Calculating Gravitational Force Between Celestial Bodies
<p>To solve for the gravitational force between the Sun and Proxima Centauri, we can use Newton's law of gravitation given by \( F = \frac{G \cdot m_1 \cdot m_2}{r^2} \) where \( F \) is the gravitational force, \( G \) is the gravitational constant, \( m_1 \) and \( m_2 \) are the masses of the two bodies, and \( r \) is the distance between the centers of the two masses.</p>
<p>\( G = 6.674 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2 \) (Gravitational constant)</p>
<p>\( m_1 = 2.00 \times 10^{30} \, \text{kg} \) (Mass of the Sun)</p>
<p>\( m_2 = 2.40 \times 10^{29} \, \text{kg} \) (Mass of Proxima Centauri)</p>
<p>\( r = 4.00 \times 10^{13} \, \text{km} \) (Distance between Sun and Proxima Centauri)</p>
<p>We need to convert \( r \) from kilometers to meters:</p>
<p>\( r = 4.00 \times 10^{13} \times 10^3 \, \text{m} \) (since \( 1 \, \text{km} = 10^3 \, \text{m} \))</p>
<p>Now we substitute the values into the equation:</p>
<p>\( F = \frac{(6.674 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2) \cdot (2.00 \times 10^{30} \, \text{kg}) \cdot (2.40 \times 10^{29} \, \text{kg})}{(4.00 \times 10^{16} \, \text{m})^2} \)</p>
<p>\( F = \frac{(6.674 \times 10^{-11}) \cdot (2.00 \times 10^{30}) \cdot (2.40 \times 10^{29})}{16.00 \times 10^{32}} \)</p>
<p>\( F = \frac{(6.674 \times 2.40) \cdot 10^{48}}{16.00 \times 10^{32}} \)</p>
<p>\( F = \frac{16.0184 \times 10^{48}}{16.00 \times 10^{32}} \)</p>
<p>\( F = 1.00115 \times 10^{16} \, \text{N} \)</p>
<p>Note: The final answer has been rounded to five significant figures.</p>
