Why do we feel gravitational acceleration from the Earth and not from the Sun?

[ejacques]

The acceleration near the earth, due to the force of gravity is g. Now every particle when moving in a curve trajectory had a centripetal acceleration towards the center (say the sun) a=(v^2)/R.
If this is true why we measure weight only with the account of g?
I guess when R is big it might be neglected, but still I wonder

 

[Ibix]

You don't "feel" a gravitational force from the Sun because you feel the same acceleration the Earth does, so you accelerate the same as all your local references. So you just go around the Sun without noticing anything.

You do see variation in gravity due to the presence of the moon and sun, though. This is the cause of tides and spring tides. It's just not a very large effect on a human scale, and depends on the gradient of the gravitational field strength, not the strength itself.

 

[A.T.]

If this is true why we measure weight only with the account of g?

With a scale, we don't measure the Earth's gravitational force directly, just the force that opposes it.

But nothing opposes the Sun's gravitational force.

 

[Drakkith]

The Earth, and everything on it, is in free fall around the Sun as we move in our orbit. But we are not in free fall around the Earth. Hence you feel the Earth's surface pushing back up on you. If you could stand on a solid surface on the Sun you would absolutely 'feel' the Sun's gravity. Or if we built a giant shell around the Sun and could stand on it without moving in an orbit we would also 'feel' the Sun's gravity.

 

[PeroK]

I guess when R is big it might be neglected, but still I wonder

If you do the maths, then the gravitational acceleration of the Earth from the Sun is very small:
$$g_{s} = \frac{GM_s}{R^2} = 0.006 m/s^2$$And, using $T = \frac{2\pi R}{v}$ for the period of the Earth's circular orbit, we can rewrite the equation for centripetal acceleration:
$$a_c = \frac{4\pi^2 R}{T^2} = 0.006 m/s^2$$

 

[A.T.]

If you do the maths, then the gravitational acceleration of the Earth from the Sun is very small:

And no matter how large it would be, a scale on Earth would only be affected by its gradient, as @Ibix noted.