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Four uniform spheres, with masses \(m_{\Lambda}=40 \mathrm{~kg}, m_{B}=35 \mathrm{~kg}\), \(m_{C}=200 \mathrm{~kg}\), and \(m_{D}=50 \mathrm{~kg}\), have \((x, y)\) coordinates of \((0,50 \mathrm{~cm})\) \((0,0),(-80 \mathrm{~cm}, 0)\), and \((40 \mathrm{~cm}, 0)\), respectively. In unit-vector notation, what is the net gravitational force on sphere \(B\) due to the other spheres?

Short Answer

Expert verified
The net gravitational force on sphere B is the vector sum: \[-\frac{G m_B m_C}{0.64}\hat{i} + \frac{G m_B m_D}{0.16}\hat{i} - \frac{G m_B m_\Lambda}{0.25}\hat{j}\].

Step by step solution

01

Understanding the problem

We need to calculate the net gravitational force exerted on sphere B due to the other three spheres: \( \Lambda \), C, and D. We will use the formula for the gravitational force \( F = \frac{Gm_1m_2}{r^2} \) where \( G \) is the gravitational constant, \( m_1 \) and \( m_2 \) are the masses of the two bodies, and \( r \) is the distance between them.
02

Calculate force between B and \( \Lambda \)

Sphere B is at the origin (\(0,0\)), and sphere \( \Lambda \) is at \((0, 50\text{ cm})\). Therefore, the distance \( r_{B\Lambda} = 0.5 \text{ m} \). Under gravitational force \( \mathbf{F}_{B\Lambda} = -\frac{G m_B m_\Lambda}{(0.5)^2} \hat{j}\), where \(-\hat{j}\) indicates the downward direction (toward B from \( \Lambda \)).
03

Calculate force between B and C

Sphere C is at \((-80\text{ cm}, 0)\), so the distance \( r_{BC} = 0.8\text{ m} \). Apply the equation \( \mathbf{F}_{BC} = \frac{G m_B m_C}{0.8^2} (-\hat{i})\), where \(-\hat{i}\) indicates leftward direction from B to C.
04

Calculate force between B and D

Sphere D is at \((40\text{ cm}, 0)\), giving a distance \( r_{BD} = 0.4\text{ m} \). Calculate the force \( \mathbf{F}_{BD} = \frac{G m_B m_D}{0.4^2} \hat{i}\), where \(+\hat{i}\) indicates the right direction from B to D.
05

Sum up the forces

Add up the forces obtained from steps 2, 3, and 4. \( \mathbf{F}_{B} = \mathbf{F}_{B\Lambda} + \mathbf{F}_{BC} + \mathbf{F}_{BD} = (-\frac{G m_B m_\Lambda}{0.25} \hat{j}) + (-\frac{G m_B m_C}{0.64} \hat{i}) + (\frac{G m_B m_D}{0.16} \hat{i}) \).
06

