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(1) the mass of the Earth is \( 6 \times 10^{24} \mathrm{~kg} \). 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 \( 1,9 \times 10^{20} \mathrm{~N} \). what is the approximate mass of the moon (2) calculate the value of \( g \) at a point \( 8,10^{6} \mathrm{~m} \) from the center of the earth use \( 6 \times 10^{24} \mathrm{~kg} \) as the maer of the earth (3) use the following data to calealate the mass of the earth: gis equal to \( 9,8 \mathrm{~m} .5^{-2} \) \( r=6 \times 10^{6} \mathrm{~m} \) \( G=6,7 \times 10^{-11} \mathrm{~N} . \mathrm{m}^{2} \) per \( \mathrm{kg}^{2} \)

Ask by Schneider Tran. in South Africa
Mar 14,2025

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- **Mass of the Moon**: Approximately \( 3.7 \times 10^{22} \, \text{kg} \) - **Value of \( g \) at \( 8.1 \times 10^{6} \, \text{m} \) from Earth's center**: Approximately \( 0.2457 \, \text{m/s}^2 \) - **Mass of the Earth**: Approximately \( 1.301 \times 10^{4} \, \text{kg} \)

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To calculate the approximate mass of the Moon using the gravitational force formula \( F = G \frac{m_1 m_2}{r^2} \), where \( F \) is the gravitational force, \( G \) is the gravitational constant, \( m_1 \) is the mass of the Earth, \( m_2 \) is the mass of the Moon, and \( r \) is the distance between them, we can rearrange the equation to find \( m_2 \) (the mass of the Moon). Plugging in \( F = 1.9 \times 10^{20} \mathrm{~N} \), \( G = 6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^2/\mathrm{kg}^2 \), \( m_1 = 6 \times 10^{24} \mathrm{~kg} \), and \( r = 3.9 \times 10^{8} \mathrm{~m} \) will lead you to calculate \( m_2 \approx 7.34 \times 10^{22} \mathrm{~kg} \). For calculating the value of \( g \) at a distance of \( 8 \times 10^{6} \mathrm{~m} \) from the center of the Earth, the formula is \( g = G \frac{m}{r^2} \), where \( m \) is the mass of the Earth, \( G \) is the gravitational constant, and \( r \) is the distance from the center of the Earth. Given \( m = 6 \times 10^{24} \mathrm{~kg} \) and \( G = 6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^2/\mathrm{kg}^2 \), you can compute \( g \approx 8.9 \mathrm{~m/s}^2 \) for that distance. When it comes to finding the mass of the Earth using the known value of \( g \) at the surface, we can rearrange the formula \( g = G \frac{m}{r^2} \) to \( m = g \frac{r^2}{G} \). Substituting \( g = 9.8 \mathrm{~m/s^2} \), \( r = 6 \times 10^{6} \mathrm{~m} \), and \( G = 6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^2/\mathrm{kg}^2 \) will lead to \( m \approx 6 \times 10^{24} \mathrm{~kg} \), confirming the mass of the Earth.

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