Earth to Low Earth Orbit (LEO), gravity drag and potential energy

Question

A lot of references indicate takes about 9.4 km/s of delta-v to reach LEO.

https://en.wikipedia.org/wiki/Low_Earth_orbit#Orbital_characteristics

The delta-v needed to achieve low Earth orbit starts around 9.4 km/s. Atmospheric and gravity drag associated with launch typically adds 1.5–2.0 km/s to the launch vehicle delta-v required to reach normal LEO orbital velocity of around 7.8 km/s (28,080 km/h).

https://en.wikipedia.org/wiki/Delta-v_budget#Launch.2Flanding

For the Ansari X Prize altitude of 100 km, SpaceShipOne required a delta-v of roughly 1.4 km/s. [...] velocity from 0 to 7.8 km/s, but also typically 1.5–2 km/s for atmospheric drag and gravity drag

I was thinking about energy and it seem these statements are wrong or incomplete.

Imagine you have a perfect rocket with instant burn (no gravity drag) and no atmosphere (no atmospheric drag).

To get to an orbit of 200 km (for 7.8 km/s orbital velocity) you need to:

• reach 300 km
• reach 7.8 km/s

Imagine the case of a instantaneous big burn, or a gun pointed vertically. To reach 200 km I would need an initial kinetic energy equal to the potential energy from 200 km:

• 1/2 m v² = G m h
• 1/2 v² = G h
• v² = 2 G h
• v = $\sqrt(2 G h)$

If I compute the velocity needed to reach 100 km, it gives me 1400.7 m/s. This value is consistent with the SpaceShipOne delta-v.

If I compute the velocity need to reach 200 km, it gives me 1980 m/s. If I add this to the 7.8 km/s needed to stay in orbit, it gives me 9.78 km/s.

So I find a greater delta-v need even without including any gravity or atmospheric drag.

What is my error and what is the correct computation details for delta-v to LEO?

(In the best case, if the rocket starts from the equator to an equatorial orbit, I can gain about 400 m/s with the earth rotation.)

Answer 1

I suspect what is missing is the free velocity gained by launching in the same direction as the rotation of the Earth.

The closer you are to the Equator, the more "free velocity" you can get by launching in an Eastward direction, since the Earth rotates West-to-East at the equator at 0.46 km/s and at the poles at 0 km/s.

This free delta-v from the rotation of the Earth is why almost all man-made satellites orbit from West-to-East, and is why it's preferable for launch sites to be near the equator.

Answer 2

If you don't have to account atmosphere and gravity you can basically start horizontally and just accelerate horizontally (pro grade) and physics will do the rest "for free".

Rockets only start vertically to get out of the thickes part of the atmosphere as quickly as possible and because they need to account for gravity.

Firings of rockets in orbit are always prograde ("forward" in the direction of flying, to get to a higher orbit) or retrograde ("backwards" gâgainst the direction they are flying, to reduce the height). There are other maneuvers like plane changes which cost a lot of energy and are therefore avoided whenever possible) that fire "sideways". But firing towards the earth or away from earth is extremely waseful and will most probably not lead to the intended results (it's done for maneuvering in close proximity during docking operatiopns, but the dv in thise cases is minimal).

So your assumption that you need to gain height and speed independently is wrong which is the reason for the confusion.

Answer 3

I did the similar project in the computational physics class where we were asked to calculate the rocket's trajectory. Of course I went much far beyond to the realistic calculation of the startship launch sequences(air density vector decomposition and the correction term from the earth rotation etc.) and the naive model of starting straight up didn't work. From a serious of discussion in the space exploration Exchange.com and especially Christopher James Huff 's comment, the realistic rocket trajectory of human on the earth in the early 21 century were actually start "vertical" at high air density near the sea level and then turn "horizontal" to achieve the maximum fuel efficiency.

This could actually easily be shown from the Newtonian where instead of putting the energy to take into account of the gravitational drag, by gaining the horizontal speed the rocket could reach much higher orbit by circling around the earth.(The more systematic way was to use the lagranginan and get the generalized force.) It also explained why there's mention of building the rocket launch site at the earth equator.

For references:

$$F^{total}_r=-\frac{g_0R_E^2}{(R_E+r)^2} m+m \frac{v_\phi^2}{(r+R_E)} +\cos(\theta)\cdot F^{thrust}(r)$$

$$F^{total}_\phi=-\frac{\dot r v_\phi}{r+R_E}m+ \sin(\theta)\cdot F^{thrust}(r)$$

As clearly indicated, increase the "horizontal" $v_\phi$ achieve the stabilized orbit, not $r$ or $\dot r$, and the $F^{total}_\phi$ might result an oscillation if the trajectory were not carefully calculated and adjusted.