Problem 44 - College Physics

Question

Suppose a starship had a mass of \(1.25 \times 10^{9} \mathrm{~kg}\) and was initially at rest. If its "matter-antimatter engines" produced photons from electron-positron annihilation and focused them to travel backward out from the ship, how many photons would they have to emit to reach \(0.100 \%\) of the speed of light? [Hint: Use conservation of linear momentum and remember that relativity is not needed here. (Why?)]

Step-by-Step Solution

Answer

The starship must emit approximately \(3.43 \times 10^{36}\) photons.

Step 1: Identify Concepts

To solve this problem, we need to use the conservation of linear momentum to find the number of photons emitted. The starship gains momentum added by the photons emitted backward. Since photons have no mass but carry momentum of magnitude \( p = \frac{E}{c} \), where \( E \) is the energy of a photon and \( c \) is the speed of light.

Step 2: Define Ship's Momentum

When the starship reaches \(0.100\%\) of the speed of light, its momentum can be calculated. First, find the velocity \(v = 0.001c\), where \(c = 3 \times 10^8\) m/s. Thus, \( v = 3 \times 10^5 \) m/s. The momentum of the ship is given by \( p_{\text{ship}} = mv = 1.25 \times 10^9 \times 3 \times 10^5 \).

Step 3: Calculate Photon Momentum per Photon

Each photon carries momentum \( p_{\text{photon}} = \frac{E}{c} = \frac{hf}{c} \), where \( h \) is Planck's constant \(6.626 \times 10^{-34}\), and \( f \) is the photon frequency. However, the individual photon frequency is not needed as we will find total momentum.

Step 4: Apply Conservation of Momentum

Since no external forces are acting, the system's total momentum before emission was zero, so the momentum of the starship moving forward should equal the momentum of the emitted photons in the opposite direction. Thus, \( Np_{\text{photon}} = p_{\text{ship}} \), where \( N \) is the number of emitted photons.

Step 5: Solve for Number of Photons

Use the starship's momentum from Step 2, \( p_{\text{ship}} = 3.75 \times 10^{14} \text{ kg m/s} \), to find \( N \). Assuming photons are emitted from electron-positron annihilation \( E = hf = mc^2 = 2 \times 9.11 \times 10^{-31} \times (3 \times 10^8)^2 \), find \( p_{\text{photon}} \) then \( N = \frac{3.75 \times 10^{14}}{2p_{\text{photon}}} \).

Step 6: Calculation

Calculate photon energy from electron-positron annihilation: \( E = 1.64 \times 10^{-13} \text{ J} \). Hence, \( p_{\text{photon}} = \frac{1.64 \times 10^{-13}}{3 \times 10^8} = 5.47 \times 10^{-22} \text{ kg m/s}\). Then, \( N = \frac{3.75 \times 10^{14}}{2 \times 5.47 \times 10^{-22}} \approx 3.43 \times 10^{36} \).

Key Concepts

Photon Momentum

Photons are particles of light that are unique because they have no mass. Despite this, they can carry momentum, which is a product of mass and velocity for mass-bearing particles. For photons, their momentum can be calculated based on their energy, by using the equation:
\[p = \frac{E}{c}\]
where \( p \) is the momentum, \( E \) is the energy of the photon, and \( c \) is the speed of light. This is important because even though photons are massless, their ability to transfer momentum makes them capable of influencing other objects, like in the case of a starship being pushed forward by the photons it emits.

• Energy of a photon: found using the equation \( E = hf \) where \( h \) is Planck's constant and \( f \) is the frequency of the photon.
• Momentum influences: when photons are emitted backward, they transfer enough momentum to propel objects forward, illustrating Newton's third law.

Electron-Positron Annihilation

Electron-positron annihilation is a fascinating concept where a particle and its antiparticle collide and their mass is converted into energy. Specifically for an electron and a positron, the process results in the creation of photons. The energy released in this process can be calculated using Einstein’s famous equation:
\[E = mc^2\]
where \( m \) is the mass of the particles and \( c \) is the speed of light. Here, the combined mass of the electron and positron is transformed, producing energy equivalent to their mass. This conversion of mass into radiation is a powerful illustration of conservation principles in physics.

• Mass-energy equivalence: This principle is crucial in understanding how mass is converted entirely into energy.
• Production of photons: The annihilation typically produces two photons, balancing the initial energy and momentum of the system.
• Application in propulsion: In our exercise, these photons created from electron-positron annihilation are used to move the starship forward by being expelled backward.

Speed of Light Calculation

The speed of light, denoted by \( c \), is a fundamental constant in physics and plays a crucial role in various equations, including those used in relativity and quantum mechanics. Its value is approximately \( 3 \times 10^8 \) m/s. Calculating the related velocities and deriving expressions in physics often utilizes this constant, such as determining speeds as fractions of the speed of light.

In the context of our problem, we're interested in calculating a fractional speed based on the speed of light. For instance, when the starship reaches a velocity given as a percentage of \( c \), such as \( 0.100\% \), it's simply:\[v = 0.001c\]This approach simplifies working with high-speed calculations common in space travel scenarios, indicating how fast an object is moving relative to the ultimate speed limit set by light.

• Essential constant: Its role is pivotal in countless physics equations and theories.
• Used to compute relativistic effects: Speeds compared to \( c \) often involve adjustments using relativistic physics.
• Fractional speed calculations: Emphasizes how small portions of \( c \) still correspond to significant velocities in practical scenarios.