Part 1: Find the minimum kinetic energy needed for a 4.20 x10^4 kg rocket to esc
ID: 1483206 • Letter: P
Question
Part 1: Find the minimum kinetic energy needed for a 4.20 x10^4 kg rocket to escape the moon if it is launched from the moons surface. (Express answers in 3 sig figs) Mass of moon = 7.35 x10^22 kg. Radius of Moon = 1740 km
Part 2: How much work is done by the moons gravitational field as a 995 kg meteor comes in from outer space and impacts on the moon's surface? Assume that far from the moon the meteor is at rest. (express using 3 sig figs)
Part 3: show that the minimum speed needed for an object to "break free" from the gravitational attraction of the moon is ve= square root 2GMm/r. In other words, derive the equation for escape velocity. Hint: utilize the conservation of energy.
Explanation / Answer
escape velocity of moon = sqrt(2 * G * mass of moon / radius of moon)
escape velocity of moon = sqrt(2 * 6.676 * 10^-11 * 7.35 * 10^22 / 1740000)
escape velocity of moon = 2374.88 m/s
kinetic energy = 0.5 * mass * velocity
kinetic energy = 0.5 * 4.2 * 10^4 * 2374.88^2
kinetic energy required to escape the moon = 118441155302 J
initial potential energy of meteor = 0 J
final potential energy of meteor = G * mass of moon * mass of meteor / radius of moon
final potential energy of meteor = 6.67 * 10^-11 * 7.35 * 10^22 * 995 / 1740000
final potential energy of meteor = 6.67 * 10^-11 * 7.35 * 10^22 * 995 / 1740000
final potential energy of meteor = 2803412500 J
work done by moon's gravitational field = change in potential energy
work done by moon's gravitational field = 2803412500 J