(1) the mass of the Earth is \( 6 \times 10^{24} \mathrm{ky} \). If the centres of the earth and moon are \( 3,9 \times 10^{8} \mathrm{~m} \) apart, the force of gravitational attraction between them is \( 1,9 \times 10^{20} \mathrm{~N} \). whert is the approximate mass of the moon (2) Calculate the value of \( g \) at a point \( 8,10 \mathrm{~m} \) from the center of the earth use \( 6 \times 10^{24} \mathrm{~kg} \) as the mass of the earth (3) Use the following data to calculate the mass of the earth: \( g \) is equal to \( a, 8 \mathrm{~m} \cdot \mathrm{~s}^{-2} \) \[ \begin{array}{l} r=6 \times 10^{6} \mathrm{~m} \\ G=6,7 \times 10^{-11} \mathrm{~N} \cdot \mathrm{~m}^{2} \text { per } \mathrm{kg}^{2} \end{array} \]

The approximate mass of the moon is \(8.600893 \times 10^{10} \mathrm{kg}\). ### Problem 2: Calculate the value of \( g \) at a point \(8.10 \mathrm{~m}\) from the center of the Earth Given: - Mass of the Earth, \( m = 6 \times 10^{24} \mathrm{kg} \) - Distance from the center of the Earth, \( r = 8.10 \mathrm{~m} \) We can use the formula for gravitational acceleration: \[ g = \frac{G \cdot m}{r^2} \] Plugging in the values: \[ g = \frac{6.7 \times 10^{-11} \mathrm{N} \cdot \mathrm{m}^2 \mathrm{kg}^{-2} \times 6 \times 10^{24} \mathrm{kg}}{(8.10 \mathrm{~m})^2} \] \[ g = \frac{4.02 \times 10^{14} \mathrm{N} \cdot \mathrm{m}^2}{65.61 \mathrm{~m}^2} \] \[ g \approx 6.14 \times 10^{12} \mathrm{~m/s}^2 \] simplify answer: The value of \( g \) at a point \(8.10 \mathrm{~m}\) from the center of the Earth is approximately \(6.14 \times 10^{12} \mathrm{~m/s}^2\). ### Problem 3: Calculate the mass of the Earth using the given data Given: - Gravitational acceleration, \( g = 1.8 \mathrm{~m/s}^2 \) - Radius of the Earth, \( r = 6 \times 10^{6} \mathrm{~m} \) - Gravitational constant, \( G = 6.7 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^2 \mathrm{kg}^{-2} \) We can rearrange the gravitational acceleration formula to solve for mass: \[ g = \frac{G \cdot m}{r^2} \] \[ m = \frac{g \cdot r^2}{G} \] Plugging in the values: \[ m = \frac{1.8 \mathrm{~m/s}^2 \times (6 \times 10^{6} \mathrm{~m})^2}{6.7 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^2 \mathrm{kg}^{-2}} \] \[ m = \frac{1.8 \times 3.6 \times 10^{13} \mathrm{~m}^2}{6.7 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^2 \mathrm{kg}^{-2}} \] \[ m = \frac{6.48 \times 10^{13} \mathrm{~m}^2}{6.7 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^2 \mathrm{kg}^{-2}} \] \[ m \approx 9.66 \times 10^{23} \mathrm{kg} \] simplify answer: The mass of the Earth is approximately \(9.66 \times 10^{23} \mathrm{kg}\).