# Why astronauts during launch fly faster than jets but experience less Gs? [closed]

The NASA Space Shuttle used to lunch astronauts into spaces at high speeds, supersonic speeds. When you google whats the Gs force they experienced while departing or simply “Gs forces astronauts” You get 3gs. I watched an interview of an astronaut he confirmed it, it was about 3Gs. Lets say 4Gs. Why is it that even though they were experiencing higher than a jet aircraft speeds they were only experiencing 3-4gs, while jets fly much slower but experience many more Gs?

I think it has to be with the rate of acceleration? Like how fast the velocity vector is changing, correct? In magnitude or in direction or both (maybe)?

So would you experience the same amount of Gs if moving in a straight line either vertically or horizontally (assuming both have the same rate of acceleration change)? Or no Gs at all?

Maybe Im confusing inertia like when you accelerate (or break) in your car in the freeway your body either move forward or backwards, right? But if thats the case, why do astronauts experience Gs at all (assuming they are moving straight up, is it because they are not?)?

*Also, I do understand how Gs forces are formed while flying in a turn, less vertical lift more total lift needed, then more L to W ratio. But I guess Im confused whats going on while moving in a straight path.

• Basically you are asking for a complete explanation of the concept of G-loading, and a specific question about the G-load felt on the space shuttle vs on airplanes? That:s pretty broad. Oct 11, 2022 at 5:35

You are confusing velocity and acceleration. Flight speed is velocity (meters persecond, miles per hour, knots, etc) while "Gs" are acceleration (meters per second per second, ).

I think it has to be with the rate of acceleration? Like how fast the velocity vector is changing, correct? In magnitude or in direction or both (maybe)?

You kind of have the right idea there in the second part of your statement. Acceleration is the rate at which the velocity vector is changing. The change in rate of acceleration is something else completely called a jerk/jolt.

Imagine two cars at a stop light. When the light turns green, one car floors the gas for a couple seconds and reaches 40 miles per hour. When that happens, you feel slammed into the back of your seat (higher Gs). The other car presses the gas halfway but holds it there for a minute and reaches 200 mph. With this car, you feel much less force pushing you back in your seat (lower Gs). The first car had a high acceleration but lower speed while the second car had a lower acceleration and a higher speed. The first car is like the fighter jet and the second is like the rocket.

Astronauts experience a G force due to the acceleration as the space craft gains speed. There is very little change in direction.

Pilots also experience a small amount of G force due to acceleration when they gain speed, but they experience far more g Forces when changing direction. The G force is experienced when the vector changes. When accelerating the vector gets longer. When turning the vector's orientation changes. Very high G force is only encountered in aircraft during very tight turns at high speed when the vectors are changing very rapidly.

Imaging a 3 meter long stick. Add 1 meter to the end of it making it 4 meter long. This a change of 1 meter. Now take the 3 meter stick and rotate it 180 degrees so it points in the opposite direction. This is a 6 meter change.

So would you experience the same amount of Gs if moving in a straight line either vertically or horizontally (assuming both have the same rate of acceleration change)?

No you would not.

For example, if you feel a G-load of 10 G vertical (in the "positive" direction, i.e. the direction associated with upward acceleration), then you are accelerating a little more slowly than if you feel a G-load of 10 G in a horizontal direction. Or to put it another way, if the magnitude of the actual acceleration is the same in both cases, the "felt" G-loading is higher when the object is accelerating vertically upward than when the object is accelerating horizontally. For example, if the actual rate of acceleration is 10 X 9.8 m/s/s, the resulting net G-loading is 11 G if the object is accelerating vertically upward, while it is only 10.05 G (the vector sum of 10 x 9.8 m/s/s horizontal acceleration and 1 x 9.8 m/s/s vertical acceleration) if the object is accelerating horizontally.1

To understand why, please read on -- we'll try to make a bit of sense of all this!

(And please note that this apparent "discrepancy" between G-loading and actual acceleration, when we are dealing with a vertical acceleration, works in the opposite direction from what would be needed to help explain your intuition that the G-loading "felt" by astronauts during a rocket launch is abnormally low, compared to the G-loading felt by the pilot of a maneuvering jet fighter!)

G-loading is related to acceleration, but it is not exactly the same as acceleration.2 G-loading is the vector sum of a body's (e.g. a vehicle's) actual acceleration vector, minus the 1- G earthward acceleration vector caused by gravity.

Here's another way to say that -- G-loading is what we get when we take the vector sum of all the forces acting on a body other than gravity, and then divide by weight. In the context of flight, this means that G-loading is what we get when we take the vector sum of all the aerodynamic and thrust forces acting on a body, and then divide by weight.

If an astronaut feels a G-loading of 3 G on launch, he is only accelerating upward at 2 x 9.8 m/s/s , not 3 x 9.8 m/s/s. The third G is used to cancel gravity, or if you prefer to put it the other way, the third G is cancelled by gravity, and thus does not contribute to the actual acceleration of the body. The 3 G figure represents the net aerodynamic and thrust forces acting on the vehicle, divided by weight, while the actual net acceleration, due to the vector sum of all forces acting on the vehicle (including gravity), is only 2 G.

Likewise in straight and level constant-speed flight-- or at rest on the ground-- the crew (and the G-meter) "feel" 1 G, but their actual acceleration due to the vector sum of all forces (including gravity) is zero. For the in-flight case, the 1-G figure is due to the upward lift generated by the wings. Since thrust and drag are equal and opposite, this also represents the net vector sum of all the aerodynamic and thrust forces acting on the aircraft. At rest on the ground, the 1-G figure represents the upward force of the earth on the tires, transmitted to the bodies of the occupants through the landing gear, aircraft structure, and seat cushions.

