The initial velocity cannot be 0. Unless how are you gonna send the rocket up? It is velocity at the maximum height that should be zero.
i was working on a differential equation to find the initial velocity of a rocket to get to height h.
so i have, which i got and its correct cause its the solution in the book, v^2 = 2g(R/(R+h))-2gR+V(0)^2
v = velocity
g = acceleration of gravity at earths surface
R= radius of earth
h = height
and i assume v(0) is the velocity v at height=0 so v(0)=0?
Problem is when i keep getting a negative number. so i assume I'm making an amateur mistake.
What am i doing wrong? ive tried converting etc. im using km
v^2 = 2g((R^{^2})/(R+h))-2gR+V(0)^2
v_{0}^2 = 2gR - 2g((R^{^2})/(R+h)) + v^2
so v^2 is 0 cause at the velocity of h the rocket stops. and now solve for V_{0}
v_{0} = √(2gR) - √(2g((R^{^2})/(R+h)))
R= 6378.1 km
h= 450 km
g= 9.80665 m/s^{2} = 127094.184 km/hr^{2}
v_{0} = √((2)(127094.184)(6378.1)) - √(2(127094.184)((6378.1^{^2})/(6378.1+450)))
v_{0} = 1328.1287 km/hr = 368.9246 m/s
Did i solve this correctly?
when i try to do this in metres instead of kiometers using the variables
R= 6378100 m
h= 450000 m
g= 9.80665 m/s^{2}
i get v_{0} = 374.7107 m/s
is this more or less accurate?