rocket eqn

DgrayMan · **Joined:** Sat, 21 Jan 2012 03:59:22 UTC **Posts:** 182

sorry trying to delete

outermeasure · **Posted:** Sun, 22 Apr 2012 18:21:50 UTC

DgrayMan wrote:

I am trying to find initial velocity to reach a certain height

So here's what i got

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? are my units of measurement correct when i plug them in the equation?

No!
$v_0^2=2gR^2\left(\dfrac{1}{R}-\dfrac{1}{R+h_{max}}\right)$
does not mean
$v_0=\sqrt{2gR^2}\left(\sqrt{\dfrac{1}{R}}-\sqrt{\dfrac{1}{R+h_{max}}}\right).$

DgrayMan · **Joined:** Sat, 21 Jan 2012 03:59:22 UTC **Posts:** 182

you are right!

Shadow · **Posted:** Sun, 22 Apr 2012 18:33:45 UTC

outermeasure wrote:

DgrayMan wrote:

I am trying to find initial velocity to reach a certain height

So here's what i got

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? are my units of measurement correct when i plug them in the equation?

No!
$v_0^2=2gR^2\left(\dfrac{1}{R}-\dfrac{1}{R+h_{max}}\right)$
does not mean
$v_0=\sqrt{2gR^2}\left(\sqrt{\dfrac{1}{R}}-\sqrt{\dfrac{1}{R+h_{max}}}\right).$

But then that might mean that $\sqrt{a+b}\ne \sqrt{a}+\sqrt{b}$ ....what will we tell the freshmen?

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rocket eqn

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