Why do some people think that the gravitational constant g is decreasing?

Fallacy. The formula for calculating the gravitational constant shows that as long as the mass and radius of the earth do not change, g will not change, so don't be misled.

If the gravitational constant g increases, then the action of gravitational force becomes stronger. Newtonian mechanics is mainly applicable to low-speed motion and weak gravitational fields, but not to high-speed motion close to the speed of light or strong gravitational fields, which need to be described by the theory of relativity.

When the gravitational constant g increases, the gravitational force becomes stronger, then in some cases, the scope of application of Newtonian mechanics may become smaller. For example, in the motion of celestial bodies that could have been described by Newtonian mechanics, if gravity becomes stronger, then the relativistic effect may become more pronounced, causing Newtonian mechanics to no longer apply. However, it depends on the situation, as sails and ridges are in many everyday situations, and an increase in the gravitational constant g may not be sufficient to invalidate Newtonian mechanics.

In general, an increase in the gravitational constant g may lead to a smaller range of application of Newtonian mechanics.

The gravitational constant is very small, indicating that the mass ratio of the gravitational force is very small, so it is negligible is wrong? For example, the gravitational acceleration gm effect produced by the mass of an artificial satellite is only a few thousand kilograms, and the gravitational acceleration gm ratio of the earth with a mass of 10 to the 24th power is naturally negligible. The radius of the static orbit and the mass calculated by Li Mengqing's period belong to the Earth.

However, the gravitational acceleration of the Moon, which has one-eighty-one Earth's mass, cannot be ignored, so the mass calculated by lunar orbit observations is the sum of the Earth and the Moon.

It follows that Newton's formula for gravitation, f gmm r mg m, is only suitable for calculating negligible approximations of mass and Earth, such as artificial satellites, and for the Moon, the natural formula of mutual gravitation should be used

f＝mg﹙m＋m﹚／r² ＝mg﹙mm﹚。

f=gmm/r^2

Although g is small, as long as the mass is large, the product mm will be large and f will be large, so gravitational force cannot be ignored in any case because g is small. Only the banquet may be ignored when the mass is not too large or the old man is relatively small. Concealed.

High school physics considers it constant.

The gravitational constant g does not change. After measuring the gravitational force between some objects and calculating the gravitational constant g, the gravitational force between various objects was measured, and the results obtained were the same as those obtained by using the gravitational constant g according to the law of gravitation. Therefore, the universality of the gravitational constant becomes a witness to the correctness of the law of gravitation.

It is said to be constant, how can it be changed.

The answer is C

The gravitational constant g measured that the predecessors also had to rely on gravity to stand on the earth, which is also of practical value, A is wrong.

The magnitude of the gravitational constant g is invariant and has nothing to do with the mass of the object, b is wrong from the gravitational formula, c is right.

The magnitude of its value is related to the choice of the unit system, d is wrong.

So the answer is C

ab is clearly wrong.

C is right. The first half of the sentence d is correct, the second half of the sentence is a bit problematic, the unit system changes, the actual size of g does not change, and the value changes. I think the person who wrote the question meant that it was human, and this sentence is right.

t=24h=24*60*60=86400sw=2π/t=1/43200rad/s

Earth's surface: mg=gmm r2

gm=gr^2

Geostationary satellites: m'w^2(r+h)=gmm'(r+h) 2(r+h) 3=gm w 2=gr 2 w 2h = (gr w ) to the 3rd power root -r

It's too hard to calculate, so you can calculate the results yourself.

Answer: GM r 2=g so GM=gr 2, angular velocity w=2 t, let the mass of the geosynchronous satellite be m, then gmm r 2=mr(2 t) 2 [t is the period of rotation of the earth] can be found r, v can be found from v=rw, and h can be obtained from the radius r-r from the center of the earth

gmm r2=mv2 r know, then replace gm=gr2 with **, use v=root number gr, that is, v2=gr, v uses (2 factions t) r, get (2 factions t) square r=g, get r, and then use r to subtract the radius of the earth to get the height h

From mg=gmm r2, gm=gr2, geostationary satellite: gm r2=4 r 2 t2 to solve r is the distance from the geostationary satellite to the earth's spherical star, the height r-r, with the radius behind it is simple.

The angular velocity of a geostationary satellite is known, w=2 |t t=24h followed by just a formula.

A. Formula f=gmmr

where g is the gravitational constant, which is measured experimentally by Cavendish and therefore a is wrong;

b. When the distance between two objects approaches 0, the two objects cannot be regarded as particles, and the gravitational formula is no longer applicable, so B is wrong;

c、m1m2

The gravitational force between them is an interacting force, so it is always equal in magnitude, and it has nothing to do with whether the masses of m1m2 are equal, but it has to do with the product of their masses so c is correct;

d. For the distance between the centers of the spheres with uniform mass distribution, r can be regarded as the distance between the centers of the spheres, so d is correct;

Selection: CD