# What if the Shuttle went straight up?

It seems like Space Shuttle vehicles ascend on a diagonal in order to enter orbit. However, what would happen if they just kept going straight up?

• – uhoh Mar 16 at 2:39
• After thinking about it a bit, I realized this is not a duplicate of Does the space shuttle fly straight “up” when leaving Earth? since the OP points out in the question that it takes a turn, but asks instead what would happen if it didn't. This can be answered. – uhoh Mar 16 at 2:47
• Any chance you can put it im as most simple terms as possible – Ryan Lally Mar 16 at 2:53
• It took me longer than I thought, it's an interesting question! – uhoh Mar 16 at 5:38
• This may help what-if.xkcd.com/58 – Craig Constantine Mar 16 at 10:41

...what would happen if it just keep going straight up...?

It would fall to Earth or possibly glide back down to Earth and land when the fuel ran out.

It wouldn't go much higher than 1200 km, and g-force would even drop slightly below 1 g for about a minute when the SRBs were ejected, until it burned off some fuel.

I did a somewhat similar calculation in this answer to How much Delta-V would be required to escape the Gravitational Influence of the Earth without entering orbit? but the Shuttle is a sufficiently different situation that it deserves its own answer.

Simple Shuttle model

             m0 (kg)    mdot (kg/s)   t_burn (s)   v_exhaust (m/s)   thrust (N)
SRB(each)      570,000       4032          126          2640         10,644,480
main           760,000       1460          480          4460          6,511,600


Remember there are two SRBs.

When you run these approximations out and external tank and spent solid boosters have been jettisoned the remaining mass (shuttle plus payload/crew/supplies plus some remaining propellants) is about 200,000 kg. That's about 10% of the gross weight before launch.

I did a calculation below, that ignores the Earth's atmosphere completely. It's a 1D and so rotation of the Earth is ignored also. It says that on an airless Earth the shuttle would

• experience less than 1-g for about a minute after the SRBs are ejected
• burn out at about 8 minutes, 630 km and 2,960 m/s
• peak at about 14.8 minutes 1220 km
• impact at about 24.4 minutes and -4,480 m/s

But that's on an airless world. With Earth's atmosphere the drag would slow things down a bit, and since the Shuttle is a glider, it is conceivable that it could glide back to Earth and land.

But landing should be addressed by someone who knows more about the Shuttle than I do, so I'll leave it here. def m_and_T(t):
m_srb     = (m0_SRB  - mdot_SRB * min(t, t_burn_SRB))*(t<t_burn_SRB)
m_main    = (m0_main - mdot_MAIN * min(t, t_burn_main))*(t<t_burn_main)
mass      = m0_shuttle  + m_main + 2 * m_srb
mdot_srb  = mdot_SRB * (t<t_burn_SRB)
mdot_main = mdot_MAIN * (t<t_burn_main)
thrust    = mdot_main * v_ex_main + 2 * mdot_srb * v_ex_SRB
return mass, thrust

def deriv(X, t):
x, v = X
mass, thrust = m_and_T(t)
ath  = +thrust / mass
aGM  = -GMe / x**2
return np.hstack((v, aGM + ath))

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

halfpi, pi, twopi = [f*np.pi for f in (0.5, 1, 2)]
GMe    = 3.986E+14    # m^3/s^2
Re     = 6378137.    # meters
g0     =  9.80665    # m/s^2

m0_SRB,  mdot_SRB,  t_burn_SRB,  v_ex_SRB  = (570000, 4032, 126, 2640) # kg, kg/s, sec, m/s
m0_main, mdot_MAIN, t_burn_main, v_ex_main = (760000, 1460, 480, 4460) # kg, kg/s, sec, m/s
m0_shuttle = 200000.

seconds  = np.arange(25*60.)    # seconds
minutes  = seconds/60.

X0     = np.array([Re, 0])

answer, info = ODEint(deriv, X0, seconds, rtol=1E-09, full_output=True)

h    = x - Re

i_engineout = 481
h_engineout = h[i_engineout]
v_engineout = v[i_engineout]
minutes_engineout = minutes[i_engineout]

print minutes_engineout, h_engineout, v_engineout

i_maxh = np.argmax(h)
h_max  = h[i_maxh]
v_max  = v[i_maxh]
minutes_max = minutes[i_maxh]

print minutes_max, h_max, v_max

i_impact = np.argmax(h<0)
t_impact = seconds[i_impact]
minutes_impact = minutes[i_impact]
v_impact = v[i_impact]

print minutes_impact, 0, v_impact

if True:
plt.figure()

plt.subplot(6, 1, 1)
plt.plot(minutes, 0.001*h)
plt.ylabel('altitude (km)', fontsize=16)
plt.plot(minutes, np.zeros_like(minutes), '-k')

plt.subplot(6, 1, 2)
plt.plot(minutes, v)
plt.ylabel('velocity (m/s)', fontsize=16)
plt.plot(minutes, np.zeros_like(minutes), '-k')

mass, thrust = zip(*[m_and_T(t) for t in seconds])
dt           = seconds - seconds
dv_dt        = (v[1:] - v[:-1])/dt
gabs         = GMe/x[1:]**2 + dv_dt
gratio       = gabs/g0

plt.subplot(6, 1, 3)
plt.plot(minutes[1:], dv_dt)
plt.ylabel('dv/dt (m/s^2)', fontsize=16)
plt.plot(minutes, np.zeros_like(minutes), '-k')

plt.subplot(6, 1, 4)
plt.plot(minutes[1:], gratio)
plt.ylabel('g-force (g0)', fontsize=16)
plt.plot(minutes, np.ones_like(minutes), '-k')

plt.subplot(6, 1, 5)
plt.plot(minutes, mass)
plt.ylabel('mass (kg)', fontsize=16)

plt.subplot(6, 1, 6)
plt.plot(minutes, thrust)
plt.ylabel('thrust (N)', fontsize=16)

plt.xlabel('time (minutes)', fontsize=16)

plt.suptitle('Simplistic Shuttle Vertical Launch', fontsize=18)

plt.show()

