What is the escape of the planets? What is the escape velocity of the planets of the solar system?

The planet is out of the gravitational range of the star in its motion around the star. Taking the solar system as an example, planetary escape means that all eight planets have left the solar system, and the fundamental reason is that the gravitational pull of the stars becomes smaller, and the planets cannot be restrained from continuing to move in their original orbits.

It's all gravitational attraction, when a planet's own momentum can cancel out the central gravitational pull.

is the second cosmic velocity, and the planets are different, and the velocities are different.

When the gravitational force between the satellite and the planet is not enough to provide the centripetal force that orbits the planet, the satellite will break away from the planet and move far away, which is called satellite escape; There is also the mutual bondage of particles in microscopic physics that is freed from each other, which is also called escape.

When it comes to escape, it is usually said that it is "the escape velocity of the planet", which means that if something is launched from the surface of the planet, it must at least reach the escape velocity of the planet in order for it to completely leave the planet without falling back; It's not about planets leaving the solar system.

For the Earth, we launch something, such as a rocket, which must be reached before it can leave the Earth and travel the solar system, and this is the escape velocity of the Earth, which is the minimum speed required to leave the Earth.

When a celestial body is freed from the restraint of another celestial body, it does not revolve around it.

Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, Neptune.

The speed of each star is different.

I see, it seems to be in the high school books.

The escape velocity of a star is the lowest velocity at which the star can escape the gravitational restraint of the astral body.

The escape speed depends on the mass of the planet. If the mass of a planet is large, its gravitational pull will be strong, and the escape velocity value will be high. Conversely, a lighter planet will have a smaller escape velocity.

The escape velocity also depends on the distance of the object from the center of the planet. The closer the distance, the greater the escape velocity.

If an object is shot vertically upwards on the surface of the planet, if the initial velocity is less than a certain value, the object will only rise a certain distance, after which the acceleration generated by the planet's gravity will eventually cause it to fall. If the initial velocity reaches a certain value, the object will completely escape the gravitational pull of the planet and fly out of the planet. The velocity at which the object needs to escape the gravitational pull of the planet is called the escape velocity.

The minimum velocity required for an object on the surface of a celestial body to free itself from the gravitational pull of that celestial body and fly into space. For example, the Earth's escape velocity is kilometers and seconds (i.e., the second cosmic velocity). Supplement.

Addendum: Calculation method An object with mass m has velocity v, then it has kinetic energy of mv 2 2. Assuming that the gravitational potential energy at infinity is zero (this assumption is reasonable because the gravitational potential energy experienced by the object at infinity from the Earth is zero), then the potential energy of an object at a distance r from the Earth is -mar (a is the gravitational acceleration of the object at that point, and the negative sign indicates that the potential energy of the object is less than the potential energy at infinity).

And because the gravitational pull of the earth on an object can be regarded as the weight of the object, there is gmm r 2=ma i.e. a=(gm) r 2So the potential energy of the object can be written as -gmm r, where m is the mass of the earth. Let the velocity of the object on the ground be v and the radius of the earth be r, then according to the law of conservation of energy, the sum of the kinetic energy and potential energy of the object on the surface of the earth is equal to the sum of the kinetic energy and potential energy at r, i.e., mv 2 2+(-gmm r)=mv 2 2+(-gmm r).

When the object is free from the gravitational pull of the earth, r can be seen as infinity, and the gravitational potential energy is zero, then the above equation becomes mv 2 2-gmm r=mv 2 2Obviously, when v is equal to zero, the required detachment velocity v is the smallest, i.e., v=2gm r root number, and because gmm r 2=mg, so v=2gr root number, in addition, it can be seen from the above equation that the escape velocity (second cosmic velocity) is exactly 2 times the root number of the first cosmic velocity. where g is the gravitational acceleration of the earth's surface, and its value is newton kilograms.

The radius of the Earth r is about 6370 km, so the final velocity of the Earth is km. Different celestial bodies have different escape velocities, and the exit velocity formula applies equally to other celestial bodies. Follow-up:

Does this mean that an object on Earth can escape gravity when it reaches kilometers and seconds? : Not necessarily the direction is also important, that's just the minimum speed.