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

AI Thread Summary
Gravitational acceleration on Earth is primarily influenced by the Earth's gravity (g), while the Sun's gravitational pull is negligible at our scale, measuring only about 0.006 m/s². We do not "feel" the Sun's gravity because both the Earth and everything on it are in free fall around the Sun, resulting in no relative acceleration felt locally. The gravitational effects from the Sun and Moon do cause variations, such as tides, but these are minor compared to the direct gravitational force we experience from Earth. Scales measure the force opposing Earth's gravity, not the Sun's, which explains why weight is calculated using g alone. Thus, while the Sun's gravity exists, its effects are not perceptible in our daily experience on Earth.
ejacques
Messages
3
Reaction score
0
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 🤔
 
Physics news on Phys.org
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.
 
ejacques said:
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.
 
  • Like
Likes sophiecentaur and vanhees71
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.
 
  • Like
Likes russ_watters and vanhees71
ejacques said:
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:
$$a_c = \frac{4\pi^2 R}{T^2} = 0.006 m/s^2$$
 
PeroK said:
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.
 
Last edited:
  • Like
Likes hutchphd and vanhees71
The rope is tied into the person (the load of 200 pounds) and the rope goes up from the person to a fixed pulley and back down to his hands. He hauls the rope to suspend himself in the air. What is the mechanical advantage of the system? The person will indeed only have to lift half of his body weight (roughly 100 pounds) because he now lessened the load by that same amount. This APPEARS to be a 2:1 because he can hold himself with half the force, but my question is: is that mechanical...
Some physics textbook writer told me that Newton's first law applies only on bodies that feel no interactions at all. He said that if a body is on rest or moves in constant velocity, there is no external force acting on it. But I have heard another form of the law that says the net force acting on a body must be zero. This means there is interactions involved after all. So which one is correct?
Let there be a person in a not yet optimally designed sled at h meters in height. Let this sled free fall but user can steer by tilting their body weight in the sled or by optimal sled shape design point it in some horizontal direction where it is wanted to go - in any horizontal direction but once picked fixed. How to calculate horizontal distance d achievable as function of height h. Thus what is f(h) = d. Put another way, imagine a helicopter rises to a height h, but then shuts off all...
Back
Top