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Physics · GRAVITATION
हिन्दीThe Moon completes one revolution around the Earth in \( 27.3\text{ days} \) in an orbit of radius \( 3.84 \times 10^8\text{ m} \). If \( G = 6.67 \times 10^{-11}\text{ N m}^2\text{ kg}^{-2} \), the mass of the Earth calculated from this data is approximately:
Correct answer · Option 1
\( 6.02 \times 10^{24}\text{ kg} \)
Explanation
\( M_E = \frac{4\pi^2 R^3}{G T^2} \). Substituting \( R = 3.84 \times 10^8\text{ m} \), \( T = 27.3 \times 86400\text{ s} \), we get \( M_E \approx 6.02 \times 10^{24}\text{ kg} \).
