As impractical as it sounds (and it's definitely impractical) I was wondering what the Delta-V required to leave Earth's influence would look like if we launched "just straight up" into the Earths atmosphere. What if we kept fighting the gravity directly without the added benefit of the Earth's rotation, perhaps even having to also fight the Earths rotation.

What would this be compared to the required Delta-V for the average escape velocity? You can assume any vessel build that you would like using available data, I just was looking for a ratio of Delta-V saved on orbital ascents compared to a "straight-up" approach. I want to say aerodynamics isn't what I'd like to focus on, if that isn't a required part for an approximate calculation.

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    $\begingroup$ +1 I think the mathematical beauty of this question is lost on some people. Launching "straight up" is probably going to be much more costly. $\endgroup$
    – uhoh
    Commented Jun 15, 2018 at 13:10
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    $\begingroup$ @uhoh I mean, the math is lost on me currently too hah. (Tangential Note: replying to you is hilarious, it's like admitting fault immediately. "uhoh! I asked because I didn't know how to do the math.") $\endgroup$ Commented Jun 15, 2018 at 13:15

4 Answers 4


I've assumed a meta-Falcon 9 Full Thrust and shot it straight up on a non-rotating Earth with no atmosphere.

                          1st stage    2nd stage    coast stage
                          ---------    ---------     ---------
       total mass (kg)      422000       128000         28000
  propellant mass (kg)      370000       108000
             Isp (sec)         300          350
exhaust velocity (m/s)        2943         3433.5
       burn time (sec)         160          400          1800
    mass flow (kg/sec)        2312.5        270

For the 1st and 2nd stage, I've integrated

\begin{align} F_{thrust} & = v_{exhaust} \times \frac{dm}{dt} \\ m(t) & = m_0 - t \frac{dm}{dt} \\ a_{thrust} & = \frac{F_{thrust}}{m} \\ a_{gravity} & = -\frac{GM}{r^2} \\ \end{align}

As I suspected:

Launching "straight up" is probably going to be much more costly.

There are so many answers in this site that explain the idea that getting to orbit is about getting moving sideways fast-enough, soon-enough. Here is what happen if you don't do that.

I've dubbed the mission "Falcon Nein" because even with zero additional payload, a Falcon 9 Full Thrust, shot straight up, will fall back to Earth, both the 1st stage, and unladen 2nd stage, be it African or European.

A half-hour of gravity, in the absence of orbit, is horribly costly in fact.

I've simulated for five "settings" of Earth gravity; 100%, 75%, 50%, 25% and zero.

enter image description here

Python script:

def deriv_1(X, t):
    x, v  = X
    F_t   = v_ex_1 * mdot_1
    m     = m_tot_1 - t * mdot_1 + m_tot_2
    acc_t = F_t / m
    acc_g = -GM / x**2
    return np.hstack((v, acc_t + acc_g))

def deriv_2(X, t):
    x, v  = X
    F_t   = v_ex_2 * mdot_2
    m     = m_tot_2 - (t-t_burn_1) * mdot_2
    acc_t = F_t / m
    acc_g = -GM / x**2
    return np.hstack((v, acc_t + acc_g))

def deriv_3(X, t):
    x, v  = X
    acc_g = -GM / x**2
    return np.hstack((v, acc_g))

import numpy as np
import matplotlib.pyplot as plt
from scipy.integrate import odeint as ODEint

GMe = 3.986E+14  # m^3/s^2
Re  = 6378137.   # m

GMs = GMe * np.linspace(0, 1, 5)

# Setup:

# first stage
m_fuel_1  = 370000.       # kg
m_tot_1   = 422000.       # kg
t_burn_1  = 160.          # sec
Isp_1     = 300.          # sec
v_ex_1    = Isp_1 * 9.81  # m/s
mdot_1    = m_fuel_1 / t_burn_1  # kg/sc

