Are the centripetal forces of satellites and objects on the Earth provided by gravity? If not, what's the difference

The object on the earth's surface is subject to two forces: the earth's gravitational force on the object and the ground's supporting force on the object. The supporting force is equal to gravity. The gravitational force minus gravity is the centripetal force of the object. Or in this way, the gravitation is divided into two parts. The first part makes the centripetal force, and the rest makes the gravity of the object

On the relationship between gravity and centripetal force between earth and satellite
There is only one universal gravitation between the artificial satellite and the earth, so the universal gravitation is equal to the centripetal force at this time, and because the centripetal force + gravity = universal gravitation, so the gravity is 0, isn't it? Why does the Internet say that gravity provides the centripetal force at this time? Why is gravity equal to the universal gravitation at high altitude? Does the artificial satellite have the centripetal force that rotates with the earth? If it is the sun and the earth, The earth moves around the sun in a circular motion at a constant speed, so is the centripetal force equal to the force of gravity?

First of all, I'm sure some of your statements are right: 1. There is only one universal gravitation between the satellite and the earth
2. The satellite circulates around the earth at a uniform speed. At this time, gravity is equal to centripetal force
Now that you have made clear the force on the satellite at this time, why do you still have the problem of gravity? This is because you are used to studying objects at rest on the earth. Objects at rest on the earth satisfy centripetal force + gravity = universal gravitation, while objects at rest on the earth have almost zero centripetal force, so gravity is equal to universal gravitation. On satellites, we still mix universal gravitation with gravity according to our habits
Universal gravitation is a popular but not too strict statement. It is similar to the static objects on the earth. It is still used for satellites and objects in space. Or gravity is the pronoun of universal gravitation

Under what circumstances does gravity = universal gravitation, centripetal force = universal gravitation, and gravity = centripetal force, there will be gravity = centripetal force = universal gravitation

At both ends of the earth's axis, gravity = universal gravitation
All satellites (Moon) centripetal force = gravitation
When the satellite flies over the pole along the meridian, gravity = centripetal force = universal gravitation

The angular velocity and linear velocity of two satellites in uniform circular motion are W1, W2 and V1, V2 respectively. The ratio of orbit radius R1: R2 = 1:2, the ratio of angular velocity and linear velocity is calculated

Because gravitation is equal to centripetal force
So ① g * MM1 / R1 squared = M1 * V1 squared / R1
We get V1 = GM / R1 under root sign
Similarly: V2 = GM / r2 under root sign
So V1 ∶ V2 = radical (R2 ∶ R1) = radical 2:1
② G * MM1 / R1 squared = M1 * W1 squared R1
We get the third power of GM / R1 under W1 = root sign
Similarly, W2 = GM / r2 cubic under root sign
So W1 ∶ W2 = root sign (R2 cubic: R1 cubic) = 2 times root sign 2:1
It's troublesome to have no sign

If the orbit radius Ra of two artificial earth satellites is known to be 2rb, the ratio of their linear velocity, angular velocity, acceleration and period is 0

VA: VB = 1: radical 2
wA:wB=1:1
aA:aB=1:4
TA:TB=1:1

The ratio of the earth's two satellites M1: M2 = 1:2, the ratio of the radius of the circular orbit R1: R2 = 1:2
The ratio of the earth's two satellites M1: M2 = 1:2, the ratio of the radius of the circular orbit R1: R2 = 1:2
Find: (1) the ratio of linear velocity;
(2) The ratio of angular velocity
(3) The ratio of operation cycle
(4) Centripetal force ratio

GMM / R ^ 2 = MV ^ 2 / R v = √ (GM / R), so V1: V2 = √ (R2 / R1) = √ 2:1
GMM / R ^ 2 = MRW ^ 2, w = √ (GM / R ^ 3), so W1: W2 = √ (R2 ^ 3: R1 ^ 3) = 2 √ 2:1
GMM / R ^ 2 = M4 π ^ 2R / T ^ 2 get t = √ (4 π ^ 2R ^ 3 / GM) T1: T2 = 1:2 √ 2
F=GMm/r^2 F1：F2=2:1

The period of a near earth satellite around the earth is 84 minutes. If the period of the moon around the earth is 30 days,
How many times the radius of the moon's axis is that of the earth

According to the formula of universal gravitation and the two laws of Newton:
F=ma=GMm/(r^2)=ma=m*w^2*r=m*(2π/T)^2*r
So R1: R2 = triple root (T1 ^ 2 / T2 ^ 2) = 64.19
So the orbit radius of the moon is 64.19 times that of the earth

What are the linear velocity, angular velocity and period of all kinds of artificial satellites (including synchronous satellites)

A: the basic principle is that gravity equals centripetal force
Calculate the linear velocity V from GMM / R & # 178; = MV & # 178 / R=
Calculate the angular velocity from GMM / R & # 178; = Mr ω & # 178=
Find the period T from GMM / R & # 178; = MR (2 π / T) &# 178=
Sometimes it is necessary to use that the gravity near the ground is equal to the universal gravitation, that is, Mg = GMM / R & # 178;, that is, the so-called Golden substitution relation GM = GR & # 178; (where G is the acceleration of gravity near the ground and R is the radius of the earth.)

The ratio of linear velocity of a and B satellites is 2:1, and the ratio of rotation radius, angular velocity, period and centripetal acceleration of the satellites are calculated

1.1:4
GMm/r^2=m*v^2/r
r=GMm/v^2
2.8:1
GMm/r^2=m*w^2*r
3.1:8
w=2π/T
4.64:1
F=ma=mw^2r

In uniform circular motion, the greater the radius, the greater the a.b.c. running cycle and the greater the centripetal acceleration

Uniform circular motion, that is, circular motion with the same linear velocity
A obviously wrong
The larger the radius is, the smaller the angular velocity is
C correct. The larger the radius, the larger the circumference, and the same linear velocity, the longer the period
The larger the radius is, the smaller the angular velocity is, and the smaller the centripetal acceleration is
The answer is C

Are the centripetal forces of satellites and objects on the Earth provided by gravity? If not, what's the difference