Problem 45 - University Physics with Modern Physics

Question

A uniform sphere with mass 50.0 kg is held with its center at the origin, and a second uniform sphere with mass 80.0 kg is held with its center at the point \(x =\) 0, \(y =\) 3.00 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 kg placed at the point \(x =\) 4.00 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) Find net force by calculating forces between Sphere 3 and each of Spheres 1 and 2, then summing components. (b) For zero force, place Sphere 3 along the line between Spheres 1 and 2, solving distances via gravitational balance.

Step 1: Identify Given Information and Convert to Coordinates

We have three spheres: Sphere 1 with mass \(m_1 = 50.0 \text{ kg}\) at \((0, 0)\), Sphere 2 with \(m_2 = 80.0 \text{ kg}\) at \((0, 3.00 \text{ m})\), and Sphere 3 with \(m_3 = 0.500 \text{ kg}\) at \((4.00 \text{ m}, 0)\). We need to find the gravitational force on Sphere 3.

Step 2: Apply Newton's Law of Gravitation

Use the formula for gravitational force: \[ F = \frac{G m_1 m_2}{r^2} \]where \(G\) is the gravitational constant \(6.674 \times 10^{-11} \text{ N}(\text{m}^2/\text{kg}^2)\), \(m_1\) and \(m_2\) are the masses of two objects, and \(r\) is the distance between their centers.

Step 3: Calculate Force Between Sphere 1 and Sphere 3

The distance \(r_{13}\) between Sphere 1 and Sphere 3 is 4.00 m. The force is directed along the x-axis.\[ F_{13} = \frac{G \times 50.0 \times 0.500}{4.00^2} \]Calculate \(F_{13}\).

Step 4: Calculate Force Between Sphere 2 and Sphere 3

The distance \(r_{23}\) between Sphere 2 and Sphere 3 is calculated using the Pythagorean theorem:\[ r_{23} = \sqrt{(4.00)^2 + (3.00)^2} = 5.00 \text{ m}\]Then calculate the force:\[ F_{23} = \frac{G \times 80.0 \times 0.500}{5.00^2} \]This force has both x and y components, with the angle \(\theta\) defined by \(\tan^{-1}(3/4)\).

Step 5: Resolve F23 into Components

The force \(F_{23}\) can be broken into components using the angle:- \(F_{23x} = F_{23} \cos(\theta)\)- \(F_{23y} = F_{23} \sin(\theta)\)

Step 6: Calculate Net Gravitational Force on Sphere 3

Add the x-components and y-components:\[ F_{net,x} = F_{13} - F_{23x} \]\[ F_{net,y} = -F_{23y} \]Then calculate the magnitude and direction:\[ F_{net} = \sqrt{(F_{net,x})^2 + (F_{net,y})^2} \]\[ \theta_{net} = \tan^{-1}\left(\frac{F_{net,y}}{F_{net,x}}\right) \]

Step 7: Find Equilibrium Point for Zero Gravitational Force

For Sphere 3 to experience zero net force, the forces from Sphere 1 and Sphere 2 must be equal and opposite. Setequations:\[ \frac{G \times 50.0 \times 0.500}{r_1^2} = \frac{G \times 80.0 \times 0.500}{r_2^2} \]Solve for the distances \(r_1\) and \(r_2\) such that the center of mass position causes forces to cancel. Typically this will place Sphere 3 on the line connecting Spheres 1 and 2.

Key Concepts

Newton's Law of Gravitation

Newton's Law of Gravitation is a fundamental principle of physics that describes the attractive force between any two masses in the universe. This force is responsible for the orbits of planets, the fall of objects on Earth, and many other phenomena.
The formula is \[ F = \frac{G m_1 m_2}{r^2} \]where:

• \(F\) is the gravitational force between two masses.
• \(G\) is the gravitational constant, approximately \(6.674 \times 10^{-11} \text{ N}(\text{m}^2/\text{kg}^2)\).
• \(m_1\) and \(m_2\) are the masses of the objects in kilograms.
• \(r\) is the distance between the centers of the two masses, measured in meters.

Understanding this law enables us to calculate how much gravitational pull two objects exert on each other. It's essential to bear in mind that the force is mutual; both objects exert an equal force on one another, regardless of their size or mass.

Coordinate System

A coordinate system helps in specifying the location of an object in a spatial arrangement, like the positions of the spheres in this exercise. The system used here is a simple two-dimensional Cartesian plane with x and y axes.

• Spheres are situated using ordered pairs like \((x, y)\).
• Sphere 1 is located at the origin \((0, 0)\), signifying that it's at the intersection of the two axes.
• Sphere 2 is at \((0, 3.00)\), meaning it is 3 meters above the origin along the y-axis.
• Sphere 3, the one we aim to analyze, is situated at \((4.00, 0)\), lying along the x-axis, 4 meters to the right of the origin.

Introducing a coordinate system allows us to calculate distances and directions between objects, create vectors, and determine angles—essentially helping in mathematical analysis of physical problems.

Vector Components

When dealing with forces in physics, it's beneficial to break them into vector components to simplify calculations of net force. The vector describes both the magnitude and direction of a force.

The gravitational force between objects can be resolved into:

• **X-components:** These lie along the horizontal direction.
• **Y-components:** These lie along the vertical direction.

For example, consider the force between Sphere 2 and Sphere 3. It is at an angle and has both x and y components. To find these components, we use trigonometric functions with the angle \(\theta\):

• The x-component is calculated as \(F_{23x} = F_{23} \cos(\theta)\).
• The y-component is \(F_{23y} = F_{23} \sin(\theta)\).

Breaking forces into components makes it easier to sum them and find resultant forces. This process is crucial for understanding the net effect on an object, especially if multiple forces act in different directions.

Equilibrium Point

An equilibrium point in the context of gravitational forces is a position where a body experiences no net force. It's as if the body is "floating," perfectly balanced by opposing forces.

To find such a point for Sphere 3 among Spheres 1 and 2, the forces exerted by the other two spheres should cancel out.Imagine placing Sphere 3 on an imaginary line between Spheres 1 and 2. By balancing the gravitational forces such that \[ \frac{G \times 50.0 \times 0.500}{r_1^2} = \frac{G \times 80.0 \times 0.500}{r_2^2} \]we can determine at what point Sphere 3 will experience equal and opposite forces.

Such a balance ensures Sphere 3 remains stationary because the forces cancel out. Practically, this involves calculating precise locations where these conditions are true.