Two stars are 1.3times 10^3 m apart. Each star has a mass of 3.8times 10^15 kg. What is the gravitational altractive force between them? h FSEtimes 10^14 8.54times 10^24 9.99times 10^37 1.1times 10^20

To find the gravitational attractive force between the two stars, we can use Newton's law of universal gravitation. The formula is:<br /><br />\[<br />F = \frac{G \cdot m_1 \cdot m_2}{r^2}<br />\]<br /><br />where:<br />- \( F \) is the gravitational force,<br />- \( G \) is the gravitational constant, approximately \( 6.674 \times 10^{-11} \, \text{N}\cdot(\text{m/kg})^2 \),<br />- \( m_1 \) and \( m_2 \) are the masses of the two stars,<br />- \( r \) is the distance between the centers of the two masses.<br /><br />Given:<br />- \( m_1 = m_2 = 3.8 \times 10^{15} \, \text{kg} \),<br />- \( r = 1.3 \times 10^{3} \, \text{m} \).<br /><br />Substitute these values into the formula:<br /><br />\[<br />F = \frac{6.674 \times 10^{-11} \cdot (3.8 \times 10^{15}) \cdot (3.8 \times 10^{15})}{(1.3 \times 10^{3})^2}<br />\]<br /><br />Calculate the numerator:<br /><br />\[<br />6.674 \times 10^{-11} \cdot (3.8 \times 10^{15})^2 = 6.674 \times 10^{-11} \cdot 1.444 \times 10^{31} = 9.632 \times 10^{20}<br />\]<br /><br />Calculate the denominator:<br /><br />\[<br />(1.3 \times 10^{3})^2 = 1.69 \times 10^{6}<br />\]<br /><br />Now divide the results:<br /><br />\[<br />F = \frac{9.632 \times 10^{20}}{1.69 \times 10^{6}} \approx 5.7 \times 10^{14} \, \text{N}<br />\]<br /><br />Therefore, the gravitational attractive force between the two stars is approximately \( 5.7 \times 10^{14} \, \text{N} \). However, this does not match any of the provided options exactly. Please check if there might be a typographical error in the options or if additional context is needed.