Question

What is the launch speed of a projectile that rises vertically above the Earth to an altitude equal to 10 REarth before coming to rest momentarily?

Answer 1

Using Energy conservation between initial and final position:

KEi + PEi = KEf + PEf

KEi = (1/2)*m*Vi^2

Vi = Initial speed of projectile = ?

m = mass of projectile

PEi = -G*m*Me/Re = potential energy at the surface of earth

PEf = -G*m*Me/h = potential energy at the altitude of 10*Re

h = Re + 10*Re = 11*Re, So

PEf = -G*m*Me/(11*Re)

KEf = 0, as at the max height projectile stops momentarily

Using these values:

(1/2)*m*Vi^2 + (-G*m*Me/Re) = 0 + (-G*m*M/(11*Re))

Vi = sqrt ((2*G*Me/Re)*(1/11 - 1)) = sqrt (20*G*Me/(11*Re))

Using known values:

Vi = sqrt (20*6.67*10^-11*5.98*10^24/(11*6.37*10^6))

Vi = 10670 m/sec = initial speed of projectile

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