Problem 10 - University Physics with Modern Physics

Question

(a) Assuming that only gravity is acting on it, how far does an electron have to be from a proton so that its acceleration is the same as that of a freely falling object at the earth's surface? (b) Suppose the earth were made only of protons but had the same size and mass it presently has. What would be the acceleration of an electron released at the surface? Is it necessary to consider the gravitational attraction as well as the electrical force? Why or why not?

Step-by-Step Solution

Answer

(a) An electron needs to be very close to a proton for Earth's free-fall acceleration; (b) Electrical forces dominate gravitational ones due to their relative strengths.

Step 1: Understanding Free-Fall Acceleration

The acceleration of an object in free fall near the Earth's surface is due to gravity and is approximately \( g = 9.81 \, \text{m/s}^2 \). This is the acceleration we want the electron to experience due to the gravitational force exerted by the proton.

Step 2: Calculating Gravitational Force Between Electron and Proton

The gravitational force between two masses \( m_1 \) and \( m_2 \) separated by a distance \( r \) is given by \[ F_g = \frac{G m_1 m_2}{r^2}, \]where \( G = 6.674 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \) is the gravitational constant. For an electron and a proton, \( m_1 = m_e = 9.109 \times 10^{-31} \, \text{kg} \) and \( m_2 = m_p = 1.673 \times 10^{-27} \, \text{kg} \).

Step 3: Setting Electron's Gravitational Acceleration Equal to Free-Fall Acceleration

The acceleration due to gravity at distance \( r \) from the proton equates to:\[ a = \frac{Gm_p}{r^2} = g, \]where \( a \) is the acceleration we want, and \( g = 9.81 \, \text{m/s}^2 \). Solve for \( r \):\[ r^2 = \frac{Gm_p}{g}, \]\[ r = \sqrt{\frac{Gm_p}{g}}. \]

Step 4: Solving for r

Substitute known values:\[ r = \sqrt{\frac{(6.674 \times 10^{-11} \, \text{N m}^2/\text{kg}^2)(1.673 \times 10^{-27} \, \text{kg})}{9.81 \, \text{m/s}^2}} \]This gives the distance \( r \).

Step 5: Calculating Earth's Modified Charge Density

If the Earth were composed purely of protons, the mass is unchanged but each particle now carries a charge. The charge of a proton is \( e = 1.602 \times 10^{-19} \, \text{C} \). The mass remains \( 5.972 \times 10^{24} \, \text{kg} \).

Step 6: Calculating Electrical Force on Electron

The electrical force on an electron at this charged Earth’s surface is given by Coulomb's Law:\[ F_e = \frac{k e^2}{r^2}, \]where \( k = 8.988 \times 10^9 \, \text{N m}^2/\text{C}^2 \), \( e \) is the electron charge, and \( r \) is Earth's radius.

Step 7: Equating Forces to Find Electron's Acceleration

For the electron, the net force is almost entirely from the electrical force, not gravitational:\[ a = \frac{k e^2}{m_e r^2}. \]

Step 8: Conclusion for Negligibility of Gravitational Force

The ratio of the electric force to the gravitational force is substantial due to the relative magnitudes of the constants involved. This means electrical forces vastly overshadow gravitational ones in this scenario.

Key Concepts

Free-Fall Acceleration

When you drop an object near the Earth's surface without any other forces acting, it accelerates downward due to gravity. This is known as free-fall acceleration and is denoted by the symbol \( g \). On Earth, \( g \) has a consistent value of about \( 9.81 \, \text{m/s}^2 \). This constant gravitational acceleration is key in problems where we need to consider the motion of objects solely influenced by Earth's gravity.
In the case of an electron experiencing the same acceleration due to a proton, it would mean that the forces acting on it mimic that of an object freely falling on Earth. Despite the incredibly small masses involved, the principles remain the same: seek a scenario where the gravitational pull from a proton causes the electron to accelerate at \( 9.81 \, \text{m/s}^2 \).
If an electron were released on Earth's surface under normal conditions, it would undergo this free-fall acceleration due to Earth's gravity. However, when evaluating forces on atomic particles, especially within a different context like the proton scenario, this simplification serves to compare gravitational versus other fundamental forces at play.

Coulomb's Law

Coulomb's Law describes the force exerted by charged objects on one another. It is expressed mathematically as: \[ F_e = \frac{k |q_1 q_2|}{r^2}, \]where:

• \( F_e \) is the electric force between two charges.
• \( k = 8.988 \times 10^9 \, \text{N m}^2/\text{C}^2 \) is Coulomb's constant.
• \( q_1 \) and \( q_2 \) are the magnitudes of the charges.
• \( r \) is the distance between the charges.

Coulomb's Law shows that electric forces increase with the charge amount but decrease with the square of the distance between them.
This law comes into play in scenarios involving charged particles such as electrons and protons. It outlines how these particles interact electrically, exerting forces of attraction or repulsion depending on their charge signs.
In situations where expressive electric fields like Earth's hypothetical proton composition are considered, the electric force can be dominant. Thus, it renders gravitational effects nearly negligible by comparison, emphasizing the potency of electromagnetic interactions at short ranges.

Gravitational Constant

The gravitational constant, denoted as \( G \), is a fundamental constant in physics representing the strength of gravity in the universe. It appears in the universal law of gravitation:\[ F_g = \frac{G m_1 m_2}{r^2} \] where:

• \( F_g \) is the gravitational force between two masses \( m_1 \) and \( m_2 \).
• \( G = 6.674 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \).
• \( r \) is the distance between the centers of the two masses.

The value of \( G \) is tiny, reflecting gravity's relatively weak influence compared to other fundamental forces like electromagnetism. Despite its small magnitude, \( G \) is crucial for calculations involving astronomical bodies and derivations of planetary orbits.
In the context of particle interactions, such as between an electron and a proton, \( G \) helps calculate the minuscule gravitational pull these particles would exert on each other. However, given their small masses, this force is generally much less significant than other forces like the electric force dominated by Coulomb's Law.

Electric Force

Electric force is a fundamental aspect of physics governing how charged particles like electrons and protons interact. According to Coulomb's Law, charged objects exert forces on each other, which can be either attractive or repulsive based on their charges.

• For like charges, the force is repulsive.
• For opposite charges, the force is attractive.

It is important to note that the electric force is significantly more powerful than the gravitational force, particularly over short distances. This powerful force influences many physical phenomena, especially at the atomic and molecular level.
In educational experiments or problem-solving situations, such as when comparing gravitational and electric forces, it is common to find the electric force predominates. This principle becomes evident when determining an electron’s acceleration because the overwhelming influence of electric force often overshadows any gravitational interaction.
This is why, when tasked to equate an electric force scenario to a gravitational one, the emphasis is on matching accelerations rather than forces themselves, as the scales at which they operate are vastly different.