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Mobile App Chapter 3: Problem 39
Two identical planets, Alpha and Beta, touch each other. Khan the Conqueror suddenly halves the density of Beta, and doubles the radius of Alpha. The gravitational force between them: (A) stays the same. (B) doubles. (C) halves. (D) goes up by a factor of \(16 / 9\). (E) goes down by a factor of \(9 / 16\).
Short Answer
Expert verified
**Answer:** The gravitational force between the planets goes up by a factor of 16/9.
Step by step solution
01
Write down the initial gravitational force formula
The gravitational force between two objects can be calculated using Newton's Law of Universal Gravitation, which is given by:
\(F = G \cdot \frac{m_1 \cdot m_2}{r^2}\)
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 their centers.
02
Express masses in terms of densities and radii
Since the planets are initially identical, let's define their initial masses as \(m\) and radius as \(r_0\). Their densities are the same, and can be defined as \(\rho\). Using the formula for mass as a function of density and volume, we have that the initial masses are:
\(m_1 = m_2 = \rho \cdot \frac{4}{3} \pi r_0^3\)
03
Calculate the initial gravitational force
Plugging in the masses in terms of densities and radii into the gravitational force formula, we get:
\(F_{initial} = G \cdot \frac{(\rho \cdot \frac{4}{3}\pi r_0^3)^2}{(2r_0)^2}\)
04
Apply changes to Beta and Alpha
Now, we need to apply the changes to the planets. For Alpha, the radius doubles, so the new radius is \(2 r_0\). For Beta, the density is halved, so the new density is \(\frac{1}{2} \rho\). We can now calculate their new masses:
\(m_{1_{new}} = \rho \cdot \frac{4}{3} \pi (2r_0)^3\)
\(m_{2_{new}} = \frac{1}{2} \rho \cdot \frac{4}{3} \pi r_0^3\)
05
Calculate the new gravitational force
With the new masses, the distance between the centers of the planets would still be \(2 r_0\). So the new gravitational force is:
\(F_{new} = G \cdot \frac{m_{1_{new}} \cdot m_{2_{new}}}{(2r_0)^2}\)
06
Calculate the ratio of the new gravitational force to the initial gravitational force
To find the correct answer, we need to compare the new gravitational force to the initial one. To do so, we can calculate the ratio \(\frac{F_{new}}{F_{initial}}\):
\(\frac{F_{new}}{F_{initial}} = \frac{G \cdot \frac{m_{1_{new}} \cdot m_{2_{new}}}{(2r_0)^2}}{G \cdot \frac{(\rho\cdot \frac{4}{3}\pi r_0^3)^2}{(2r_0)^2}}\)
After simplifying, we get:
\(\frac{F_{new}}{F_{initial}} = \frac{16}{9}\)
07
Choose the right answer
Since the ratio between \(F_{new}\) and \(F_{initial}\) is \(\frac{16}{9}\), we can conclude that the gravitational force between the planets goes up by a factor of \(16 / 9\). Therefore, the correct answer is (D).
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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
Understanding the Newton's Law of Universal Gravitation is crucial for solving many physics problems involving the force of gravity between two objects. Sir Isaac Newton formulated this law, which 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 law is mathematically expressed as:
\[F = G \cdot \frac{m_1 \cdot m_2}{r^2}\]
Where:\
\[F = G \cdot \frac{m_1 \cdot m_2}{r^2}\]
Where:\
- \(F\) is the gravitational force between the two masses,
- \(G\) is the gravitational constant (\(6.674 \times 10^{-11} \, N \cdot (m/kg)^2\)),
- \(m_1\) and \(m_2\) are the masses of the two objects,
- \(r\) is the distance between the centers of the two masses.
Mass-Density-Volume Relationship
To delve into the mass-density-volume relationship, it's important to recognize that mass is a measure of the amount of matter in an object, density is the measure of mass per unit volume, and volume defines the space occupied by an object. The relationship among these three properties is given by the equation:
\[mass = density \times volume\]
In the context of gravitational force problems, this relationship allows us to calculate the mass of an object if we know its density and volume. Volume, for spherical objects like planets, can be determined using the formula for the volume of a sphere:\[ V = \frac{4}{3} \pi r^3 \]
where \(r\) is the radius of the sphere. When we alter the density or the volume (through radius, in the case of a sphere), the mass of the object changes, and according to Newton's Law of Universal Gravitation, this will consequently affect the gravitational force between two masses.
\[mass = density \times volume\]
In the context of gravitational force problems, this relationship allows us to calculate the mass of an object if we know its density and volume. Volume, for spherical objects like planets, can be determined using the formula for the volume of a sphere:\[ V = \frac{4}{3} \pi r^3 \]
where \(r\) is the radius of the sphere. When we alter the density or the volume (through radius, in the case of a sphere), the mass of the object changes, and according to Newton's Law of Universal Gravitation, this will consequently affect the gravitational force between two masses.
Gravitational Force Calculation
The calculation of gravitational force between two masses incorporates the mass-density-volume relationship and Newton's Law of Universal Gravitation. Given the mass of two objects and the distance between their centers, we can calculate the force of gravity that they exert on each other. If any changes occur to the density or size (radius for spherical objects) of the participating objects, these changes will affect the final gravitational force.
For instance, if the radius of one object doubles, its volume increases by a factor of eight (since volume is proportional to the cube of the radius), and if the object's density remains constant, its mass will increase by the same factor of eight. Applying these changes to the formula of gravitational force calculation can help deduce how alterations in physical properties influence the gravitational force. This thorough understanding is essential when solving complex problems where multiple variables might change, as it empowers students to systematically approach the solution by considering the individual implications of each variable change on the overall force.
For instance, if the radius of one object doubles, its volume increases by a factor of eight (since volume is proportional to the cube of the radius), and if the object's density remains constant, its mass will increase by the same factor of eight. Applying these changes to the formula of gravitational force calculation can help deduce how alterations in physical properties influence the gravitational force. This thorough understanding is essential when solving complex problems where multiple variables might change, as it empowers students to systematically approach the solution by considering the individual implications of each variable change on the overall force.
