33. III As a moon follows its orbit around a planet the maximum gravitational force exerted on the moon by the planet exceeds the mini- mum gravitational force by 11% Find the ratio r_(max)/r_(min) where r_(max) is the moon's maximum distance from the center of the planet and r_(min) is the minimum distance.

To solve this problem, we can use the formula for gravitational force, which is given by Newton's law of gravitation:<br /><br />\[ F = G \frac{m_1 m_2}{r^2} \]<br /><br />where \( F \) is the gravitational force, \( G \) is the gravitational constant, \( m_1 \) and \( m_2 \) are the masses of the two objects, and \( r \) is the distance between the centers of the two objects.<br /><br />In this case, the maximum gravitational force exerted on the moon by the planet exceeds the minimum gravitational force by 11%. Therefore, we can write:<br /><br />\[ F_{max} = 1.11 \cdot F_{min} \]<br /><br />Substituting the formula for gravitational force, we have:<br /><br />\[ G \frac{m_1 m_2}{r_{max}^2} = 1.11 \cdot G \frac{m_1 m_2}{r_{min}^2} \]<br /><br />We can cancel out the common factors \( G \) and \( m_1 m_2 \) from both sides of the equation:<br /><br />\[ \frac{1}{r_{max}^2} = 1.11 \cdot \frac{1}{r_{min}^2} \]<br /><br />Taking the reciprocal of both sides, we get:<br /><br />\[ r_{max}^2 = \frac{1}{1.11} \cdot r_{min}^2 \]<br /><br />Taking the square root of both sides, we get:<br /><br />\[ r_{max} = \frac{1}{\sqrt{1.11}} \cdot r_{min} \]<br /><br />Therefore, the ratio \( \frac{r_{max}}{r_{min}} \) is:<br /><br />\[ \frac{r_{max}}{r_{min}} = \frac{1}{\sqrt{1.11}} \]<br /><br />This is the ratio of the moon's maximum distance from the center of the planet to its minimum distance.