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Short answer: Neither, but it's closer to 29.4 m/s^{2}.

Long answer: The "g factor" refers to the load felt by the spacecraft's
structure and the people inside. As such, it's equal to the sum of all
*non-gravitational* forces on the spacecraft, divided by the force
gravity would exert on the spacecraft if it were sitting on the ground.

g-factor = (sum of non-grav forces) / (m g) <-- g = 9.8 m/sWhy "non-gravitational"? The gravitational force acts on all parts of the spacecraft and its occupants equally, so it doesn't place any stress on them. (Imagine being in a rollercoaster going down a steep drop. If both you and the rollercoaster are accelerating downward at 9.8 m/s^{2}

Example 1: A space shuttle weighing 2
million kg is sitting at rest on the launchpad. Gravity is pulling the
spacecraft downward with a force of 9.8 * 2 million kg = 18.6 million N,
but the ground is supplying an equal and opposite force upward. The sum of
non-gravitational forces, divided by the spacecraft's weight, is 1, so the
craft experiences 1 g. The sum of *all* forces on the craft is
zero, so the acceleration is zero.

Example 2: The shuttle has launched, but somebody throttles the engine so that the thrust exactly equals its weight, so it just hovers at the launch pad. The situation is exactly the same as before: non-gravitational forces equal weight, so the shuttle experiences 1 g. The sum of all forces is zero, so the acceleration is zero, even though huge amounts of fuel are being burned.

Example 3: In a real launch, the shuttle's engines develop 35
million newtons of force at lift-off. Gravity is pulling down on the
space shuttle with a force of (1 g * 2 million kg) = 18.6 million newtons.
The sum of non-gravitational forces divided by the spacecraft's weight on
the ground is (35 million N / 18.6 million N) = 1.8 g. (The maximum g
loads on the shuttle do not occur at liftoff -- see below.) However, the
sum of all forces on the spacecraft is (35 million N - 18.6 million N) =
16.4 million N, so the acceleration is only 8.2 m/s^{2}. This
might lead you to suspect that your first answer is correct: that a shuttle
launching at 3 g is accelerating at 19.6 m/s^{2}. However, that's
not
always the case.

Example 4: A rocket sled of mass 1000 kg is mounted on a horizontal track
on the ground. The rocket exerts a force of 10,000 N horizontally.
Gravity is pulling down on the sled with a force of (1000 kg * 9.8) = 9,800
N. The ground is pushing up on the sled with a force equal to gravity,
9,800 N, to keep the sled from falling through the ground. Now we have to
consider the *vector sum* of non-gravitational forces. This sum has
a magnitude of about 14,000 N, and is directed at about a 45 degree angle.
The sled's g-factor is 14,000 N/9,800 N = 1.4. However, the
*net* force on the sled is 10,000 N (since the ground force and
gravity cancel), so the sled's acceleration is 10 m/s^{2}.

Finally, we come to the answer to your question. The shuttle's maximum
acceleration does not occur at liftoff, but instead at
the end of the launch, when its mass is much, much less because most of
its fuel is burned. (Its thrust is less too, because the solid rocket
boosters have separated, but the loss of mass is larger.) At this time,
the shuttle is firing its rocket engines *horizonally*, not
vertically. It is not attempting to counteract gravity (as it does at
liftoff): instead, gravity causes it to fall around
the Earth without hitting the ground. The force of the rocket engines
is the only non-gravitational force on the spacecraft: since the shuttle
weighs about 140,000 kg at this point, the rocket force is (3 * g * 140,000
kg) = 4.1 million N horizontally. However, the sum of all forces on the
orbiter is the rocket force plus the gravitational force. The grav force
is about (g*140,000 kg) = 1.4 million N downward. The vector sum of these
forces is 4.3 million N at an angle of 19 degrees from the horizontal.
This leads to an acceleration of magnitude (4.3 million N / 140,000 kg) =
31 m/s^{2}.

However, notice that the gravitational force is not making the shuttle go
any faster: it's simply allowing it to move in a circular orbit around the
Earth. If what you mean by acceleration is the "go-faster" part, that's
just the component of the acceleration due to the rocket thrust, which is
(4.1 million N / 140,000 kg) = 29 m/s^{2}.

In summary, the relationship between acceleration and "g"-loads depends on the direction in which the thrust is applied.

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