Launch Velocity of Spacecraft at ISS: Solving for v_L

[compwiz3000]

If we launch a satellite to a circular orbit around the Earth at height 357.1 km, to find the velocity needed at launch, do we just set the energies equal?:
$$- \frac {\mu}{2\left(r_E + h\right)} = \frac {v_L^2}{2} - \frac {\mu}{r_E}$$
and then solve for $$v_L$$?

$$\mu = GM$$, where $$M$$ is the mass of the Earth.
Then, in space, the mechanical energy is
$$\frac {v^2}{2} - \frac {\mu}{r}.$$
Using centripetal force, we have
$$\frac {m v^2}{r} = \frac {\mu m}{r^2} \implies v^2 = \frac {\mu}{r},$$
so the mechanical energy is
$$- \frac {\mu}{2r}.$$
Since the mechanical energy is conserved at the surface, we set it equal to that mechanical energy to find the launch velocity.

$$- \frac {\mu}{2\left(r_E + h\right)} = \frac {v_L^2}{2} - \frac {\mu}{r_E}$$
and then solve for $$v_L$$.

Is my reasoning for the launch velocity correct?

 

[Bob S]

If you gave the satellite the needed velocity at launch, the satellite would orbit the Earth with a perigee that would be several miles above the treetops. Secondly, because the Earth is rotating with an angular speed of over 1000 mph at the equator, the easterly velocity at Cape Canaveral is about 1000 cos(28 degrees) = 880 mph.

 

[compwiz3000]

What do you mean by "needed velocity at launch"?
And is my approach correct?

 

[Bob S]

You have correctly identified the vertical velocity required to reach the ISS orbit, and the ISS orbital velocity, but the actual trajectory is probably one that first reaches the top of the atmosphere (50 Km?) and an orbital velocity (~7880 kps (kilometers per sec))that would be needed to reach the minimum orbital momentum and maintain a circular orbit in case of loss of thrust. Then, by boosting the velocity to about 7960 kps, the satellite would be on an elliptical orbit that would reach the ISS orbit as an apogee. But when arriving at the ISS orbit, its velocity would be only 7610 kps, so it would have to boost its velocity to about 7700 kps to maintain a circular orbit. At this time, its period would be about 91.6 minutes. By dropping down 10 km in altitude, it could shave maybe 12 sec per revolution and catch the ISS in a few turns if necessary.

 

[compwiz3000]

The path you are describing is the Hohmann transfer orbit, right?
How much velocity boost would I be saving in your case?

It is not feasible to directly launch to the ISS?

Also, if I use the energy conservation law with an initial thrusting of $$v_i$$ and a final velocity of $$0$$, the satellite would not have enough velocity to undergo circular motion, correct?

 

[Bob S]

The path you are describing is the Hohmann transfer orbit, right??

Yes.

How much velocity boost would I be saving in your case??

Don't know.

It is not feasible to directly launch to the ISS??

For safety reasons, the satellite has to quickly reach an orbital (not vertical) velocity that will have sufficient angular momentum to circle the Earth above the atmosphere.

Also, if I use the energy conservation law with an initial thrusting of $$v_i$$ and a final velocity of $$0$$, the satellite would not have enough velocity to undergo circular motion, correct?

Probably not. Any orbit that begins on the Earth with final velocities will be on an elliptical orbit (neglecting air drag) that intersects the Earth in less than one revolution.