Question

At the Earth's surface, a projectile is launched straight up at a speed of 10.5 km/s. To what height will it rise? Ignore air resistance

Answer 1

KE = (1/2)m*v^2

Re = radius of the Earth
M = mass of the Earth
h = maximum altitude above the Earths surface
r = maximum distance from the Earths center = Re + h

PE0 = potential energy at the surface = -GM*m/Re
PE1 = potential energy at max altitude = -GM*m/r

KE = PE1 - PE0
(1/2)m*v^2 = -GM*m/r - (-GM*m/Re) = GM*m(1/Re - 1/r)
v^2/[2*GM] = 1/Re - 1/r

1/Re - 1/r = 1/Re - 1/(Re + h) = h/[Re*(Re + h)] where h = maximum altitude
Re*(Re + h)*v^2/(2*GM) = h
Re^2*v^2/(2*GM) = h[1 - Re*v^2/(2*GM)]

I am going to set K = Re*v^2/(2*GM) so we have:
h = Re*K = h(1 - K)
h = Re*K/(1 - K)

Let us now calculate K.
Re = 6.378x10^6 m
G = 6.673x10^(-11) m&^3/(kg s^2)
M = 5.9742x10^24 kg
v = 10.5 km/s = 1.05x10^4 m/s

K = (6.378x10^6)(10.5)^2/()/[2*6.673x10^(-11... = 0.7436
K/(1 - K) = 2.9
h = Re*2.9

So the maximum altitude will be 2.9 times the radius of the Earth.
Maximum altitude = 1.56x10^7 m = 15,600 km