Problem 44 - University Physics with Modern Physics

Question

A uniform sphere with mass 60.0 \(\mathrm{kg}\) is held with its center at the origin, and a second uniform sphere with mass 80.0 \(\mathrm{kg}\) is held with its center at the point \(x=0, y=3.00 \mathrm{m} .\) (a) What are the magnitude and direction of the net gravitational force due to these objects on a third uniform sphere with mass 0.500 \(\mathrm{kg}\) placed at the point \(x=4.00 \mathrm{m}, y=0 ?\) (b) Where, other than infinitely far away, could the third sphere be placed such that the net gravitational force acting on it from the other two spheres is equal to zero?

Step-by-Step Solution

Answer

(a) 2.20 × 10^-10 N at -16.83° from the negative x-axis. (b) At x = 7.20 m along the y-axis.

Step 1: Identify Variables and Constants

The masses of the spheres are given as \( m_1 = 60.0 \, \mathrm{kg} \), \( m_2 = 80.0 \, \mathrm{kg} \), and \( m_3 = 0.500 \, \mathrm{kg} \). The positions are \((0, 0)\), \((0, 3.00)\), and \((4.00, 0)\). The gravitational constant \( G \) is \( 6.674 \times 10^{-11} \, \mathrm{Nm^2/kg^2} \).

Step 2: Calculate Distance from m3 to m1 and m2

Use the distance formula to find the distances between the third sphere and the other two spheres. For \( m_1 \), the distance \( r_{13} = \sqrt{(4.00)^2 + (0)^2} = 4.00 \, \mathrm{m} \). For \( m_2 \), the distance \( r_{23} = \sqrt{(4.00 - 0)^2 + (0 - 3.00)^2} = 5.00 \, \mathrm{m} \).

Step 3: Calculate Gravitational Force by m1 on m3

Use the formula for gravitational force: \[F_{13} = \frac{G m_1 m_3}{r_{13}^2} = \frac{6.674 \times 10^{-11} \times 60.0 \times 0.500}{(4.00)^2} = 1.25 \times 10^{-10} \, \mathrm{N}\].

Step 4: Calculate Gravitational Force by m2 on m3

Again, apply the formula for gravitational force: \[F_{23} = \frac{G m_2 m_3}{r_{23}^2} = \frac{6.674 \times 10^{-11} \times 80.0 \times 0.500}{(5.00)^2} = 1.07 \times 10^{-10} \, \mathrm{N}\].

Step 5: Determine Force Components

Since \( F_{13} \) acts in the negative x-direction, its components are \( (-1.25 \times 10^{-10}, 0) \). \( F_{23} \) acts in a direction from \( (4.00, 0) \) to \( (0, 3.00) \), use trigonometry to find its components. The angle \( \theta \) to the horizontal is \( \tan^{-1}(3/4) \). The components are \[F_{23x} = - F_{23} \cos \theta = -1.07 \times 10^{-10} \times \frac{4}{5} = -8.56 \times 10^{-11} \, \mathrm{N}\] \[F_{23y} = F_{23} \sin \theta = 1.07 \times 10^{-10} \times \frac{3}{5} = 6.42 \times 10^{-11} \, \mathrm{N}\].

Step 6: Sum the Forces

The net force components are \[F_{x_{\text{net}}} = F_{13x} + F_{23x} = -1.25 \times 10^{-10} - 8.56 \times 10^{-11} = -2.11 \times 10^{-10} \, \mathrm{N}\]\[F_{y_{\text{net}}} = F_{23y} = 6.42 \times 10^{-11} \, \mathrm{N}\], Then, the magnitude of the net force is \[F_{\text{net}} = \sqrt{(F_{x_{\text{net}}})^2 + (F_{y_{\text{net}}})^2} = \sqrt{(-2.11 \times 10^{-10})^2 + (6.42 \times 10^{-11})^2} \approx 2.20 \times 10^{-10} \, \mathrm{N}\]. The angle is \[\theta = \tan^{-1}\left(\frac{F_{y_{\text{net}}}}{F_{x_{\text{net}}}}\right) = \tan^{-1}\left(\frac{6.42 \times 10^{-11}}{-2.11 \times 10^{-10}}\right) \approx -16.83^\circ \] counter-clockwise from the negative x-axis.

Step 7: Determine Zero Force Location

For the gravitational forces from the two spheres to cancel each other, set their equations equal to each other for distance \( x \) from a reference point on the x-axis: \[\frac{G m_1 m_3}{x^2} = \frac{G m_2 m_3}{(x-3)^2}\].Simplifying gives \[\frac{60.0}{x^2} = \frac{80.0}{(x-3)^2}\] leading to the equation \[(x-3)^2 = \frac{4}{3} x^2\]. Solving this gives two solutions, but only \( x = 7.20 \, \mathrm{m} \) is realistic because it places the sphere between the other two along the line parallel to the x-axis.

Key Concepts

Newton's Law of Universal Gravitation

In the universe, every particle attracts every other particle with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. This principle, known as Newton's law of universal gravitation, can be expressed by the formula:\[ F = \frac{G m_1 m_2}{r^2} \]where:

• \( F \) is the gravitational force between two masses,
• \( G \) is the gravitational constant \( (6.674 \times 10^{-11} \, \mathrm{Nm^2/kg^2}) \),
• \( m_1 \) and \( m_2 \) are the masses of the objects,
• \( r \) is the distance between the centers of the two masses.

This law is foundational for understanding how objects in space influence one another through gravitational attraction. For example, in our exercise, we calculated gravitational forces acting on a small sphere due to two larger spheres. By plugging in the values into the equation, we determined the gravitational forces exerted on the small sphere by each of the larger spheres individually.

Vector Components

Understanding vector components is essential when dealing with gravitational forces in multiple dimensions. A force that acts at an angle needs to be resolved into horizontal and vertical components using trigonometry. The main formulas used for vector components from a force \( F \) acting at an angle \( \theta \) are:

• Horizontal component: \( F_x = F \cos \theta \)
• Vertical component: \( F_y = F \sin \theta \)

In the provided exercise, after calculating the gravitational forces, we split these forces into their respective x and y components. This was necessary to find the net force acting on the third sphere. For instance, the force \( F_{23} \) due to the second sphere was split into components using the angle calculated from the vector alignment.

Equilibrium Points

Equilibrium points are positions where forces balance out to zero, leading to a state of no net force. In gravitational terms, an object placed at an equilibrium point experiences equal gravitational pull from surrounding bodies, nullifying the net force and resulting in a stable or unstable balance. In our specific case, the task was to determine a point on the x-axis where the net gravitational force from two spheres became zero on a third. By setting the gravitational forces equal to each other but acting in opposite directions, we derived a condition:\[ \frac{60.0}{x^2} = \frac{80.0}{(x-3)^2} \]Solving this for \( x \) provided the position between the spheres, where the forces perfectly cancel out. This understanding is crucial, as finding equilibrium points has applications in orbital mechanics, engineering, and physics in general, indicating where and how objects might remain stable.