Final answer:
The kinetic energy to rest energy ratio of a particle traveling at 0.97c is calculated using relativistic physics. The Lorentz factor must be determined first, which will result in a ratio much greater than 1.5. Without the precise calculation, the exact ratio is not presented here.
Step-by-step explanation:
The question involves applying the principles of relativistic physics to calculate the kinetic energy of a particle traveling at relativistic speeds. When an object moves at a speed close to the speed of light (denoted as c), classical mechanics no longer applies, and instead, one must use the relativistic energy-momentum relation.
The total energy E of a particle is given by the equation E = √((m_0c^2)^2 + (pc)^2), where m_0 is the rest mass of the particle, p is its momentum, and c is the speed of light. The kinetic energy (KE) is the total energy minus the rest energy m_0c^2, so KE = E - m_0c^2. The momentum p can be found using p = γm_0v, where γ (gamma) is the Lorentz factor γ = 1/√(1-(v/c)^2) and v is the velocity of the particle.
For a particle traveling at v = 0.97c, γ is significantly greater than 1, resulting in the kinetic energy being several times the rest energy. Without the exact value of γ, we cannot provide the exact ratio, but we can say it will be much greater than the ratio of 1.5 mentioned for velocities around 0.9c. To find the exact ratio, one would calculate γ for v = 0.97c and then use it to find the total energy and subtract the rest energy to find the kinetic energy. The ratio of the kinetic energy to rest energy would then be KE to m_0c^2.