational constant.
On substituting the known values,
[tex]\begin{gathered} v=\sqrt[]{\frac{6.67\times10^{-11}\times7.3\times10^{22}}{1840\times10^3+122\times10^3}} \\ =1.58\times10^3\text{ m/s} \end{gathered}[/tex]Thus the orbital speed of the spacecraft is 1.58×10³ m/s
b.
The time it takes for the spacecraft to orbit once is called the time period of the spacecraft.
The time period of the spacecraft is given by,
[tex]T=2\pi\sqrt[]{\frac{(R+h)^3}{GM}}[/tex]On substituting the known values,
[tex]\begin{gathered} T=2\pi\times\sqrt{\frac{(1840\times10^3+122\times10^3)^3}{^{}6.67\times10^{-11}\times7.3\times10^{22}}} \\ =7.83\times10^{13}\text{ s} \end{gathered}[/tex]Converting the period to hours,
[tex]\begin{gathered} T=\frac{7.83\times10^{13}}{3600} \\ =2.18\times10^{10}\text{ hr} \end{gathered}[/tex]Thus the time it takes for the spacecraft to orbit once is 7.83×10¹³ s, that is 2.18×10¹⁰ hr
