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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 gravitational force magnitude is \(\frac{G M^2}{4R^2}\).

Step by step solution

01

Understand the Gravitational Force Formula

The force of gravitational attraction between two masses is given by Newton's Law of Universal Gravitation: \[ F = \frac{G \cdot m_1 \cdot m_2}{r^2} \]where:- \(F\) is the gravitational force,- \(G\) is the gravitational constant (\(6.674 \times 10^{-11} \, \text{m}^3 \, \text{kg}^{-1} \, \text{s}^{-2}\)),- \(m_1\) and \(m_2\) are the masses of the two objects,- \(r\) is the distance between the centers of the two masses.
02

Define Given Variables and Conditions

The problem states that each sphere has a mass \(M\) and radius \(R\). Since the spheres are touching each other, the distance \(r\) between the centers of the two spheres is the sum of their radii: \[ r = R + R = 2R \].
03

Substitute and Calculate Gravitational Force

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

Present the Final Formula

The magnitude of the gravitational force of attraction between the two spheres is:\[ F = \frac{G M^2}{4R^2} \].

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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 a fundamental interaction between two masses. This force was first articulated by Sir Isaac Newton and is described by his Law of Universal Gravitation. It states that every mass attracts every other mass in the universe 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.
To express this mathematically:
  • The formula is: \[ F = \frac{G \cdot m_1 \cdot m_2}{r^2} \]
  • \(F\) represents the gravitational force measured in Newtons (N).
  • \(m_1\) and \(m_2\) are the masses of the two objects.
  • \(r\) is the distance from the center of one mass to the center of the other.
Gravitational force is responsible for keeping planets in orbit around stars, moons orbiting planets, and even affects the trajectory of comets and asteroids. It is a force that decreases with increasing distance, making it relatively weaker compared to other fundamental forces like electromagnetism. Despite its apparent weakness, gravitational force dominates on large scales, maintaining the structure of our universe.
Uniform Spheres
Uniform spheres refer to spherical objects that have a consistent density throughout. This means their mass is evenly distributed, giving the sphere uniform gravitational properties. A uniform sphere can be thought of as a perfect sphere in mathematical problems to simplify calculations of gravitational interactions.
When dealing with uniform spheres:
  • The center of mass is located at the geometric center of the sphere. This helps in simplifying the calculations by treating the entire mass as if it is concentrated at the sphere's center.
  • They are often used in physics problems like this one to apply Newton's gravitational law easily.
In the problem at hand, two uniform spheres of radius \(R\) and mass \(M\) touch. When spheres touch, the distance \(r\) between their centers, which is necessary for calculating gravitational force, is simply the sum of their radii, or \(2R\). This characteristic simplifies establishing the required variables for using Newton’s Law of Universal Gravitation.
Gravitational Constant
The gravitational constant, denoted by \(G\), is a key component in the equation for gravitational force. This constant of proportionality allows us to quantify the strength of the gravitational force between two masses.
Details about the gravitational constant:
  • Its value is approximately \(6.674 \times 10^{-11} \, \text{m}^3 \, \text{kg}^{-1} \, \text{s}^{-2}\).
  • It is a very small number which illustrates why gravity is a relatively weak force compared to others, like electromagnetism, at the level of atoms.
The gravitational constant is universal, meaning it has the same value throughout the universe. Its constancy is crucial for ensuring that the law of gravitation holds under any circumstances, from the movement of planets to the gravitational interactions of galaxies. In the computation of gravitational forces between uniform spheres in our exercise, the gravitational constant \(G\) allows us to calculate the force of attraction in an exact manner.

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Most popular questions from this chapter

The acceleration due to gravity at the north pole of Neptune is approximately 10.7 \(\mathrm{m} / \mathrm{s}^{2}\) . Neptune has mass \(1.0 \times 10^{26} \mathrm{kg}\) and radius \(2.5 \times 10^{4} \mathrm{km}\) and rotates once around its axis in about 16 \(\mathrm{h}\) . (a) What is the gravitational force on a \(5.0-\mathrm{kg}\) 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.)

Calculate the earth's gravity force on a \(75-\mathrm{kg}\) astronaut who is repairing the Hubble Space Telescope 600 \(\mathrm{km}\) above the earth's surface, and then compare this value with his weight at the earth's surface. In view of your result, explain why we say astronauts are weightless when they orbit the earth in a satellite such as a space shuttle. Is it because the gravitational pull of the earth is negligibly small?

Planets are not uniform inside. Normally, they are densest at the center and have decreasing density outward toward the surface. Model a spherically symmetric planet, with the same radius as the earth, as having a density that decreases linearly with distance from the center. Let the density be \(15.0 \times 10^{3} \mathrm{kg} / \mathrm{m}^{3}\) at the center and \(2.0 \times 10^{3} \mathrm{kg} / \mathrm{m}^{3}\) at the surface. What is the acceleration due to gravity at the surface of this planet?

The planet Uranus has a radius of \(25,560 \mathrm{km}\) and a surface acceleration due to gravity of 11.1 \(\mathrm{m} / \mathrm{s}^{2}\) at its poles. Its moon Miranda (discovered by Kniper in 1948 ) is in a circular orbit about Uranus at an altitude of \(104,000 \mathrm{km}\) above the planet's surface. Miranda has a mass of \(6.6 \times 10^{19} \mathrm{kg}\) and a radius of 235 \(\mathrm{km}\) . (a) Calculate the mass of Uranus from the given data, (b) Calculate the magnitude of Miranda's acceleration due to its orbital motion about Uranus. (c) Calculate the acceleration due to Miranda's gravity at the surface of Miranda. (d) Do the answers to parts (b) and (c) mean that an object released 1 \(\mathrm{m}\) above Miranda's surface on the side toward Uranus will fall up relative to Miranda? Explain.

There are two equations from which a change in the gravitational potential energy \(U\) of the system of a mass \(m\) and the earth can be calculated. One is \(U=m g y\) (Eq. 7.2\()\) . The other is \(U=-G m_{\mathrm{E}} m / r(\mathrm{Eq} .12 .9)\) As shown in Section \(12.3,\) the first equation is correct only if the gravitational force is a constant over the change in height \(\Delta y .\) The second is always correct. Actually, the gravitational force is never exactly constant over any change in height, but if the variation is small, we can ignore it. Consider the difference in \(U\) between a mass at the earth's surface and a distance \(h\) above it using both equations, and find the value of \(h\) for which Eq. \((7.2)\) is in error by 1\(\% .\) Express this value of \(h\) as a fraction of the earth's radius, and also obtain a numerical value for it.

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