Providing the centripetal force through gravitational force, it can be seen that the larger the radi

No. When a satellite is moving in a uniform circular motion around a certain orbit, if it wants to go to an orbit with a larger radius, it needs to overcome the earth's gravitational pull and provide energy. Hey, why is that?

We integrate the gravitational force formula. w':f:

gmm r 2, get the gravitational work fx:w:-gmm r;;; Suppose the object has a velocity of V at height r and rises from height r to 4r.

v':v 2 is obtained from vvm r:gmm r 2.

Change from the conservation of energy ek to gravity w:x. Equal to [v 2 2-(v 2) 2 2] [gmm r-(-gmm 4r)]:

x;(3mv 2 2)-3gmm 4r:x because v 2:gm r.

So x:[(3 2)-3]gm r is also equal to -3 2mv 2... What does a negative value of x mean???

It shows that the kinetic energy of the object plus x is equal to the work done by gravity. That's what it takes to rise and absorb energy.... For some reason, the phone cannot type an equal sign, but can only use a colon':

Instead, understand....

The orbital velocity and the launch velocity are not the same.

The bigger the what, the bigger the bigger (smaller), this is called proportionality in life or in physics. On the contrary, proportionality requires certain conditions - the control variable method! Simple example:

The higher the voltage, the greater the current – this statement is false. Because it doesn't have a control variable resistance that doesn't change! That is, there is no mention of how the resistance changes!

So it's wrong. Straight to the point!

You said: "The larger the radius, the smaller the angular velocity". Its condition is: the centripetal force in the circular motion is certain before it is established!

f=mω²r

In addition, the larger the radius, the smaller the (linear) velocity, and its condition is: the angular velocity is certain or constant, and it is true.

v= r, if the conditions are different, then this ratio will not hold.

The motion of an artificial satellite (celestial body): the centripetal force is provided by gravity: f=gmm r =mv r to deform the equation:

gm=v r, i.e. v r is a fixed value. That is, v is inversely proportional to r.

To increase the radius, the linear velocity must be reduced.

Your statement that "if the celestial bodies move off orbit, the speed will be greater" is correct. It turns out that the satellite orbits the earth, and the gravitational force just provides the centripetal force, and when the speed increases, a greater centripetal force must be needed, so the original centripetal force is not enough, then the satellite will do centrifugal motion, that is, deorbit "outward". When the required position is reached, if you want the satellite to move in a circle again with a new radius, then you have to reduce the speed !!

Re-reach an equilibrium gm=v' r' where: v'r。

Deorbiting requires high speed. First of all, you have to understand that the deorbit is from near to far, that is, the radius is from small to large, so to overcome the gravitational pull with a small radius, you need a large speed, the greater the velocity, the greater it is to a certain extent (the first cosmic velocity), the gravitational pull of the earth is not enough to bind the object, and the object will be out of orbit. There is also the first.

Second, the third cosmic velocity, which will not be explained in detail here, is similar in principle.

To get out of orbit is to get rid of gravity, and to get rid of gravity is to get rid of gravity, you need a lot of speed. Just like launching a rocket, a rocket needs to get rid of the gravitational pull of the Earth in order to get into space. Without getting rid of gravity, the larger the radius, the slower the speed when moving in an orbit bound by gravity.

Not contradictory! First, two, three, the cosmic velocity, the gravitational force calculates the core, the mass of the object is different! If you do a rough calculation.

The first speed with the earth.

The second speed with the sun.

Third speed with the Milky Way.

The law is established in every quality center system!

What is the orbital equation for the motion of a particle in a gravitational field.

In the gravitational field, the trajectory of the particle can be described by the Kepler equation of motion: r = frac, where r is the distance between the particle and the center of gravity (i.e., the sagittal diameter from the particle to the center of gravity), a is the semi-axis of the elliptical orbit, e is the eccentricity of the elliptical orbit, is the true perisis angle of the particle in the orbit (i.e., the angle between the line between the particle and the center of gravity and the line between the pericentric point and the center of gravity), and 0 is the reference angle of the mountain key orbit. This equation can be used to calculate the trajectory of the ridge in the gravitational field.

