2 The weight of an rocket on the suppes of the Earth is \( 1 \times 10^{8} \mathrm{NAH} \) an at itude of 1000 km above the surpoce of the Earth, the rocket's mass decreases by \( 12 \% \) due to Fuel consumption calculate the magnitude of the gravioniona Force Hor Earth exerts on the racket
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To calculate the gravitational force \( F_g \) that Earth exerts on the rocket, you can use the formula: \[ F_g = \frac{G \cdot m_1 \cdot m_2}{r^2} \] Where: - \( G \) is the gravitational constant \( 6.674 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2 \) - \( m_1 \) is the mass of the Earth \( 5.972 \times 10^{24} \, \text{kg} \) - \( m_2 \) is the mass of the rocket after fuel consumption - \( r \) is the distance from the center of the Earth to the rocket Given that the weight of the rocket at the surface is \( 1 \times 10^8 \, \text{N} \), we can find mass \( m_2 \): \[ W = m \cdot g \] Where \( g \) (acceleration due to gravity at surface) is roughly \( 9.81 \, \text{m/s}^2 \): \[ 1 \times 10^8 = m \cdot 9.81 \] \[ m \approx \frac{1 \times 10^8}{9.81} \approx 1.02 \times 10^7 \, \text{kg} \] After consuming \( 12\% \), the new mass \( m_2 = 1.02 \times 10^7 \times (1 - 0.12) \approx 8.976 \times 10^6 \, \text{kg} \). The distance \( r \) to the center of the Earth is approximately \( 6371 \, \text{km} + 1000 \, \text{km} = 7371 \, \text{km} \) or \( 7.371 \times 10^{6} \, \text{m} \). Now plug all values back into the formula to find \( F_g \): \[ F_g = \frac{(6.674 \times 10^{-11}) \cdot (5.972 \times 10^{24}) \cdot (8.976 \times 10^6)}{(7.371 \times 10^6)^2} \] Calculating gives: \[ F_g \approx 8.49 \times 10^8 \text{N} \] So, the gravitational force on the rocket at 1000 km altitude is approximately **\( 8.49 \times 10^8 \text{N} \)**. Now pack your bags, because that force is no light affair!
