Differential equation for velocity

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

Re: Differential equation for velocity

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.

Re: Differential equation for velocity

so should i be trying to solve for v(0) instead of v? i guess you're saying the velocity goes to 0 when we reach the height we want to achieve right?

Re: Differential equation for velocity

That's right. In your question it self has asked to find the initial velocity hasn't it?

Re: Differential equation for velocity

Yes, to get to a certain height. i will work on it, then post what i have.

Re: Differential equation for velocity

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?

Re: Differential equation for velocity

Now your main equation is,

$\displaystyle v^2=2g\left(\frac{R^2}{R+h}\right)-2gR+v_0^2$ right...

with v=0, we have,

$\displaystyle v_0^2=2g\left[R-\left(\frac{R^2}{R+h}\right)\right]$

$\displaystyle v_0=\sqrt{2g\left[R-\left(\frac{R^2}{R+h}\right)\right]$ not $\displaystyle v_0=\sqrt{2gR}-\sqrt{2g\left(\frac{R^2}{R+h}\right)}$

Re: Differential equation for velocity

OMG you are right! i feel a fool. im getting matching answers now. of 2.8713 km/s and 2871.2917 m/s