# second stage
m_fuel_2  = 108000.       # kg
m_tot_2   = 128000.       # kg
t_burn_2  = 400.          # sec
Isp_2     = 350.          # sec
v_ex_2    = Isp_2 * 9.81  # m/s
mdot_2    = m_fuel_2 / t_burn_2  # kg/sc

# coast stage
t_coast   = 1800.         # sec

# Go!
trajectories = []

for GM in GMs:

    traj = []

    # first stage
    X0_1   = np.array([Re, 0.0])
    time_1 = np.linspace(0, t_burn_1, 101)

    answer_1, info = ODEint(deriv_1, X0_1, time_1, full_output=True)
    x_1, v_1 = answer_1.T

    traj.append((time_1, (x_1, v_1)))

    # second stage
    X0_2   = answer_1[-1]
    time_2 = np.linspace(0, t_burn_2, 101) + time_1[-1]

    answer_2, info = ODEint(deriv_2, X0_2, time_2, full_output=True)
    x_2, v_2 = answer_2.T

    traj.append((time_2, (x_2, v_2)))

    # coast stage
    X0_3   = answer_2[-1]
    time_3 = np.linspace(0, t_coast, 201) + time_2[-1]

    answer_3, info = ODEint(deriv_3, X0_3, time_3, full_output=True)
    x_3, v_3 = answer_3.T

    traj.append((time_3, (x_3, v_3)))


if True:

    for traj in trajectories:

        for (time, (x, v)), color in zip(traj, ('-b', '-g', '-r')):

            plt.subplot(2, 1, 1)
            plt.plot(time, 0.001 * (x-Re), color)

            plt.subplot(2, 1, 2)
            plt.plot(time, 0.001 * v, color)

    plt.subplot(2, 1, 1)
    plt.ylabel('altitude (km)', fontsize=16)
    plt.xlabel('time (sec)', fontsize=16)
    plt.ylim(0, 15000)

    plt.subplot(2, 1, 2)
    plt.ylabel('speed (km/s)', fontsize=16)
    plt.xlabel('time (sec)', fontsize=16)

    plt.suptitle('Vertical Launch ("Falcon Nein") GM = (1, 0.75, 0.5, 0.25, 0) x GMe', fontsize=16)

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    $\begingroup$ I've been meaning to do the calculations like this, great analysis! And +1 for Falcon Nein! What factor of GM is our break-even point for F9? ie. what factor allows us to achieve the (new) escape velocity? $\endgroup$
    – Jack
    Commented Jun 21, 2018 at 7:56
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    $\begingroup$ The fact that you took rotation and atomosphere out of the equation makes this answer beautiful-- and EXACTLY what I was looking for. The other answers do a great job of talking about how you'd be fighting the atmosphere and rotation of the earth; this focuses on the theoretical situation of a direct gravity escape w/o orbit. That's exactly what I was wanted :). Possible in KSP, horrible in practice; like most things in that game hah. $\endgroup$ Commented Jun 21, 2018 at 14:00
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    $\begingroup$ @uhoh the only thing I was going to ask for was similar graphs for the actual falcon 9 launch under similar conditions (0/.25/.5/.75/1). But I didn't want to impose on your time either :). I love that you posted the code, that's super helpful as well. $\endgroup$ Commented Jun 21, 2018 at 16:24
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    $\begingroup$ @MagicOctopusUrn oh I see. I'll check back tomorrow, it's late now here. $\endgroup$
    – uhoh
    Commented Jun 21, 2018 at 16:29
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    $\begingroup$ @uhoh don't feel obligated, this is a great answer as is! $\endgroup$ Commented Jun 21, 2018 at 16:29

The difference (and it's a difference, not a ratio) between a direct ascent to escape, and going temporarily into a low parking orbit and then on to the same escape, is on the order of 100 m/s. That is the cost of raising the periapsis, which you would not need to do on a direct ascent.

