Differential equation for velocity

• Apr 21st 2012, 06:21 PM
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
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
• Apr 22nd 2012, 07:00 AM
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.
• Apr 22nd 2012, 07:49 AM
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?
• Apr 22nd 2012, 08:07 AM
Re: Differential equation for velocity
That's right. In your question it self has asked to find the initial velocity hasn't it?
• Apr 22nd 2012, 08:22 AM
Re: Differential equation for velocity
Yes, to get to a certain height. i will work on it, then post what i have.
• Apr 22nd 2012, 08:50 AM
Re: Differential equation for velocity
v^2 = 2g((R^2)/(R+h))-2gR+V(0)^2

v0^2 = 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 V0

v0 = √(2gR) - √(2g((R^2)/(R+h)))

R= 6378.1 km
h= 450 km
g= 9.80665 m/s2 = 127094.184 km/hr2

v0 = √((2)(127094.184)(6378.1)) - √(2(127094.184)((6378.1^2)/(6378.1+450)))

v0 = 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/s2

i get v0 = 374.7107 m/s

is this more or less accurate?
• Apr 22nd 2012, 09:16 AM
Re: Differential equation for velocity
$\displaystyle v^2=2g\left(\frac{R^2}{R+h}\right)-2gR+v_0^2$ right...
$\displaystyle v_0^2=2g\left[R-\left(\frac{R^2}{R+h}\right)\right]$
$\displaystyle v_0=\sqrt{2g\left[R-\left(\frac{R^2}{R+h}\right)\right]$ not $\displaystyle v_0=\sqrt{2gR}-\sqrt{2g\left(\frac{R^2}{R+h}\right)}$