The first cosmic velocity is the velocity necessary for a near-Earth satellite to move in a circle around the Earth, as long as it is a circular motion, it is satisfied that the larger the orbital radius, the smaller the linear velocity. The second and third cosmic velocities are the velocities required to get out of the earth and the solar system, and when the velocities reach that point, they no longer move in a circular motion, and naturally do not satisfy the law of circular motion.

The first cosmic velocity is the maximum operating velocity and the minimum launch velocity (launching the Earth satellite) and the second cosmic velocity is the escape velocity, which is the velocity at which the launcher leaves the Earth, the two are not the same, and the formula you said is to compare the orbital velocity (around the same central celestial body).

Centripetal motion, the radius is the radius of the circumambulation.

Basis: The centripetal force of the spacecraft moving in a uniform circular motion around the Earth is provided by the gravitational force.

Get: 4m 2r t 2 = gmm r 2

m=4π^2r^3/gt^2

f 10,000 f direction.

g*m*m r 2 m(2 t) where m is the mass of the earth and m is the mass of the spacecraft.

The mass of the Earth is m (2 t) 2*r 3 g (2 ) 2*r 3 (g*t 2 ).

In this type of problem, there are two different length values. In the case of the solar-terrestrial system:

One is, of course, the so-earth distance (the distance between the heliocentric and the geocentric, of course), and the other is the so-called radius of orbital curvature. The radius of curvature of the orbit is defined as follows: Suppose the Earth passes through the three points in the orbit in turn, abc, and since abc is not in a straight line, a circle can be determined by passing through abc.

When AC is infinitely close to B, there is a limit circle, and the radius of this limit circle is the radius of curvature of point B.

We use the radius of curvature when calculating centripetal acceleration, and the distance between the sun and the earth when calculating gravitational force. Obviously, the magnitude and direction of the two are different, then we can decompose the gravitational force into a force f1 perpendicular to the direction of velocity and a force parallel to the direction of velocity, which is the centripetal force, which plays the role of changing the direction of velocity, and f2 plays the role of changing the magnitude of velocity.

It seems that the reason for elliptical motion is that the f centripetal force is not equal to the f gravitational force. When the gravitational force is high, it goes inward (centripetal motion), and when the hour goes outward (centrifugal motion).

Of course, because the radius is the same, from the centripetal force gravitational force, we get a = (gm divided by r squared) under the quadratic root.

Yes, only by gravitational pull.

Ceres' asteroid orbits the sun, by the cow two acacia liquid, gmm1 r1 2=m1r1(2pai t1) 2, the earth orbits the sun, gmm2 r2 2=m2r2(2pai t2) 2, r2 is the distance from the earth to the sun, it takes 8 minutes to take light from the sun to the earth, the speed of light is 300,000 km s, t2 is the earth's lead jujube around the sun for 1 year, and the solution is t1=

This statement is wrong.

The formula for centripetal acceleration A v 2 r or a 2 * r is correct to say: When the magnitude of linear velocity is constant, (known by a v 2 r) centripetal acceleration is inversely proportional to the radius.

When the angular velocity is constant, the centripetal acceleration (known by a 2 * r) is proportional to the radius.

Note: When two variables have an effect at the same time, the precondition (e.g., one of the quantities is constant) must be explained clearly before the proportional relationship can be known.

Velocity is a vector, and if you say that the magnitude of velocity does not change, this sentence is true.

Centripetal acceleration is determined by the magnitude of the centripetal force and the mass of the object, and has nothing to do with velocity and radius" means that centripetal acceleration is not directly related to velocity and radius, such as the velocity is large but the centripetal acceleration is not necessarily large, a=f to m is the determinant of centripetal acceleration, and a=v 2 r= 2 r is the definition of centripetal acceleration.

1. The centripetal force is the condition that the object needs to meet for circular motion, which is related to the speed of the object and the radius of the orbit.

2. Gravitational force: One of the factors that affect the magnitude of gravitational force is the distance between two objects. There's no radius here.

3. In the motion of celestial bodies, when celestial body A moves in a circle around celestial body B (for example, the moon revolves around the earth), the orbit of celestial body A takes celestial body B as the center and the orbital radiusExactlyIt is equal to the distance between a and b.

4. There are other forms of celestial motion. For example, in a binary star system, two bodies orbit each other, with a point on the line as the center of the circle. The sum of their respective orbital radii is equal to the distance between the two bodies.