Question

A rocket, launched from the Earth, shuts off its engines when its speed is 5.40 km/s. It reaches a maximum altitude of 2820 km before it falls back to Earth. At what altitude did the rocket's engines shut off? The mass and radius of the Earth are 5.97 10^24 kg and 6.37 10^6 m, respectively.

Answer 1

Using Energy conservation on meteor:
KEi + GPEi = KEf + GPEf
KEi = 0.5*m*Vi^2 = Initial kinetic energy at 'H' height,
Vi = 5.40 km/sec = 5400 m/sec.
KEf = 0.5*m*Vf^2
Vf = final speed of meteor = 0
GPEi = -G*m*Me/Ri = Initial Gravitational Potential Energy at 'H' height
GPEf = -G*m*Me/Rf = Final GPE at maximum altitude
Ri = (6.37*10^6 + H) m
Rf = (6.37*10^6 + 2.82*10^6) m = 9.19*10^6 m
Me = mass of earth = 5.97*10^24 kg
G = Gravitational Constant = 6.674*10^-11

H = altitude where rocket's engines shut off = ??
So, Using these values
0.5*m*Vi^2 - G*m*Me/Ri = 0.5*m*Vf^2 - G*m*Me/Rf
0.5*5400^2  - (6.674*10^-11)(5.97*10^24)/(6.37*10^6 + H) = 0.5*0 - (6.674*10^-11)(5.97*10^24)/(9.19*10^6)

0.5*5400*5400/((6.674*10^-11)*(5.97*10^24)) = [1/(6.37*10^6 + H) - 1/(9.19*10^6)]

3.66*10^-8 = [((9.19-6.37)10^6 - H)/((6.37*10^6 + H)(9.19*10^6))

(3.66*10^-8)(9.19*10^6)[(6.37*10^6 + H)] = 2.82*10^6 - H

2.14*10^6 + 0.336*H = 2.82*10^6 - H

1.336*H = 0.68*10^6

H = (0.68/1.336)*10^6

H = 509 km