- The orbital speed of the Earth's moon is \(1.02\times 10^{3} m/s\).
- The period of the Earth's moon is \(2.37\times 10^{6}\) seconds.
Given:
- Mass of the Earth = M = \(5.97\times 10^{24}kg\)
- The average distance between the center of the Earth and the center of its moon = R = \(3.84\times 10^8 m\)
To find:
The orbital speed and period of Earth’s moon.
Solution:
Mass of the earth = M = \(5.97\times 10^{24}kg\)
The average distance between the center of the Earth and the center of its moon = R = \(3.84\times 10^8 m\)
The orbital speed of moon = v
The orbital speed of a satellite is given by:
\(v=\sqrt{\frac{G\times M}{R}}\\\\v=\sqrt{\frac{6.67430 \times 10^{-11}Nm^2 kg^{-2}\times 5.97\times 10^{24}kg}{ 3.84\times 10^8 m}}\\\\v=1.02\times 10^3 m/s\)
The orbital speed of the Earth's moon is \(1.02\times 10^{3} m/s\).
The period of the Earth's moon = T
The period of a satellite is given by:
\(T=\sqrt{\frac{4\pi^2 r^3}{GM}}\\\\T=\sqrt{\frac{4\3.14^2 (3.84\times 10^8 m)^3}{6.67430 \times 10^{-11}Nm^2 kg^{-2}\times 5.97\times 10^{24}kg}}\\\\T=2.37\times 10^{6} s\)
The period of the Earth's moon is \(2.37\times 10^{6}\) seconds.
Learn more about a satellite's orbital speed and period here:
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