# (SR) Rocket dynamics tutorial

Delta_V = Isp * g * ln(Mo/Mf)

with a rocket engine of specific ISP and initial mass M0 to final mass Mf (due to propellant loss)

The rocket accelerates using that Isp to altitude of 110km straight up from the ground, perpendicular to the ground...or is that not going to happen?
given enough thrust, you could do this. It would take a delta_v of 1.459 km/sec. All this Delta_v will have been used up just fighting gravity to reach that altitude with none left over. It will have 0 vertical velocity upon reaching 110 km.
Once the rocket is at that altitude, will it just fall to the ground slowly? or will it circle the Earth?

(assume perfect scenario)

Please bear with me, I need to understand it.

It will fall back to Earth. As noted before, it would take a velocity of over 7 km/sec to maintain an orbit at that altitude.

Now your rocket will have some horizontal velocity that it picked up from the rotation of the Earth, but even if it was launched at the equator this only amounts to 0.469 km/sec; not nearly enough to maintain an orbit.

Janus to orbit the Earth x number of times, would it be better (less thrust needed) to go to a specified altitude which will have horizontal velocity picked up from the rotation of the Earth or would it be better to go to a lower altitude and use thrust for the horizontal motion?

given enough thrust, you could do this. It would take a delta_v of 1.459 km/sec.

you just used the vf^2=vi^2+2ad equation for that, right? :bugeye:

And do you think directional-vector thrusting should be used for the horizontal velocity? or should the thing use wings sort of like Pegasus rocket to take advantage of the air drag?

Also...how did you arrive to the 0.469 km/sec gain from Earth rotational motion?

Janus to orbit the Earth x number of times, would it be better (less thrust needed) to go to a specified altitude which will have horizontal velocity picked up from the rotation of the Earth or would it be better to go to a lower altitude and use thrust for the horizontal motion?

A lower orbit will always take less delta_v. Besides, you are going to have that velocity picked up from the Earth no matter what the altitude of your orbit.

you just used the vf^2=vi^2+2ad equation for that, right? :bugeye:

No I used

$$\frac{mv^2}{2}= GMm \left(\frac{1}{r_1}-\frac{1}{r_1+h} \right)$$

Where r1 is the radius of the Earth and h is the altitude the rocket reaches.

Solve for v.

Also...how did you arrive to the 0.469 km/sec gain from Earth rotational motion?

The Earth rotates once every 24hrs (86400 secs).

A point on the equator traces a circle with circumference of

$$2 \pi 6378000 = 40074156$$meters per rotation.

$$\frac{40074156}{86400} = 464 m/s$$
(when a copied the number the first time from paper, I misread the 4 as a 9. What can I say, I have bad hand writing.)

Nice post...Thank you very much for your efforts!!