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Anonymous users2024-02-12The gravitational acceleration of the earth's surface, I also have doubts, when the earth's rotation is not considered, that is, the centripetal force is not considered, then the gravitational force is equal to the gravitational force of the object, or the gravitational force acts as gravity entirely. However, if we consider the rotation of the Earth, take the object at the equator as an example, the gravitational acceleration is equal to the acceleration due to gravity + centripetal acceleration. So I think that the gravitational acceleration of the Earth's surface should be equal to the gravitational acceleration minus the centripetal force acceleration.
However, in some questions, rotation is mentioned, that is, rotation cannot be ignored, but in the answer, the gravitational acceleration on the surface is equal to the gravitational acceleration, so I also have a big doubt, I don't know when to ignore it and when not to ignore it! Or gravitational acceleration refers to the gravitational acceleration of a near-Earth satellite, as you say, but that's not what it says in some reference books! So I hope that the high school physics teacher will answer this question!
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Anonymous users2024-02-11"Gravitational acceleration on the Earth's surface" refers neither to near-Earth satellites nor to any point on Earth. This concept refers to the acceleration of an object on the Earth's surface (at any height, but close enough to the center of the Earth so that the curvature and air resistance of the Earth can be ignored).
Without considering the rotation of the Earth, the gravitational acceleration is the entire gravitational acceleration experienced by the object. However, if the rotation of the Earth is considered, for an object at the equator, its gravitational acceleration is equal to the gravitational acceleration minus the centripetal force acceleration.
Thus, the magnitude and direction of gravitational acceleration on the Earth's surface will vary depending on the geographical location and the rotation state of the Earth. At the same time, it is not directly related to the gravitational acceleration of near-Earth satellites.
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Anonymous users2024-02-10According to the f-quote = mg (g is the acceleration around the celestial body, and the earth's surface is the acceleration due to gravity), the larger the circumference radius, the smaller the acceleration velocity of the circumference around the celestial body.
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Anonymous users2024-02-09The radius of the track is larger.
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Anonymous users2024-02-08Let the gravitational acceleration of the Earth's surface be g, mg=gmm r 2, and let the centripetal acceleration of the geosynchronous satellite be a, ma=gmm r+h) 2a=gm r+h) 2 g=gm r 2 visible, a
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Anonymous users2024-02-07Summary. Hello This is a very good question, I need a little time to answer, please be patient.
Regarding the gravitational acceleration of near-Earth satellites, it is urgent
Hello This is a very good question, I need a little time to answer, please be patient.
The centripetal force of the near-Earth satellite in a circular motion is provided by the gravitational pull of the Earth on the satellite. Therefore, it is certain that g=gm r 2 can be confirmed by the substitution formula of **
Is substitution only available on the surface of a central object? Only when there is gravity on the surface, it is approximate equal to gravity, and the near-Earth satellite belongs to the gravitational force, and all provide centripetal force, so you can't use this.
It can be used. So the gravitational acceleration of a near-Earth satellite is less than the gravitational acceleration at the equator? But the centripetal acceleration on the near-Earth satellite is greater than the centripetal acceleration on the Earth, ma=mg, isn't this a conflict?
gmm/rr=mv^2 /r=m w^2 r=ma
It's still equal to MG.
Do you have a question, show me.
He said that the acceleration of near-Earth satellites is greater than that of the equator, but if so, isn't ma equal to mg? Why is the gravitational acceleration of a near-Earth satellite greater than that above the equator? Near-Earth satellites have a larger radius.
Hey, it's been a long time since I've forgotten it.
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Anonymous users2024-02-06The gravitational acceleration of the earth is the same at different locations on the earth. （）
a.That's right. b.Mistake to let Yun Tan.
Correct Answer: Quietly Lap B
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Anonymous users2024-02-05v 2 gm(2 r-1 a) g is the gravitational constant, m is the mass of the earth, r and v are the position and velocity of the artificial object with respect to the center of the earth, respectively, and a is the semi-long diameter of the orbit of the artificial object.
