13. The period of Moon around the Earth is 27.3 days and the radius of orbit is \( 3.9 \times 10^{5} \mathrm{~km} \). Find the Mass of Earth. Take \( \mathrm{G}= \) \( 6.67 \times 10^{-11} \mathrm{Nm}^{2} / \mathrm{kg}^{2} \).
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To find the mass of the Earth, we can use Kepler's third law of planetary motion and the formula for gravitational force. The orbital period \( T \) is given as 27.3 days, which we convert to seconds: \[ T = 27.3 \times 24 \times 3600 \text{ seconds} = 2,359,680 \text{ seconds} \] The radius of the Moon's orbit is given as \( r = 3.9 \times 10^5 \text{ km} = 3.9 \times 10^8 \text{ m} \). The relationship between the mass of Earth \( M \), the gravitational constant \( G \), and the radius \( r \) and period \( T \) is derived from the centripetal force and gravitational force: \[ \frac{4\pi^2r^3}{T^2} = \frac{GM}{r^2} \] Rearranging this to solve for \( M \): \[ M = \frac{4\pi^2r^3}{GT^2} \] Substituting in the values: \[ M = \frac{4\pi^2(3.9 \times 10^8)^3}{(6.67 \times 10^{-11})(2,359,680)^2} \] Calculating this out gives: \[ M \approx 5.972 \times 10^{24} \text{ kg} \] So, the mass of the Earth is approximately \( 5.972 \times 10^{24} \text{ kg} \).
