\( F=G \frac{m_{1} m_{2}}{r^{2}} \) Calculate the force of gravity between the Moon (mass \( 7.4 \times 10^{22} \mathrm{~kg} \) ) and the Sun (mass \( 2.0 \times 10^{30} \mathrm{~kg} \) ). The average Sun-Moon distance is \( 1.5 \times 10^{13} \mathrm{~m} \). Express your answer with the appropriate units. \( F= \) Value ?

To calculate the force of gravity between the Moon and the Sun, we can use Newton's law of universal gravitation: \[ F = G \frac{m_{1} m_{2}}{r^{2}} \] Where: - \( G = 6.674 \times 10^{-11} \mathrm{~N \cdot m^2/kg^2} \) (the gravitational constant) - \( m_{1} = 7.4 \times 10^{22} \mathrm{~kg} \) (mass of the Moon) - \( m_{2} = 2.0 \times 10^{30} \mathrm{~kg} \) (mass of the Sun) - \( r = 1.5 \times 10^{13} \mathrm{~m} \) (distance between the Sun and the Moon) Now substitute the values into the equation: \[ F = (6.674 \times 10^{-11}) \frac{(7.4 \times 10^{22})(2.0 \times 10^{30})}{(1.5 \times 10^{13})^2} \] First, calculate \( (1.5 \times 10^{13})^2 \): \[ (1.5 \times 10^{13})^2 = 2.25 \times 10^{26} \] Now compute the numerator: \[ (7.4 \times 10^{22})(2.0 \times 10^{30}) = 14.8 \times 10^{52} = 1.48 \times 10^{53} \] Now substitute these values into the force equation: \[ F = (6.674 \times 10^{-11}) \frac{1.48 \times 10^{53}}{2.25 \times 10^{26}} \] Now, calculate the division: \[ \frac{1.48 \times 10^{53}}{2.25 \times 10^{26}} \approx 6.5778 \times 10^{26} \] Now multiply by \( G \): \[ F \approx (6.674 \times 10^{-11}) (6.5778 \times 10^{26}) \] Calculating this gives: \[ F \approx 4390490797062400000 \, \text{N} \approx 4.39 \times 10^{18} \, \text{N} \] Thus, the force of gravity between the Moon and the Sun is approximately: \[ F \approx 4.39 \times 10^{18} \, \text{N} \]