The gravitational pull of the sun on the planets is experimental

Gravitational force exists objectively and has nothing to do with anything subjective. Only its law is recognizable, and its law is option b and a should be equal to the gravitational value. c Obviously, the gravitational force of the sun has nothing to do with experiments, and gravity exists whether to do experiments or not; d is inherently wrong because it is not moving at a uniform speed, and it forgets Newton.

What kind of problem is this?! It's shit, and at one point it's circular motion, and Kepler's law is elliptical motion. Newton's law of universal gravitation is rumored to have been figured out by a smashed head.

In fact, at that time, he thought that there would be a force called gravity. And then I wanted to generalize it to the moon and the sun, and in order to conform to Kepler's laws, I invented calculus. The law of gravitation was then derived from Kepler's law using calculus.

So B is correct.

A: The centripetal force of the planet's circular motion around the sun is provided by the gravitational force of the sun on the planet, so the gravitational force of the sun on the planet is equal to the centripetal force of the planet doing uniform circular motion. A correct.

b: The centripetal force of the planet is provided according to the gravitational force of the sun on the planet, and the gravitational force between the sun and the planet is provided (this cannot be written above.) This results in direct proportion to the mass of the Sun and the mass of the planet, and inversely to the distance between the planet and the Sun. b Not true.

C: The gravitational pull of the sun and the planets is derived. C is not true.

D: The gravitational attraction of the sun to the planets is derived from Kepler's law and the uniform circular motion of the planets around the sun. d correct.

In class, the teacher is talking about... About the particle refers to the time when the object is in motion, note that it is in motion, and in order to facilitate calculation and observation, the object can be regarded as a particle.

So B's point is correct.

Because these planets all happened gravitationally, they disappeared on their own before they could get close to the Sun.

Because the gravitational pull of the Sun is not particularly strong, and these planets do not pass through the Sun when they are moving, it is impossible for them to fall into the Sun.

The reason for the occurrence of universal repulsion force (brief description): When the high-energy particles running in the universe pass through the matter (celestial body or particle), some of the high-energy particles that are intercepted and absorbed interact with the matter (celestial body or particle), and realize the energy exchange according to the law of conservation of energy conversion mv2=e=mv2, and radiate the corresponding material energy, which is formed in the surrounding space"Eggplant energy stacks", i.e., the strong energy zone. This strong energy zone is"Repulsion"Range.

Due to the prevalence of this kind in the cosmic space"Bury the Nachali", so it is also called"Repulsion"。

There will be some interaction forces between the planets, through which this will not happen.

The rotation of the earth has a centrifugal force that balances the gravitational pull of the sun!!

The gravitational force is not only proportional to the mass of the object, but also proportional to the gravitational potential where the object is located, although the mass of the two is different, but due to the gravitational force of different objects, the gravitational potential is also different, and the multiplication is exactly the same, which is where you ignore.

Specifically, for a planet, its mass is m, which is subject to the gravitational pull of the sun, and the gravitational potential where it is located is proportional to the mass of the sun, which is gm r 2, so the gravitational force is gmm r 2; For the Sun, its mass is m, which is subject to the gravitational pull of the planet, and the gravitational potential is proportional to the mass of the planet, which is gm r 2, so the gravitational force is also gmm r 2. The gravitational pull experienced by both is exactly the same.

It is related to the ride of mm, and has nothing to do with the quality of m or m. Besides, the sun to the planets and the planets to the sun are a pair of interaction forces, of course, equal.

related, rather than "only" proportional to it (you neglected to consider the object itself), you know. f=gmm r 2, the other side is gravitationally related regardless of which side it is.

Gravitation is famous. Newton famously found a quantitative relationship between gravity and the mass and distance of an object, where gravity is inversely proportional to the square of the distance between them. And the key, or the difficulty, is how to understand the square.

Because the gravitational superposition between objects is not a photon-like primerQuantitysuperimposed, but the primer propagationSpeed, resulting in a gravitational superposition that the primer propagates twice as fast as the gravitational velocity. Since both gravitational force and kinetic energy have the concept of unit time, the magnitude of the gravitational force is proportional to the square of the propagation speed of the primer, so that the gravitational force between two objects is not the sum of the gravitational forces of the two objects, but the square of the product of the gravitational force of the two objects, so it is divided by the square of the distance.

A simple answer is to choose a planet in the solar system such as the Earth, which forms a binary star system with the sun, and the ratio of the mass of the two is about 1.4 million times, of course, the sun is heavier, which means that the two will make a circular elliptical motion centered on the center of mass of the two in the gravitational force between the two at the same time, and the ratio of the distance between the two and the center of mass is equal to the inverse proportion of mass, so the center of mass is extremely close to the sun.

