Chapter 18: Problem 26
Suppose astronomers discover a \(3-M_{\text {sun }}\) black hole located a few light-years from Earth. Should they be concerned that its tremendous gravitational pull will lead to Earth's untimely demise?
Short Answer
Expert verified
No, the gravitational pull of a 3-solar-mass black hole a few light-years away is too weak to pose a threat to Earth.
Step by step solution
01
Understand the Gravitational Force
First, recall that the gravitational force between two masses is given by Newton’s law of gravitation: \[ F = G \frac{m_1 m_2}{r^2} \]Where \( G \) is the gravitational constant, \( m_1 \) and \( m_2 \) are the masses of the objects and \( r \) is the distance between them.
02
Calculate Earth-Black Hole Distance in Meters
Assume the black hole is 4 light-years away (a few light-years). Convert light-years to meters: \[ 1 \text{ light-year} \approx 9.46 \times 10^{15} \text{ meters} \]Therefore, \[ 4 \text{ light-years} \approx 4 \times 9.46 \times 10^{15} = 3.784 \times 10^{16} \text{ meters} \]
03
Plug in the Values
Plug in the mass of the black hole \( 3M_{\text{sun}} \), mass of Earth \( M_{\text{Earth}} \), distance \( r \), and gravitational constant \( G \) into the formula: \[ F = 6.67 \times 10^{-11} \frac{(3 \times 1.989 \times 10^{30})(5.972 \times 10^{24})}{(3.784 \times 10^{16})^2} \]
04
Simplify and Solve
Calculate the force step by step:1. Multiply the masses: \[ 3 \times 1.989 \times 10^{30} = 5.967 \times 10^{30} \]2. Multiply with Earth’s mass: \[ 5.967 \times 10^{30} \times 5.972 \times 10^{24} = 3.562 \times 10^{55} \]3. Square the distance: \[ (3.784 \times 10^{16})^2 = 1.4310 \times 10^{33} \]4. Multiply by G and simplify: \[ F = 6.67 \times 10^{-11} \frac{3.562 \times 10^{55}}{1.4310 \times 10^{33}} = 1.66264 \times 10^{12} \text{ Newtons} \]
05
Understand the Result
Compare the result to typical gravitational forces experienced by Earth. The Sun's gravitational force on Earth is roughly: \[ F_{\text{Sun}} = G \frac{(1.989 \times 10^{30})(5.972 \times 10^{24})}{(1.496 \times 10^{11})^2} = 3.54 \times 10^{22} \text{ Newtons} \]The black hole's force of \( 1.66264 \times 10^{12} \text{ Newtons} \) is significantly smaller than the Sun’s.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Newton’s Law of Gravitation
Newton’s Law of Gravitation is a crucial principle in understanding the forces that celestial objects exert on each other. It 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. The mathematical representation is: \[ F = G \frac{m_1 m_2}{r^2} \]Here, \( F \) represents 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 them. This law helps us calculate the gravitational force exerted by a black hole on Earth. Using this understanding, we can compare this force with the gravitational influence we experience from the Sun.
Mass
Mass is a fundamental property of matter, indicating how much matter an object contains. In Newton’s law, mass is a critical variable because the gravitational force depends directly on the masses of the two objects involved. When dealing with celestial bodies like black holes and planets:
- The mass of the black hole is often given in terms of the Sun’s mass, for example, \( 3M_{\text{sun}} \), meaning three times the Sun's mass.
- The mass of Earth is a well-known constant, approximately \( 5.972 \times 10^{24} \text{ kg} \).
Distance
Distance is another crucial factor in Newton’s Law of Gravitation. The force of gravity decreases rapidly as the distance between two objects increases because the gravitational force is inversely proportional to the square of the distance \( r \) between them. This means that if the distance between two objects doubles, the gravitational force is reduced to a quarter. In our example:
- The black hole is four light-years away from Earth.
- Each light-year is approximately \( 9.46 \times 10^{15} \text{ meters} \).
Light-Year
A light-year is a unit of distance used in astronomy to express the vast distances between stars and other celestial bodies. It is defined as the distance that light travels in one year. For reference:
- One light-year is about \( 9.46 \times 10^{15} \text{ meters} \).
- So, four light-years is approximately \( 3.784 \times 10^{16} \text{ meters} \).
Gravitational Constant
The gravitational constant, denoted as \( G \), is a fundamental physical constant involved in the calculation of gravitational force between two bodies. Its value is approximately \( 6.67 \times 10^{-11} \text{ N(m}^2\text{/kg}^2 \). The gravitational constant makes sure that our units of mass and distance (in kilograms and meters, respectively) balance correctly to give us a force in Newtons. In our calculations:
- Using \( G \) ensures that the gravitational force between Earth and the black hole can be accurately determined.
- Without \( G \), we wouldn’t be able to relate the mass and distance to produce a meaningful force.