# How is space ship's acceleration perceived if the acceleration is perpendicular to the velocity?

Spacecraft in orbit around the Earth are constantly accelerated by the gravitational field of Earth. That's why the spacecraft ($m \ll M$) is in an (elliptical) orbit around the centre of gravity of the Earth, accelerated by

$\mathbf{g}=-{G M \over r^2}\mathbf{\hat{r}} \, .$

Plugging in the numbers for a spacecraft orbiting at roughly 420 km (such as the International Space Station), this gives:

$\mathbf{g} = {- 3.986 \cdot 10^{14} \over (6371 + 420) \cdot 10^3 m} = - 8.6 \, \mathrm{m}/\mathrm{s}^2$

So, an astronaut on-board the ISS is in a reference frame that is constantly accelerating at an acceleration of $8.6 \, \mathrm{g}/\mathrm{m}^2$. Yet unlike the astronauts featured in this related question, they do in fact feel no gravitational acceleration at all; at most they may feel some centrifugal pseudoforce, but this is considerably less than the gravitational acceleration, and at most at microgravity levels.

Is acceleration only perceived when it changes the magnitude of the velocity, as opposed to the direction? What is the fundamental reason for this? Or am I misunderstanding something?

• Note: your acceleration has the wrong units. Acceleration is length/time^2. – Michael Brown Feb 11 '13 at 13:12
• @MichaelBrown Thanks for pointing out my typo, fixed it now (handicap from daily working with data that has units $\textrm{g}/\textrm{m}^2$...) – gerrit Feb 11 '13 at 13:14

In the case of the freefall orbit the Earth's gravity pulls on the ship. But it also pulls on the astronaut and everything else inside the ship equally, so there is no force from the floor on the astronauts feet. This is because the force of gravity is proportional to the mass $F=mg$, and by Newton's law $F=ma$, so $m a = m g$ and the mass cancels out. Everything accelerates equally. This is the (weak) equivalence principle.