Second cosmic velocity v2: At this time, the orbital radius of the artificial object is , it gets rid of the gravitational field of the earth and flies away from the earth.
v2 2=2gm r is calculated: v2=
The third cosmic velocity v3: the flat ** velocity of the earth's motion around the sun is in the earth's orbit, and the escape velocity of the artificial celestial body from the sun's gravitational field is when it is consistent with the direction of the earth's motion, and the earth's motion speed can be fully utilized, in this case, the velocity required by the artificial celestial body after leaving the earth's gravitational field is only the difference between the two v0 = set on the earth's surface and the emission velocity is v3, and the two vitality formulas are listed respectively and are combined:
v3 2-v0 2=gm(2 r-2 d) where d is the radius of the earth's gravitational force, and since d is much larger than r, the term 2 d can be ignored compared with the term 2 r, from which it can be calculated:
v3=, which is the third cosmic velocity.
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Anonymous users2024-02-04Gravitational force acceleration on a near-Earth satellite.
The gravitational velocity of du and objects on Earth is not the same. Since the NEO satellite is moving in a circular motion around the Earth (without considering the influence of friction) and is in a stable state (not falling), it means that the gravitational force is provided to the centripetal force, and theoretically there is no gravitational force on the NEO, so the gravitational acceleration is zero. (Probably microgravity, actually).
The velocity of the rotation of objects on Earth with the Earth is very small, and the gravitational force is mostly supplied to gravity (all of which is provided to gravity at the poles), and only a small part acts as a centripetal force for circular motion, which can even be ignored.
As a result, the gravitational acceleration of an object on Earth is much greater than that of a near-Earth satellite.
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Anonymous users2024-02-03The gravitational acceleration of an object on the surface of the earth is the gravitational force, and there is no supporting force.
The gravitational force formula f=gmm r 2
The centripetal force formula is f=m2 r=mw 2r
gmm/r^2=mw^2r.gm r 2=a can be obtained, so the larger the radius, the smaller the centripetal force.
And even if it is the a=rw 2 of the satellite you want, this w is the w of the satellite, not the w of the earth
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Anonymous users2024-02-02Since a=w 2*r; For the centripetal force there are: f with f surface. There is t the same to get w to be the same.
That is, the centripetal force required for geostationary satellites is greater than that required by surface objects.
At the same time, for surface objects there are:
The gravitational pull from the earth is manifested by two forces: first, the centripetal force required for circular motion; Second, it is manifested as the gravitational force of the object. And: at this time f to f million mg
The centripetal acceleration of a geostationary satellite is: a=gm (r+h) 2....h is the height of the satellite above the earth's surface.
The acceleration due to gravity on the Earth's surface is: g=gm r 2....m is the mass of the Earth; g is the gravitational constant.
It's clear: a g
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Anonymous users2024-02-01Did you learn the first, second, and third cosmic velocity? It is easy to understand that first of all, in the first, then the satellite will not exceed the Earth-Moon system, and the synchronous satellite is the same as the Earth-Moon rotation speed, if it exceeds the first speed and is between the second speed, then it will be between the Earth-Moon and the solar system, the second super will be in the Sun and the Milky Way, and the third super will fly out of the Milky Way.
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Anonymous users2024-01-31Let the gravitational acceleration of the Earth's surface be g, mg=gmm r 2, and let the centripetal acceleration of the geosynchronous satellite be a, ma=gmm r+h) 2a=gm r+h) 2 g=gm r 2 visible, a
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Anonymous users2024-01-30Compared with the earth's surface, satellites are farther away from the earth, and the gravitational force is smaller, and the gravitational force to mass ratio is called smaller.
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Anonymous users2024-01-29According to the gravitational force formula, the gravitational force at a distance is smaller, and the centripetal acceleration is smaller.
There are five types of terrain on land, including plains, hills, plateaus, mountains, basins.
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