The orbital period of the moon in Earth days is 26 days.
The given parameters;
The centripetal force experienced by the planet is given as;
[tex]F_c = \frac{mv^2}{r} \\\\v^2 = \frac{F_cr}{m} \\\\v= \sqrt{\frac{F_cr}{m} } \\\\v = \sqrt{\frac{1.1 \times 10^{19} \times 1.5\times 10^8}{9.4\times 10^{21}} } \\\\v = 418.97 \ m/s[/tex]
The orbital period of the moon in Earth days is calculated as follows;
[tex]v = \frac{2\pi r}{T} \\\\T = \frac{2\pi r}{v} \\\\T = \frac{2\times \pi \times 1.5\times 10^8}{418.97} \\\\T = 2,249,803.09 \ s\\\\T = 2,249,803.09 \ s \times \frac{1 \ day}{86,400 \ s} \\\\T = 26 \ days[/tex]
Thus, the orbital period of the moon in Earth days is 26 days.
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The orbital period of moon in Earth days is 26 days.
Given data:
The mass of moon is, [tex]m=9.4 \times 10^{21} \;\rm kg[/tex].
The radius of orbit is, [tex]r = 1.5 \times 10^{8} \;\rm m[/tex]/.
The gravitational force is, [tex]F =1.1 \times 10^{19} \;\rm N[/tex].
The expression for the orbital period is,
[tex]T=\dfrac{2 \pi r}{v}[/tex]
Here, v is the orbital velocity.
The gravitational force provides the necessary centripetal force. Then,
[tex]F =F_{c}\\F=\dfrac{mv^2}{r} \\1.1 \times 10^{19}=\dfrac{9.4 \times 10^{21}\times v^2}{1.5 \times 10^{8}}\\v=\sqrt \dfrac{1.1 \times 10^{19}\times 1.5 \times 10^{8}}{9.4 \times 10^{21}}}\\v=418.96 \;\rm m/s[/tex]
Then orbital period is,
[tex]T=\dfrac{2 \pi r}{v}\\T=\dfrac{2 \pi \times 1.5\times 10^{8}}{418.96}\\T=2249565.104 \;\rm\\\\T=\dfrac{2249565.104 }{86400} \\T \approx 26 \;\rm days[/tex]
Thus, the orbital period of moon in Earth days is 26 days.
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