11. If the gravitational force is 60 G , what would it be if the distance is tripled? What is changing? How is it changing? What does this do to gravity? 12. If the gravitational force is 15 G , what would it be if the mass is quadrupled? What is changing? How is it changing? What does this do to gravity?

To solve these problems, we will use the formula for gravitational force, which is 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 objects, - \( r \) is the distance between the centers of the two masses. ### Problem 11 **Given:** - Initial gravitational force \( F_1 = 60G \) - Distance \( r \) is tripled, so the new distance \( r_2 = 3r \) **What is changing?** - The distance between the two masses is changing (it is being tripled). **How is it changing?** - The distance \( r \) is increased from \( r \) to \( 3r \). **What does this do to gravity?** - We can calculate the new gravitational force \( F_2 \) using the formula: \[ F_2 = \frac{G \cdot m_1 \cdot m_2}{(3r)^2} = \frac{G \cdot m_1 \cdot m_2}{9r^2} = \frac{1}{9} \cdot \frac{G \cdot m_1 \cdot m_2}{r^2} = \frac{1}{9} F_1 \] Now substituting \( F_1 = 60G \): \[ F_2 = \frac{1}{9} \cdot 60G = \frac{60G}{9} = \frac{20G}{3} \] ### Problem 12 **Given:** - Initial gravitational force \( F_1 = 15G \) - Mass \( m_1 \) is quadrupled, so the new mass \( m_2 = 4m_1 \) **What is changing?** - The mass of one of the objects is changing (it is being quadrupled). **How is it changing?** - The mass \( m_1 \) is increased from \( m_1 \) to \( 4m_1 \). **What does this do to gravity?** - We can calculate the new gravitational force \( F_2 \) using the formula: \[ F_2 = \frac{G \cdot (4m_1) \cdot m_2}{r^2} = 4 \cdot \frac{G \cdot m_1 \cdot m_2}{r^2} = 4F_1 \] Now substituting \( F_1 = 15G \): \[ F_2 = 4 \cdot 15G = 60G \] ### Summary of Results 1. For the first problem, when the distance is tripled, the new gravitational force becomes \( \frac{20G}{3} \). 2. For the second problem, when the mass is quadrupled, the new gravitational force becomes \( 60G \). These calculations show how gravitational force is inversely proportional to the square of the distance and directly proportional to the product of the masses involved.