The above only considers the ideal situation, that is, there is only one planet and no other external forces, as for the reality, the position of all the planets is not in a straight line, and it is not on the same side of the sun, some of the forces cancel each other, and the mass of the sun is huge, so the movement of the sun is not obvious.

Moreover, the description of motion is relative to the frame of reference, and the solar system is centered on the sun, and the sun does not move.

No, the sun is actually moving. But because the sun is large, the balance point of gravity is in the interior of the sun.

So it looks like the Earth is revolving around the Sun.

You can refer to the relationship between the moon and the earth.

The centripetal force is the same, both are the gravitational force between the two, and this pair of gravitational force is the action force and the reaction force, and the magnitude is equal.

You judge by :mv r that there's clearly a problem. First of all, they have different speeds. Secondly, r is also different, for the sun and the planets, both are moving in a circle around a point on their line, and their orbital radius r is not the distance between them.

You are talking about relative motion. is wrong. Because both the planets and the sun are moving in a circular motion and both have acceleration, they are not inertial frames of reference, so when studying, using one of them as a reference to study the motion of the other, Newton's second law is not valid.

However, for the Sun, his acceleration is small and can be considered stationary, so when studying the motion of the planets, we can think of the Sun as an inertial frame of reference. To study the movement of the sun, one cannot use the planets as a reference. He is not an inertial frame.

I have a similar question :

In the free fall motion near the ground, the ground (i.e., the earth) is used as the reference frame, and the acceleration of the falling body can be g=gm r 2, where g is the gravitational constant, m is the mass of the earth, and r is the radius of the earth. Branches.

Conversely, taking the falling body as the reference frame, the Earth accelerates closer to the falling body with an acceleration of g'=gm r 2, where m is the mass of the falling body.

In the transition frame of reference, the acceleration g and g' of the falling body and the earth should be the same size, but according to the above equation, the two are hardly equal. So why did this happen?Is the above inference wrong?

If you can answer my question, I believe you will understand your problem.

The first thing to do is to familiarize yourself with the law of gravitation.

According to this law, gravity can be derived as long as the mass and distance of the sun and planets are known. But none of these three can be measured directly. Therefore, we usually have to rely on Kepler's three laws, derive the planetary cycle from astronomical observations, and then derive the solution.

The law of gravitation is derived from Kepler's law as follows:

If the orbits of the planets are approximately circular, the angular velocity of the planets is certain from Kepler's second law, i.e.,

2 t (cycle).

If the mass of the planet is m, the distance from the Sun is r, and the period is t, then the magnitude of the force on the planet is .

mrω^2=mr（4π^2）/t^2

In addition, it can be obtained from Kepler's third law.

r3 t 2 = constant k'

Then the force along the direction of the sun is.

mr（4π^2）/t^2=mk'（4π^2）/r^2

From the relationship between the action and reaction forces, it can be seen that the sun is also subjected to the same magnitude of the above-mentioned forces. From the perspective of the Sun, the mass of the Sun m) (k''）（4π^2）/r^2

It is the sun that experiences a force in the direction of the planets. Since they are forces of the same magnitude, it can be seen from the comparison of these two equations that k'contains the mass of the Sun m,k''contains the mass m of the planet. From this, it can be seen that these two forces are proportional to the product of the masses of the two celestial bodies, which is called gravitational force.

If a new constant (called the gravitational constant) is introduced, and the masses of the Sun and the planets are considered, as well as the previously derived 4· 2, then it can be expressed as.

Gravitational force = (gmm) (r 2) The gravitational force between two ordinary objects is so small that we are not aware of it and can be disregarded. For example, if two people with a mass of 60 kilograms are meters apart, the gravitational force between them is less than one millionth of a newton, and an ant drags a stalk of grass with 1,000 times the force of this gravitational force! However, in the celestial system, gravitational attraction plays a decisive role due to the large mass of the celestial body.

The Earth, which is still relatively small in the mass of celestial bodies, already has a huge effect on the gravitational pull of other objects, it binds humans, the atmosphere and all terrestrial objects to the Earth, and it makes the Moon and artificial Earth satellites revolve around the Earth without leaving.

When moving in a circle on the surface of a planet, the gravitational force can be regarded as gravity, both mg = (gmm) (r 2) and gm = g (r 2), which is the substitution formula for **. and there is mr 2 = mr (4 2) t 2 = mg. (This conclusion is only for the surface of the planet).

A. A is briefly explained as follows:

aThe sun and the planets interact with each other, so they are interacting forces.

b The interaction force between the sun and the planets is equal to the magnitude.

The centripetal force is the resultant effect force, and the directional validity is limited to the radius direction.

d The closer an object is to the central celestial body, the faster the velocity and the smaller the period.