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What must the separation be between a \(5.2 \mathrm{~kg}\) particle and a \(2.4 \mathrm{~kg}\) particle for their gravitational attraction to have a magnitude of \(2.3 \times 10^{-12} \mathrm{~N} ?\)

Short Answer

Expert verified
The separation must be approximately 18.94 meters.

Step by step solution

01

Understanding the problem

We need to find the distance between two particles with given masses, so their gravitational force magnitude equals a specified value.
02

Identify the formula

The gravitational force between two masses is given by the formula: \[ F = \frac{G m_1 m_2}{r^2} \]where \( F = 2.3 \times 10^{-12} \mathrm{~N} \), \( m_1 = 5.2 \mathrm{~kg} \), \( m_2 = 2.4 \mathrm{~kg} \), and \( G = 6.674 \times 10^{-11} \mathrm{~m^3~kg^{-1}~s^{-2}} \) is the gravitational constant.
03

Rearrange the formula

Rearrange the formula to solve for \( r \):\[ r = \sqrt{\frac{G m_1 m_2}{F}} \]
04

Plug in the values

Substitute the given values into the formula:\[ r = \sqrt{\frac{(6.674 \times 10^{-11} \mathrm{~m^3~kg^{-1}~s^{-2}})(5.2 \mathrm{~kg})(2.4 \mathrm{~kg})}{2.3 \times 10^{-12} \mathrm{~N}}} \]
05

Calculate the separation distance

Calculate the result using a calculator:\[ r = \sqrt{\frac{8.32608 \times 10^{-10} \mathrm{~m^3~s^{-2}}}{2.3 \times 10^{-12} \mathrm{~N}}} \]\[ r = \sqrt{3.6196 \times 10^2 \mathrm{~m^2}} \]\[ r \approx 18.94 \mathrm{~m} \]

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Key Concepts

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

Mass
Mass is a fundamental property of matter. It is essentially the amount of "stuff" in an object. Mass determines how much an object will resist changes to its motion when a force is applied—a concept known as inertia. It is measured in kilograms (kg) in the metric system.

In gravitational force calculations, mass is crucial because it directly affects the force exerted between two objects. The interaction between masses is governed by gravity, the force that pulls objects toward each other.
  • A larger mass means a stronger gravitational pull.
  • Two objects with more mass have a greater gravitational force between them.
  • Gravity acts even if the objects are far apart, though its effect lessens with distance.
When working with problems involving gravitational force, understanding each mass involved allows us to predict the nature of their interaction. For instance, the typical symbols used are \( m_1 \) and \( m_2 \), referring to the two masses in the formula for gravitational force.
Distance Calculation
Distance calculation is key when figuring out how strong the gravitational force is between two objects. The formula involves dividing by the square of the distance (\( r \)) between the centers of the two masses. As the distance increases, the gravitational force decreases sharply.

This is because in the gravitational force formula:
  • Force is inversely proportional to the square of the distance. This means doubling the distance makes the force four times weaker.
  • To determine the distance using the gravitational force formula, rearrange and solve for \( r \).
  • The formula becomes: \[ r = \sqrt{\frac{G m_1 m_2}{F}} \]
So, understanding and calculating distance is vital for predicting gravitational interactions. In practical problems, make sure to substitute correct values and use a calculator for precision.
Gravitational Constant
The gravitational constant, denoted as \( G \), is a key factor in calculating gravitational force. It serves as a proportionality factor in the universal law of gravitation formula. Without \( G \), we wouldn't be able to quantify the gravitational force accurately.

Here are a few key points about the gravitational constant:
  • \( G \) is approximately equal to \( 6.674 \times 10^{-11} \mathrm{~m^3~kg^{-1}~s^{-2}} \).
  • It is used universally for any pair of masses, no matter where they are located in the universe.
  • The small magnitude of \( G \) indicates that gravitational forces are relatively weak compared to other forces, like electromagnetic forces.
Having \( G \) in calculations ensures that the force results are in the right unit and scale, reflecting the true nature of gravitational interactions between any two objects. Make sure you understand \( G \) as it allows gravitational force formulas to be used for practical scenarios and experimental calculations.

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Most popular questions from this chapter

Certain neutron stars (extremely dense stars) are believed to be rotating at about 1 rev/s. If such a star has a radius of \(20 \mathrm{~km}\), what must be its minimum mass so that material on its surface remains in place during the rapid rotation?

Mile-high building. In 1956, Frank Lloyd Wright proposed the construction of a mile-high building in Chicago. Suppose the building had been constructed. Ignoring Earth's rotation, find the change in your weight if you were to ride an elevator from the street level, where you weigh \(600 \mathrm{~N}\), to the top of the building.

The Martian satellite Phobos travels in an approximately circular orbit of radius \(9.4 \times 10^{6} \mathrm{~m}\) with a period of \(7 \mathrm{~h} 39 \mathrm{~min}\). Calculate the mass of Mars from this information.

Assume a planet is a uniform sphere of radius \(R\) that (somehow) has a narrow radial tunnel through its center (Fig. 13-7). Also assume we can position an apple anywhere along the tunnel or outside the sphere. Let \(F_{R}\) be the magnitude of the gravitational force on the apple when it is located at the planet's surface. How far from the surface is there a point where the magnitude is \(\frac{1}{2} F_{R}\) if we move the apple (a) away from the planet and (b) into the tunnel?

An object lying on Earth's equator is accelerated (a) toward the center of Earth because Earth rotates, (b) toward the Sun because Earth revolves around the Sun in an almost circular orbit, and (c) toward the center of our galaxy because the Sun moves around the galactic center. For the latter, the period is \(2.5 \times 10^{8} \mathrm{y}\) and the radius is \(2.2 \times 10^{20} \mathrm{~m} .\) Calculate these three accelerations as multiples of \(g=9.8 \mathrm{~m} / \mathrm{s}^{2}\)

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