Re: spaceship accelearation eventual velocity

Calculating the time taken by a space ship theoretically is complicated 
as shown by the following equations:

Since the mass(fuel consumption,detached parts while in flight) of the 
space ship varies continuosly,its acceleration depends on the
first rocket equation:
              ma = R*v(relative)
                   where m is the mass of the rocket, a is its 
acceleration ,R is the fuel consumption rate,v(relative velocity) is the 
rocket velocity relative to the exhaust gas
               R*V(relative) is called as Thrust.
        This equation gives acceleration at that instant.

Also the velocity of the rocket is given by second rocket equation
               v=v(relative)ln(M1/M2)   M1 and M2 are mass of rocket at 
different instances,M1 being earlier.

The moving rocket is acted upon by Air resistance and gravity drag.
    T-mg-F(air)=ma        where T is the thrust,mg is due to gravity 
 A low accelerating rocket cannot go much high unless its upward force is 
more than downward force.
The gravitation force will gets lesser with altitude but is never zero.

This requirement is expressed in terms of Delta-V. 
Delta-V = Integrate from initial to final time}[Thrust/Instantaneous mass]

For example it is found that to reach a speed of 7.8 km/s orbit requires 
a delta-v of between 9 and 10 km/s.

If a rocket travels a straight path,then the time taken to reach a 
certain distance is
                 t2=[(v2-v1)/a]+t1    v1 and v2 are initial and final 
velocities of the rocket respectively
                                      t1 is the instantaneous time of 
takeoff.
But since the rocket is to be in orbit,it has to take a curved path 
gradually.
Hence the equation of path is found by resolving it into horizontal and 
vertical motions to find horizontal and vertical range.

While the time taken to reach a certain distance is one stage,maintaining 
the orbit is the second stage.
To maintain the orbit naturally,the centripetal force of the orbiting 
rocket should be balanced by gravitation of the earth.
Hence the rocket needs some time to accelerate to the speed to achieve 
this purpose.
                       Mg=mv*v/r        r=radius of curvature of the   
orbit.

If the launch is from space(neglecting gravity,air resistance) and the 
motion is straight line,
    then a=(v2-v1)/(t2-t1)=v2/t2 for v1=0 and t1=0
        
           1.5*9.81=14.715=v2/t2

       substitute v2=d/t2    d=distance travelled

       14.715=d/(t2*t2)
       
       If d=250 miles=250*1.6093*1000 meters = 402325 meters,
     then t2=square root(402325/14.715)
          t2=165 seconds.

 v2=(d/t2)=(402325/165)=2438.33 m/s.


References:http://www.phy.ntu.edu
           Newton.dep.anl.gov