Simplify the vector sum

Combine the \( \hat{i} \) components: \(-\frac{G m_B m_C}{0.64} + \frac{G m_B m_D}{0.16} \). The \( \hat{j} \) component remains: \(-\frac{G m_B m_\Lambda}{0.25} \). Substitute in values to calculate the actual force values using \( G = 6.674 \times 10^{-11} \text{ Nm}^2/\text{kg}^2 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Newton's Law of Gravitation
One of the foundational principles of physics is Newton's Law of Gravitation. This law states that every particle of matter in the universe 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 their centers. This is represented by the equation:\[F = \frac{G m_1 m_2}{r^2}\]where:
  • \( F \) is the gravitational force between two masses
  • \( G \) is the gravitational constant, approximated as \(6.674 \times 10^{-11} \text{ Nm}^2/\text{kg}^2\)
  • \( m_1 \) and \( m_2 \) are the masses of the two objects
  • \( r \) is the distance between the centers of the two masses
This law explains why planets orbit stars and why we feel the force of gravity on Earth. It highlights the important relationship between mass and distance in determining the force of gravity. Recalling this law helps in understanding how multiple objects exert gravitational forces on each other, as seen in the original problem where multiple spheres exert forces on sphere B.
Vector Addition
In physics, many quantities are vector quantities, meaning they have both a magnitude and a direction. Forces are one such type of quantity. When multiple forces act on an object, we use vector addition to determine the net effect. Vector addition involves combining both the magnitude and direction of all involved vectors.For example, when calculating the gravitational force on sphere B from the other spheres, each force has a specific direction:
  • Force from sphere \( \Lambda \) is downward, represented as \(-\hat{j}\)
  • Force from sphere C is leftward, represented as \(-\hat{i}\)
  • Force from sphere D is rightward, represented as \(+\hat{i}\)
After computing these individual forces, we sum them up vectorially. This means adding the components pointing in the same directions separately. For instance, the forces in the \( \hat{i} \) direction (left and right) are combined, and the force in the \( \hat{j} \) direction (up/down) is considered independently.The result is a net force that effectively represents the total gravitational influence on sphere B by all the other spheres.
Mass and Distance in Physics
Mass and distance play crucial roles in physics, especially in the context of gravitational forces. Larger masses exert stronger gravitational forces, which is why objects like Earth have such significant gravitational pulls compared to smaller objects.Meanwhile, the distance between objects affects the strength of the gravitational force exponentially because it follows an inverse-square law. This means that if the distance between two masses is doubled, the gravitational force they exert on each other becomes one-fourth as strong. This principle underscores the importance of both the scale of objects and their spatial arrangement in understanding gravity and other physical interactions.In the problem provided, the calculation of forces between spheres relies heavily on accurately determining their masses and precise distances. Sphere B is affected by the proximity and massiveness of spheres \( \Lambda \), C, and D. With their different masses and positions, these spheres apply varying gravitational effects on sphere B, showcasing how intricately mass and distance define physical forces.

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Most popular questions from this chapter

Hunting a black hole. Observations of the light from a certain star indicate that it is part of a binary (two-star) system. This visible star has orbital speed \(v=270\) \(\mathrm{km} / \mathrm{s}\), orbital period \(T=1.70\) days, and approximate mass \(m_{1}=6 M_{s}\), where \(M_{s}\) is the Sun's mass, \(1.99 \times\) \(10^{50} \mathrm{~kg}\). Assume that the visible star and its companion star, which is dark and unseen, are both in circular orbits (Fig. \(13-47\) ). What integer multiple of \(M_{s}\) gives the approximate mass \(m_{2}\) of the dark star?

Two Earth satellites, \(A\) and \(B\), each of mass \(m\), are to be launched into circular orbits about Earth's center. Satellite \(A\) is to orbit at an altitude of \(6370 \mathrm{~km} .\) Satellite \(B\) is to orbit at an altitude of \(19110 \mathrm{~km}\). The radius of Earth \(R_{E}\) is \(6370 \mathrm{~km}\). (a) What is the ratio of the potential energy of satellite \(B\) to that of satellite \(A\), in orbit? (b) What is the ratio of the kinetic energy of satellite \(B\) to that of satellite \(A\), in orbit? (c) Which satellite has the greater total energy if each has a mass of \(14.6 \mathrm{~kg} ?\) (d) By how much?

A satellite, moving in an elliptical orbit, is \(360 \mathrm{~km}\) above Earth's surface at its farthest point and \(180 \mathrm{~km}\) above at its closest point. Calculate (a) the semimajor axis and (b) the eccentricity of the orbit.

An object of mass \(m\) is initially held in place at radial distance \(r=3 R_{E}\) from the center of Earth, where \(R_{E}\) is the radius of Earth. Let \(M_{E}\) be the mass of Earth. A force is applied to the object to move it to a radial distance \(r=4 R_{E}\), where it again is held in place. Calculate the work done by the applied force during the move by integrating the force magnitude.

The orbit of Earth around the Sun is almost circular: The closest and farthest distances are \(1.47 \times 10^{8} \mathrm{~km}\) and \(1.52 \times 10^{8} \mathrm{~km}\) respectively. Determine the corresponding variations in (a) total energy, (b) gravitational potential energy, (c) kinetic energy, and (d) orbital speed. (Hint: Use conservation of energy and conservation of angular momentum.)

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