I've heard that because Earth's atmosphere is so thick that some spacecraft can 'bounce' off the atmosphere if they don't enter at the correct angle. This leads me to believe that shallower angles have more air resistance than steep angles.

If an object comes into an atmosphere directly perpendicular, is there more or less resistance, and why?

  • 1
    $\begingroup$ Note that all the answers assume little-to-no retro thrust. If you had unlimited retro power & fuel, you could enter at any desired angle and use your engines to control the max-g undergone. $\endgroup$ Dec 6 '19 at 14:22
  • 2
    $\begingroup$ You might be interested in the question Under what practical conditions would Earth's atmosphere be a dilatant? on the Physics SE as it explores a similar area. $\endgroup$ Dec 6 '19 at 15:46

“Bouncing off the atmosphere” is a misleading turn of phrase.

When returning to the Earth from the Moon, a spacecraft is on an elliptical orbit with the high end somewhere around the moon’s altitude and the low end just grazing the top of Earth’s atmosphere. The concern around a too-shallow reentry angle is that it won’t slow the spacecraft enough for a prompt reentry. In such cases it wouldn’t “bounce off into space and be lost forever” as is so often suggested or inferred; it would continue on an elliptical orbit -- depending on how much it slowed down in the initial pass, it could return again to Earth days or even weeks later. This would be disastrous for an Apollo-type mission because the command module only has power for a few hours of independent operation.

Atmospheric lift and atmospheric drag are closely related; without thrust, a vehicle that's experiencing lift is being slowed down at the same time. As a spacecraft orbit is slowed by drag, it bends downward, bringing it into denser air, which slows it further in a rapid feedback loop. What lift the spacecraft achieves makes the difference between a short, high-g reentry (like the early, zero-lift Mercury capsules) and a longer reentry with gentler g-forces but more total heating of the spacecraft to deal with, like the space shuttle.

Returning straight down through the atmosphere would demonstrate the most air resistance and produce the highest g-force. It would not be survivable without some sort of exotic crew protection system.

  • 2
    $\begingroup$ Thank you so much for the explanation of the misleading phrase “Bouncing off the atmosphere” $\endgroup$
    – Uwe
    Dec 5 '19 at 22:40
  • 10
    $\begingroup$ I'll clarify "more total heating of the spacecraft". In practice, it's not a total energy problem because no one cares how much the air has been heated after the spacecraft passes through it. For engineering purposes, what matters is how much heat is transferred to the spacecraft. That's the big advantage of blunt-body reentry, in fact. $\endgroup$ Dec 6 '19 at 4:14
  • 2
    $\begingroup$ @nick012000 Well, falling from a complete standstill from 1000km height, you already hit the atmosphere at over 4km/s. That results in an average (round numbers) 10G to slow down, with much higher (deadlier) peaks. $\endgroup$
    – Mr47
    Dec 6 '19 at 9:39
  • 9
    $\begingroup$ @nick012000 I was talking about returning from the moon, but even coming to a halt at 200km altitude and then free-falling, in a capsule like an Apollo CM, incurs a brief peak of about 6.5g. You accumulate a lot of speed falling before the atmosphere starts to slow you. After the peak the capsule does come to a terminal velocity equilibrium, where air resistance and gravity are balanced at 1g. From 2000 km altitude it's a 67g hit. $\endgroup$ Dec 6 '19 at 10:21
  • 2
    $\begingroup$ The book Coming Home: Reentry and Recovery from Space describes these factors and the history of research on this subject in great detail. A link to this book might make for a good addition to this answer. $\endgroup$
    – Sander
    Dec 6 '19 at 11:01

Going straight down takes a lot more fuel for deceleration.

In order for a spacecraft to stay in orbit, it needs a substantial velocity perpendicular to the direction of the vector of gravitational force. This velocity is referred to as your orbital velocity; for a near-Earth orbit about 200 miles up, you need an orbital velocity of 7.79 km/s. In order for your rocket to fall straight down, you need to first cancel out that velocity by thrusting with your rocket, which would consume a considerable amount of fuel.

By contrast, if you go down at an angle, you only need to cancel out a small portion of that velocity with your rocket's thrust, since the drag from the atmosphere can take care of the rest. This is why the space shuttle gets so hot during re-entry - the kinetic energy of its motion is converted into heat energy by the air.

  • 3
    $\begingroup$ This is only true for near-earth orbit. The further away you are, the smaller the difference. At the moon's orbit the difference (calculated as orbital speed minus Hohmann to 100km) is 0.185 km/s. $\endgroup$
    – Fax
    Dec 6 '19 at 12:59
  • 3
    $\begingroup$ @Fax: By the time you've coasted from the moon down to 100 km, the earth's gravity will have accelerated you to a speed considerably greater than the orbital speed for a circular low earth orbit, so, by starting higher up, you've made your problem worse, not better. $\endgroup$
    – Vikki
    Dec 6 '19 at 23:09
  • 6
    $\begingroup$ @Sean That's just not true. If your craft can survive arbitrarily intense deceleration and heating, de-orbiting from the Moon's altitude costs less delta V than from LEO. Consider a more extreme example. Imagine orbiting at a high enough altitude that your orbital velocity is only 10m/s. It would only cost 10m/s delta V to get an Earth intercept, as opposed to the hundreds it costs from LEO. $\endgroup$
    – Ryan_L
    Dec 7 '19 at 0:45

The resistance is proportional to the density of the air. If you come in at a shallow angle you decelerate more gently as you spend more time at a higher altitude.


The air resistance does not depend on the angle at which a spacecraft enters the atmosphere. The angle does determine how long a spacecraft is in the upper regions of the atmosphere where the air is much less dense. A spacecraft needs to be there long enough to loose most of it's speed.

The angle at which a spacecraft can approach the earth is called the re-entry corridor, see figure below. (Note that the figure is schematic, the atmosphere is only about 40 km thick, hardly a visible dot on this scale)
enter image description here
Re-entry corridor

The re-entry corridor is a narrow region in space that are-entering vehicle must fly through. If the vehicle strays above the corridor, it may skip out.If it strays below the corridor, it may burn up and/experience excessive g-forces.

A spacecraft may skip out on purpose, this is called a skipped re-entry. The maneuver may also be used to loose speed to get a spacecraft in to an orbit. If a spacecraft skips out of the atmosphere again it will come back unless it has enough speed to escape the gravitational field.

The angle at which a spacecraft approaches a planet is called re-entry flight-path angle, γ, see figure below.

enter image description here
Flight path angle

At around 20 to 30 km altitude the density of the atmosphere starts to increase significantly:

enter image description here
Density vs Altitude from North Carolina Climate Office

The resistance in air is calculated as D=Cd*0.5*density*velocity^2. It is proportional to the density, and proportional to the velocity squared. At velocities in the order of several km/s, the drag is enormous when a spacecraft comes in to the denser atmosphere with high speed. To avoid high drag and therefore high g-forces and heating, a spacecraft needs to stay long in regions with a low densty so it has enough time to shed most of its speed there without high g-forces

If the flight path angle increases, the maximum g-forces and and heat input increase rapidly:

G-force during re-entry
G-force during re-entry

Heating rate during re-entry
Heating rate during re-entry

The maximum g-force a person can survive in good health is around 10g, therefore small flight path angles are usually chosen for manned re-entry.

All figures and part the explanation are from -FAA Advanced Aerospace Medicine On-line > Section III - Space Operations > Chapter 4 Basic Concepts of Manned Spacecraft Design > 4.1.7 Returning from Space: Re-entry:

FAA Advanced Aerospace Medicine On-line:


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.