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Two uniform spheres, each with mass \(M\) and radius \(R\), touch each other. What is the magnitude of their gravitational force of attraction?

Short Answer

Expert verified
The magnitude of gravitational force is \( \frac{G \cdot M^2}{4R^2} \).

Step by step solution

01

Understand the Gravitational Force Formula

The gravitational force between two masses is determined using Newton's Law of Universal Gravitation: \[F = \frac{G \cdot m_1 \cdot m_2}{r^2}\]where:- \( F \) is the force of attraction between the masses,- \( G \) is the gravitational constant \( 6.674 \times 10^{-11} \text{ Nm}^2/\text{kg}^2 \),- \( m_1 \) and \( m_2 \) are the masses, and- \( r \) is the distance between the centers of the two masses.
02

Identify the Given Values

In this exercise, each sphere has mass \( M \) and touches each other, meaning the distances from center to center is equal to their sum of radii. Therefore, \( r = 2R \). You'll need \( G \), \( M \), and \( r = 2R \) to solve the gravitational force.
03

Substitute Values into the Formula

Substitute the known values into the gravitational force formula:\[F = \frac{G \cdot M \cdot M}{(2R)^2} = \frac{G \cdot M^2}{4R^2}\]
04

Simplify the Expression

The force of attraction simplifies to:\[F = \frac{G \cdot M^2}{4R^2}\]This expression is the gravitational force between the two touching spheres.

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Key Concepts

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

Gravitational Force
Gravitational force is an attractive force that acts between any two masses in the universe. It is a part of Newton's Law of Universal Gravitation, which gives us the equation:\[F = \frac{G \cdot m_1 \cdot m_2}{r^2}\]This formula helps us calculate the gravitational pull two objects have on each other. Here, \( F \) represents the force of gravity, \( m_1 \) and \( m_2 \) are the masses of the two objects, and \( r \) is the distance between their centers. The gravitational force is essential because it explains various natural phenomena like the motion of planets and the tides on Earth.

Some important aspects of gravitational force include:
  • It is always attractive and never repulsive.
  • The force decreases as the distance between the two masses increases.
  • The greater the mass, the stronger the gravitational force.
Understanding gravitational force is crucial for solving problems related to the motions of celestial bodies and analyzing the interaction between objects.
Universal Gravitation Constant
The universal gravitation constant, denoted by \( G \), is a key component in calculating gravitational force. It is a constant value used in Newton's law to quantify the strength of gravitational interaction in free space, given by:\[ G = 6.674 \times 10^{-11} \text{ Nm}^2/\text{kg}^2 \]This remarkably small number indicates that gravity is a relatively weak force among the four fundamental forces of nature. Despite this, it plays a dominant role in large-scale structures because of its long range and the massive bodies involved, such as planets and stars.

Some notable characteristics of the universal gravitation constant include:
  • It remains the same throughout the universe, allowing consistent calculations.
  • The smallness of \( G \) explains why we do not experience noticeable gravitational forces from everyday objects.
  • It's crucial for understanding and estimating the behavior of large astronomical bodies.
Learning about \( G \) helps appreciate why celestial movements occur and why gravity is perceived the way it is in our daily experiences.
Spheres in Contact
When considering two spheres in contact, their centers are separated by a distance equal to the sum of their radii. This is an important concept when calculating gravitational forces between such objects. In our exercise, each sphere has mass \( M \) and radius \( R \), meaning that when they touch:- The distance \( r \) between their centers is \( 2R \).Substituting this into the gravitational force formula allows us to solve for the force of attraction between the two spheres:\[F = \frac{G \cdot M^2}{(2R)^2} = \frac{G \cdot M^2}{4R^2}\]This scenario demonstrates how the positioning of spherical objects affects the gravitational force they exert on each other.

Understanding spheres in contact involves:
  • Grasping how distance relates to gravitational interactions.
  • Using geometry to determine key variables like the distance \( r \).
  • Seeing real-world applications of gravitational force calculations in simple setups.
Analyzing these problems strengthens conceptual clarity about how mass, distance, and geometry intertwine under the laws of gravity.

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Most popular questions from this chapter

The point masses \(m\) and 2\(m\) lie along the x-axis, with \(m\) at the origin and 2\(m\) at \(x\) \(=\) \(L\). A third point mass \(M\) is moved along the \(x\)-axis. (a) At what point is the net gravitational force on \(M\) due to the other two masses equal to zero? (b) Sketch the \(x\)-component of the net force on \(M\) due to \(m\) and 2\(m\), taking quantities to the right as positive. Include the regions \(x < 0\), \(0 < x < L\), and \(x > L\). Be especially careful to show the behavior of the graph on either side of \(x = 0\) and \(x = L\).

The acceleration due to gravity at the north pole of Neptune is approximately 11.2 m/s\(^2\). Neptune has mass 1.02 \(\times\) 10\(^{26}\) kg and radius 2.46 \(\times\) 10\(^4\) km and rotates once around its axis in about 16 h. (a) What is the gravitational force on a 3.00-kg object at the north pole of Neptune? (b) What is the apparent weight of this same object at Neptune's equator? (Note that Neptune's "surface" is gaseous, not solid, so it is impossible to stand on it.)

A typical adult human has a mass of about 70 kg. (a) What force does a full moon exert on such a human when it is directly overhead with its center 378,000 km away? (b) Compare this force with the force exerted on the human by the earth.

For a satellite to be in a circular orbit 890 km above the surface of the earth, (a) what orbital speed must it be given, and (b) what is the period of the orbit (in hours)?

In March 2006, two small satellites were discovered orbiting Pluto, one at a distance of 48,000 km and the other at 64,000 km. Pluto already was known to have a large satellite Charon, orbiting at 19,600 km with an orbital period of 6.39 days. Assuming that the satellites do not affect each other, find the orbital periods of the two small satellites \(without\) using the mass of Pluto.

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