• Although the accel goes to zip when the SRBs are freed, the vel is still quite positive, so I wouldn't call that a 'hover'. – amI Mar 16 at 6:22
• @amI oh geez I'm off by a derivative! Thanks for catching that; I've made a change to the plot noting that $g-g_0$ drops below zero a little bit. – uhoh Mar 16 at 6:38
• @amI done, all instances of hov* removed. thanks again. – uhoh Mar 16 at 6:49
• Even in the real world the g's went <1 after SRB sep qph.fs.quoracdn.net/main-qimg-51b11c252d34ae666f9007d59b387b50 You left out the thrust bucket though. – Organic Marble Mar 16 at 13:25
• @OrganicMarble oh that's a great plot! Nice (approximate) validation for such a simple model too, thanks! – uhoh Mar 16 at 13:41

Nominal shuttle reentry is done nearly-horizontally, over about an hour. There is not enough margin in the thermal protective elements of the Shuttle to allow it to perform a vertical descent any way but catastrophically.

The Kármán line is 100km above the surface of the earth. According to uhoh's answer, the Shuttle will have a velocity of about 4.5km/s when it arrives at that imaginary boundary. That means it has to decelerate from 4.5km/s to a 111m/s glide, in less than 44 seconds. (That works out to an average G-force of 10.3g.)

The extreme acceleration and aerodynamic load causes weaker pieces of the Shuttle, like the wings, to snap off.

The thermal protection tiles are not designed for this violent of a reentry burn. The tiles burst into a fiery ball of plasma, or are torn away, followed by the remainder of the shuttle, payload, and (likely already unconscious from the constant 10 G's) crew. The plasma-debris cloud splashes violently into the Gulf of Mexico.

It's actually even worse, because I didn't take the varying density of the atmosphere into account. Half of the mass of the atmosphere is in the bottom 5.6km of the troposphere. There's very little deceleration until the top of the troposphere. Then, too much. For the few seconds before it makes a splash, the Shuttle is a flaming asteroid.

• I'm glad you looked into this! +1 Yes, there's no available way to turn 4.5 km/s down into 4.5 km/s sideways. This should be the accepted answer. – uhoh Mar 18 at 5:20

The shuttle (or any space-bound rocket) doesn't exactly ride into space on a diagonal. These rockets follow a curved trajectory which represents a trade-off/optimization among a number of factors, including:

• Earth's gravity
• Earth's atmosphere and aerodynamic effects (drag, heating)
• Target trajectory (bound for an orbit around Earth or on a path somewhere beyond)
• Acceptable acceleration (structural limitations, human factors)
• Fuel efficiency
• Specific impulse of the propellants

The engineering goal is to get to space with the least propellant necessary, because the more you have to carry the bigger the engines you need, so the more propellant you have to carry and so on. Developing bigger and bigger rockets is difficult and expensive, so you try to do the most with the least.

Without doing any math, this leads us to the intuitive answer that straight up isn't optimal, and with only about 8 or so minutes of propellant, the vehicle won't go very far before fuel runs out and gravity pulls it back to Earth. See Uhoh's answer for a calculated estimate.

The reason the Shuttle (or Falcon, or Soyuz, or anything else) follows the trajectory it does is basically as follows (an intuitive explanation):

The first goal is to get off the ground, so you aim all your thrust straight down to push you straight up. For as long as you are not in orbit, you are fighting gravity, so you want to get into orbit quickly. That means you need to tilt the rocket in the direction of your desired orbit and start building up speed in that direction while still fighting gravity. Orbital speed is very fast - too fast to travel while in the atmosphere so you need to get above it before you get get going really fast. So, you do most of your initial acceleration upward to get out the atmosphere quickly. Once you are up in thinner air, you can tilt more to really accelerate along your orbital trajectory. Not pushing straight up so much, gravity will slow your vertical ascent, but that's okay because as you get higher the air gets thinner so you can go faster and get closer to orbital speed, which decreases the effect of gravity pulling you down, and you don't need to be travelling "up" so much as "sideways" anyway. You eventually get to the point where you are in orbit and you can turn off your engines, unless you intend to escape Earth to go somewhere else.

It would fall back down again.

To estimate the turn around point we use the conservation of energy. The kinetic energy of the shuttle after the engines shut off is given by mv^2/2 with a speed of approximately 30,000km/h.

If the shuttle is pointing straight up its kinetic energy will be converted to potential energy which gets it up to around 8,500 km above the Earth assuming a starting point of 200 km (the mass of the shuttle cancels out).