So the cost actually isn't all that much, compared to the ~9 km/s to get into orbit.

This is assuming an efficient direct ascent, which still looks like going into orbit, except for not raising periapsis. You want your rocket firing to be as close to Earth as possible to maximize the Oberth effect. So you wouldn't go "just straight up", unless you could magically complete your burn just outside the atmosphere.

  • $\begingroup$ Gravity losses matter a lot more than this. $\endgroup$ Commented Jun 16, 2018 at 2:07
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    $\begingroup$ @LorenPechtel I ignored the "straight up" part, since you would never do that. $\endgroup$
    – Mark Adler
    Commented Jun 16, 2018 at 20:01
  • $\begingroup$ Playing around with Kerbal Space Program I actually did it a few times--I was trying to model a Jules Verne approach and get into orbit with as little delta-v as possible after the initial push. (All I learned is that even MechJeb can't control it accurately enough to get any meaningful data. My intent was to boost to just under escape velocity, at apoapsis raise the periapsis to 60km and then treat it as an aerocapture maneuver.) $\endgroup$ Commented Jun 17, 2018 at 14:52
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    $\begingroup$ A typical parking orbit would be at 200 km, and with no deliberate circularization, you might end up with a periapsis at –100 km (that's 100 km below the surface of the Earth). If you intend to continue your parking orbit for at least 180°, you can see the imperative of raising the periapsis. When you're at the apoapsis of 200 km, the $\Delta V$ to raise that periapsis from –100 km to 200 km is about 90 m/s. This is all very approximate, but gives the general order of magnitude. $\endgroup$
    – Mark Adler
    Commented Jun 17, 2018 at 21:03
  • 1
    $\begingroup$ As I said, this is not assuming "straight up". $\endgroup$
    – Mark Adler
    Commented Jun 21, 2018 at 5:20

According to Wikipedia:

For a spherically symmetric, massive body such as a star, or planet, the escape velocity for that body, at a given distance, is calculated by the formula:

$v_e = \sqrt{\frac{2GM}{r}}$

where where G is the universal gravitational constant, M the mass of the body to be escaped from, and r the distance from the center of mass of the body to the object.

If you enter the values for the surface of Earth, you get 11.186 km/s.

This formula assumes a non-rotating earth. If you want to make maximum use of Earth's rotation, you would have to launch straight east from the equator. This would theoretically save you 465 m/s. If you would launch straight west, you would need 465 m/s more.

But one thing this formula does not account for is the energy you lose through aerodynamic friction while still in the atmosphere. The atmosphere makes a straight east launch obviously infeasible. That's why real world rocket launches are a compromise: You first launch straight up (relative to surface) while in the lower atmosphere, and then turn east when the atmosphere becomes thinner.

It also assumes an immediate acceleration. The longer you need to reach that speed, the more acceleration you lose through falling.

But these two factors are difficult to generalize, because they depend on the construction of your vessel.


The biggest issue here is gravity loss. If you could somehow magically come up with that 11.186 km/sec all at once (and ignoring the atmosphere) that would get you into an escape trajectory.

However, in the real world it takes several minutes to build up that velocity. While you're doing that you're also expending fuel on keeping from falling back. That's pure waste.

While you are heading straight up you're expending 9.8m/s^2 on fighting gravity. (This will decline as you get higher up, but the effect is small within the realm that you're operating.) Thus you tip your rocket over as soon as you can--you want to pile on as much horizontal velocity as possible as soon as possible. While it takes orbital velocity to cancel all of your weight any amount of horizontal velocity cancels some of it. Every m/s you save this way is a m/s your rocket doesn't need to provide. You have to balance drag losses from being too low with additional gravity loss by being too high.

There is also normally the issue of fairings. They're big and heavy, you want to get rid of them as soon as possible but you have to be high enough that your now not-aerodynamic rocket isn't going to be a problem with them gone. This causes rockets to fly a bit higher than they otherwise would.


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