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a. How does quadrupling the distance between two objects affect the gravitational force between them? b. Suppose the Sun were somehow replaced by a star with twice as much mass. What would happen to the gravitational force between Earth and the Sun? c. Suppose Earth were moved to one-third of its current distance from the Sun. What would happen to the gravitational force between Earth and the Sun?

Short Answer

Expert verified
a. The gravitational force becomes 1/16th of its original value. b. The gravitational force doubles. c. The gravitational force becomes nine times its original value.

Step by step solution

01

Understanding Gravitational Force Formula

The gravitational force between two objects is described by the formula \( F = \frac{G \cdot m_1 \cdot m_2}{r^2} \), where \( F \) is the gravitational force, \( G \) is the gravitational constant, \( m_1 \) and \( m_2 \) are the masses of the two objects, and \( r \) is the distance between the centers of the two objects.
02

Step 2(a): Quadrupling the Distance

If the distance \( r \) is quadrupled, it becomes \( 4r \). Substitute \( 4r \) into the gravitational force formula: \( F = \frac{G \cdot m_1 \cdot m_2}{(4r)^2} = \frac{G \cdot m_1 \cdot m_2}{16r^2} \). This means the gravitational force will become one-sixteenth of its original value.
03

Step 2(b): Doubling the Mass of the Star

If the mass of the Sun is doubled, \( m_2 \) becomes \( 2m_2 \). Substitute \( 2m_2 \) into the gravitational force formula: \( F = \frac{G \cdot m_1 \cdot (2m_2)}{r^2} = 2 \cdot \frac{G \cdot m_1 \cdot m_2}{r^2} \). This means the gravitational force is doubled.
04

Step 2(c): Changing Distance to One-Third

If the distance is reduced to one-third, the new distance is \( \frac{r}{3} \). Substitute \( \frac{r}{3} \) into the gravitational force formula: \( F = \frac{G \cdot m_1 \cdot m_2}{(r/3)^2} = \frac{G \cdot m_1 \cdot m_2}{\frac{r^2}{9}} = 9 \cdot \frac{G \cdot m_1 \cdot m_2}{r^2} \). This means the gravitational force becomes nine times its original value.

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Key Concepts

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

Newton's Law of Universal Gravitation
Newton's Law of Universal Gravitation is a foundational principle in physics that explains the attractive force between two bodies. This law states that every point mass attracts every other point 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. This is mathematically expressed as:\[ F = \frac{G \cdot m_1 \cdot m_2}{r^2} \]Here, \( F \) represents the gravitational force, \( G \) is the gravitational constant, approximately \( 6.674 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2 \), \( m_1 \) and \( m_2 \) are the masses of the two objects, and \( r \) is the distance between the centers of the two objects. This equation underlines that:- Larger masses exert a greater gravitational pull.- The further apart two objects are, the weaker the gravitational force they experience.This fundamental concept has not only guided scientific understanding of celestial movements but also has been critical in space exploration and understanding orbits.
Inverse Square Law
The Inverse Square Law is a powerful concept that describes how a quantity or property, such as gravitational force, decreases with the square of the distance from the source. In the context of gravity, this means that as two objects become further apart, the gravitational attraction between them decreases quite rapidly because it is divided by the square of the distance.For instance, if you increase the distance between two objects by a factor of 4, the gravitational force becomes one-sixteenth of its original value:- Original distance: \( F \propto \frac{1}{r^2} \)- Increased distance: \( F \propto \frac{1}{(4r)^2} \) which simplifies to \( \frac{1}{16r^2} \)This decrease highlights why celestial bodies that are very far apart exert insignificant gravitational forces on each other compared to those that are closer together. The Inverse Square Law is exemplified not only in gravity but also in other phenomena such as light and sound intensity, illustrating a broad application in physical sciences.
Effects of Mass and Distance on Gravity
The effects of mass and distance on gravitational forces are key to understanding how objects interact in our universe. Gravitational force increases with mass and decreases with distance.
  • Effect of Mass: As the mass of one or both objects increases, the gravitational force between them increases. For example, if you double the mass of an object, the gravitational attraction doubles, given other conditions remain constant. This is because mass is directly proportional to gravitational force, as shown in the formula: \( F \propto m \).
  • Effect of Distance: Distance plays a more complicated role due to the inverse square relationship. When you decrease the distance between two objects, the gravitational force increases, becoming significantly stronger. For instance, moving the Earth to one-third of its distance to the Sun increases the gravitational force by a factor of nine.
Understanding these effects helps in predicting the motion of planets and satellites, planning space missions, and comprehending tides and other phenomena influenced by Earth's gravity.

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Most popular questions from this chapter

Decide whether the statement makes sense (or is clearly true) or does not make sense (or is clearly false). Explain clearly; not all of these have definitive answers, so your explanation is more important than your chosen answer. I used Newton's version of Kepler's third law to calculate Saturn's mass from orbital characteristics of its moon Titan.

Be sure to show all calculations clearly and state your final answers in complete sentences. Calculate the escape velocity from each of the following. a. The surface of Mars (mass \(=0.11 M_{\text {Earth }},\) radius \(=\) \(\left.0.53 R_{\text {Earth }}\right)\) b. The surface of Mars's moon Phobos \(\left(\text { mass }=1.1 \times 10^{16} \mathrm{kg}, \text { radius }=12 \mathrm{km}\right)\) c. The cloud tops of Jupiter (mass \(\left.=317.8 \mathrm{M}_{\text {Earth }}, \text { radius }=11.2 \mathrm{R}_{\text {Earth }}\right)\) d. Our solar system, starting from Earth's orbit (Hint: Most of the mass of our solar system is in the Sun; \(\left.M_{\mathrm{Sun}}=2.0 \times 10^{30} \mathrm{kg} .\right)\) e. Our solar system, starting from Saturn's orbit

Be sure to show all calculations clearly and state your final answers in complete sentences. Using Newton's Version of Kepler's Third Law. a. Find Earth's approximate mass from the fact that the Moon orbits Earth in an average time of 27.3 days at an average distance of 384,000 kilometers. (Hint: The Moon's mass is only about \(\frac{1}{80}\) of Earth's.) \(\mathbf{b}\). Find Jupiter's mass from the fact that its moon Io orbits every 42.5 hours at an average distance of 422,000 kilometers. c. You discover a planet orbiting a distant star that has about the same mass as the Sun, with an orbital period of 63 days. What is the planet's orbital distance? d. Pluto's moon Charon orbits Pluto every 6.4 days with a semimajor axis of 19,700 kilometers. Calculate the combined mass of Pluto and Charon. e. Calculate the orbital period of a spacecraft in an orbit 300 kilometers above Earth's surface. f. Estimate the mass of the Milky Way Galaxy from the fact that the Sun orbits the galactic center every 230 million years at a distance of 27,000 lightyears. (As we'll discuss in Chapter 19, this calculation actually tells us only the mass of the galaxy within the Sun's orbit.)

Define momentum and force. What do we mean when we say that momentum can be changed only by a net force?

You and Earth attract each other gravitationally, so you should also be subject to a tidal force resulting from the difference between the gravitational attraction felt by your feet and that felt by your head (at least when you are standing \(.\) Explain why you can't feel this tidal force.

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