A lot of reference indicate it need 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, Space Ship One 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 an orbit with 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 instant big burn, or a gun pointed verticaly. To reach 200 km i need a inital 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 needs to reach 100 km it give me 1400.7 m/s. This value is consitent with the Space Ship One delta-v.

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

So i find a greater delta-v need even without integrate 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 start from the equator to a equatorial orbit, I can gain about 400 m/s with the earth rotation)

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.)