Cindy invites you to go on a journey with her into space in a small spaceship wi
ID: 1315667 • Letter: C
Question
Cindy invites you to go on a journey with her into space in a small spaceship with a mass of 1000kg. Using thermodynamics, she states that stellar bodies can be assumed to have an average mass of 1030kg (mass of the sun), and to move with a random velocity of 10 km/s. She wants to know the average velocity the rocket will reach after a long period of time. You ask her, "what happens if we fall into a star?" She tells you to neglect the possibility of that happening. After finding the answer, Cindy asked you to calculate the temperature of this space in Kelvin.
Explanation / Answer
In a gas in thermal equilibrium, all molecules have equal average KE. In this problem, consider the stars to be the "molecules" flying around together with a lighter "molecule" which is the spaceship. The kinetic energy of a star would be
1/2 M V^2, because the "low" speed in relativistic perspective allows us to use the non-relativistic formula.
The spaceship however, given the enormous kinetic energy it should also have when equilibrated, must be treated relativistic. The relativistic expression for kinetic energy is
T = mc^2 /sqrt(1-v^2/c^2) - mc^2.
So, equating the T of the spaceship to 1/2 M V^2 of a star allows us to solve for v:
mc^2 /sqrt(1-v^2/c^2) - mc^2 = 1/2 M V^2
1-v^2/c^2 = 1 / ( 1 + MV^2/(2mc^2))^2
v = c sqrt( 1- 1/( 1 + MV^2/(2mc^2))^2)
You can plug in the numbers but it will be very close to light speed c...
The temperature follows from the thermodynamic formula 1/2 M V^2 = 3/2 k T:
T = M V^2 /(3 k)
= 10^30 kg * 10^8 m^2/s^2 /(3*1.38*10^-23 kg m^2/s^2 /K)
= more than 10^70 Kelvin
Clearly, stars cannot be considered to be in thermodynamic equilibrium with each other...