The fundamental reason3 for this somewhat odd state of affairs is that unlike aerodynamic and thrust forces, which act externally on a body, gravity "acts from within". It exerts an equal acceleration on every molecule of a body, and thus creates no stresses or strains of any kind4, and so can't be "felt", and can't be detected by a G-meter.5 It has no tendency to pull a pilot down into the seat or sideways against the side of an aircraft. That's the secret to understanding weightlessness. In the arcing trajectories flown by the famous "Vomit Comet" astronaut training aircraft, aerodynamic lift is reduced to zero, and thrust is set to be equal (and opposite) to drag, and therefore the net vector sum of all aerodynamic and thrust forces is zero. Gravity is still acting on (accelerating) the aircraft and crew, but gravity can't be "felt", so weightlessness results, and the occupants float freely around the aircraft, and the G-meter6 reads zero.

Likewise in actual space flight-- beyond the atmosphere, aerodynamic forces are always zero, so the G-load is always zero (and weightless always exists) whenever the engines are not producing thrust.7 In this situation a G-meter will always read zero-- which does not mean that the vehicle has travelled beyond the reach of the earth's gravitational field!

Disclaimer: all this is nuance. It would be sufficient to simply point out that the very high G-loadings (e.g. much more than 4) sometimes experienced by jet fighter pilots are always acting in the vertical (up-and-down) plane in the aircraft's reference frame, and are due to strong lift forces produced by the wings, not due to high thrust forces produced by the engines. Even a catapult launch off an aircraft carrier only averages around 3 G's of acceleration. Adjusting the numbers up or down by 1 G, to properly account for the role of gravity, doesn't change the basic conclusion here.

It is also relevant to point out that even a modest (linear) acceleration, sustained over a long time period, will eventually produce a very high velocity (speed). That's what explains the very high speeds reached by rockets, despite their relatively modest acceleration rates.8

Footnotes:

1. In this paragraph-- and throughout this answer unless stated otherwise-- we are talking about objects on or near the surface of the earth, and thus experiencing the normal gravitational pull of the earth that we feel at the surface.

2. "G's" of acceleration are not the same as G-loading. G's of acceleration are simply equal to the acceleration divided by 9.8 m/s/s. If a vehicle is accelerating purely horizontally, at a rate of (for example) 1 G, this does not mean that the actual total G-loading is only 1 G. If we say a vehicle is accelerating upward at a rate of 1 G, it's arguably a bit unclear whether we really mean that the actual net acceleration is 1 G, or we actually mean that the G-loading is one G, in which case (for an object in the earth's gravity field relatively near the surface) the actual net acceleration would be zero. The former appears to be the more correct interpretation of what is meant by "accelerating upward at a rate of 1 G."

3. We're speaking from a Newtonian perspective here. The relativistic perspective on gravity is different, and is beyond the scope of this answer.

4. Other than tidal effects, which are normally only significant on the planetary scale. Tidal effects are due to one side of a body being in a stronger gravity field than the other side of the body.

5. Obviously, given certain constraints, a G-meter will indicate the strength of the gravitational field acting on a body. For example, if the body is known to be at rest on the ground, or is known to be travelling at a constant speed in a constant direction.

6. In this answer we're referring to a hypothetical G-meter that measures acceleration in all three dimensions. The actual G-meters found on aircraft instrument panels only measure acceleration (excluding the acceleration component caused by gravity) in one dimension, the "up and down" direction in the aircraft's own reference frame. This is normally the component that is most likely to become dangerously large in a poorly executed maneuver, turbulence, etc, due to the capability of the wings to generate very high lift forces, if allowed to meet the air at too high an angle-of-attack in relation to the airspeed. In flight, a G-meter that only registers accelerations acting in the up-and-down direction in the aircraft's own reference frame, is essentially simply deflecting in proportion to the lift force generated by the wing (at least so long as the actual weight of the aircraft is held constant.) That's why a G-meter is an essential tool for aerobatic pilots who wish to avoid accidentally pulling the wings off the airplane.

7. Note that this is true even when the trajectory is completely linear, e.g. a spacecraft travelling directly toward the earth, or directly toward the sun. It's not necessary to introduce the fictitious pseudoforce called "centrifugal force", associated with a curving flight path, to explain weightlessness, regardless of whether we are talking about the arcing trajectory of the "Vomit Comet", an orbiting spacecraft, a stunt plane temporarily "unloaded" to zero G's while inverted at the top of an (improperly flown?) loop, etc.

8. "Relatively modest" meaning "compared to the max G's pulled by a fighter pilot in a high-speed, high-lift maneuver"!

• << at rest on the ground-- the crew (and the G+meter) "feel" 1 G >> and << gravity "acts from within"... and so can't be "felt", and can't be detected by a G-meter. >> These are contrasting, which is the correct one? 1G or 0G? Oct 11, 2022 at 12:44
• Understand, according to the text in your answer the g-meter would be 1 on ground, correct? Oct 11, 2022 at 13:11
• Ok, but then I don't understand why you also write that gravity cannot be detected by a g-meter... Oct 11, 2022 at 13:27
• So if I jump up in the air I do not fall back to earth, the earth actually rises up to meet me? Oct 11, 2022 at 14:26
• It was a snarky response to the remark about feeling the earth pushing up against you. While technically correct on some level, I have always felt such distinctions detract from, rather than enhance our understanding of the way things work. And may the road rise up to meet you too! Oct 11, 2022 at 18:43