Final answer:
The maximum distance the rocket travels from the earth's surface before falling back can be determined by equating the kinetic energy at burnout to the gravitational potential energy at the peak altitude, using principles of conservation of mechanical energy.
Step-by-step explanation:
The question asked deals with the dynamics of a rocket that has been fired straight upwards and then becomes free of Earth's gravitational pull at a certain altitude. The rocket in question burns out at an altitude of 250 km while traveling at 6.00 km/s. The maximum distance the rocket travels from the earth's surface before falling back is determined by using the principles of conservation of mechanical energy.
Considering the rocket is too far from the Earth's surface for gravity to have a significant impact upon burnout, the maximum altitude the rocket will reach can be found by equating the kinetic energy at burnout to the potential energy at the peak altitude. From conservation of energy:
Kinetic Energy (KE) at burnout = Potential Energy (PE) at maximum altitude
(1/2)mv^2 = GMm/R
Where m is the mass of the rocket, v is the velocity at burnout (6.00 km/s), G is the gravitational constant, M is the mass of the Earth, and R is the distance from the Earth's center to the maximum altitude of the rocket.
Solving for R will give the maximum distance from the Earth's center. To find the distance from the Earth's surface, subtract the Earth's radius from R.
However, since this is a conceptual explanation, we don't provide the actual computation here. Nonetheless, typically, one would use the given initial kinetic energy (with the velocity of 6.00 km/s at a height of 250 km), the gravitational constant (G), and known values for the mass (M) and radius of the Earth to find the maximum distance (R).