# Why is it proclaimed for astronauts to be "falling"?

My physics teacher explained that the reason why it looks like astronauts are floating is because they are "falling". However, the way I think of it is that the outwards inertial force of the astronaut as he/she orbits the earth is equal to and opposite the force of gravity from the planet. Therefore, the net force is 0. But I don't see how this means the astronaut is "falling".

• The earth is spherical with 6000 km radius. Have an astronaut float 100km from the surface. Make the origin [0,0,0] the center of the earth. Let the astronaut start at position [6100,0,0]. Give this astronaut an initial [0,8,0] km/sec velocity. The earth mass is 6 x 10^24 kg. The astronaut mass is 100 kg. The force exerted on her is F=GMm/r^2. The vector acceleration at this instant is [0, -GM/r^2, 0]. Find the vector position of the astronaut after 1 sec. Calculate the next position and repeat. You will find that she is orbiting the earth, using only this pull! It's a marvel. Commented Nov 24, 2022 at 16:43
• Commented Nov 24, 2022 at 17:48

We always need to be careful with fictitious forces, of which the centrifugal force is one. In the reference frame of the earth, there is only one force acting on an astronaut in orbit, and that is gravity.

From the earth's frame, the astronaut is falling with some acceleration (g or perhaps a little less than that if they're far away) but they never hit the earth because they're also travelling with a large velocity around the earth. (This is after all what orbiting is).

If we instead look at the forces in the frame of the astronaut we need to be careful, because the astronaut's frame is a non-inertial frame of reference, and so fictitious forces are present. In the astronauts frame there is the force of gravity, and a fictitious centrifugal force that exactly balances it.

At the end of the day whether we really call what is happening "falling" or not doesn't matter very much. Most correctly there is only one force acting on the astronaut, gravity, just like it would be if they were on earth, but the astronaut happens to have a huge horizontal velocity, so when they "fall" towards the earth, they have already moved partway around the earth.

• That is, of course, the approved Newtonian way to look at it. But it's also OK to use centrifugal force: it yields the same answer in problems like this. And in General Relativity, the force of gravity is just as fictitious as centrifugal force. Commented Nov 24, 2022 at 16:48

Falling can have slightly different experiences/observations. From the pov of a person falling freely inside some capsule/cabin, he won't experience any normal reaction from the walls of the cabin, as the whole system fall with equal acceleration. However, from the POV of the ground observer, he sees the cabin falling toward the ground.

Falling can be defined as moving under the influence of the gravity of the celestial body, whether it be an asteroid having its course changed, or rain falling.

The astronaut is said to be falling around the earth instead of falling towards the earth. This is because of horizontal velocity with respect to the ground of the cabin which keeps it going around in orbit, albeit always under influence of gravity. So the astronaut is falling, but around, not directly downwards. The equation of force can be written as:

$$F_g=F_{centripetal}$$ $$\frac{GmM}{r^2}=\frac{mv^2}{r}$$