About the method of calculating the second velocity of the universe

Suppose that the minimum launch velocity required to launch a satellite with mass m on Earth into an orbit orbiting the Sun is V; The radius of the Earth is r;

At this time, the motion of the satellite around the sun can be considered to be not subject to the gravitational pull of the earth, and the distance from the earth is infinite;

Three cosmic velocities.

Considering that infinity is the gravitational potential energy 0 potential surface and the launch velocity is the minimum velocity, then the satellite can reach infinity.

It is obtained by the conservation of mechanical energy.

mv^2/2 - gmm/r = 0

Then mv 2 2 = gmm r (and r=r) is solved to v=spr(2gr)=

It is exactly 2 times the root number of the first cosmic velocity.

Reference from Encyclopedia - Second Cosmic Velocity.

I guess you're a high school student, and after taking college calculus, the question is pretty straightforward.

The integral of the earth's gravitational force and distance is the lower limit of the integral and the infinity is the upper limit of the integral, and this anomalous integral value is the gravitational potential energy at infinity from the earth, and the velocity obtained by substituting the kinetic energy equation is the second cosmic velocity. As for the detailed calculation process, it is too cumbersome to write, so it is recommended that you find a textbook for integrals, which must have a detailed process.

1. When calculating the second cosmic velocity, the escape refers to the ability to fly to infinity (theoretically) without the action of external forces (conservation of mechanical energy). In terms of calculation, the value of the gravitational potential energy (negative value) on the surface of the sphere is calculated by taking the infinity as the potential energy 0 reference point. Then there is the idea of "non-negative kinetic energy at infinity, non-negative initial mechanical energy, minimum initial kinetic energy, minimum initial velocity (second cosmic tempo sharpness)".

2. g*m*m r 2 = m*(v 2) r g gravitational constant is matched by the celestial mass, m is surrounded by the mass of the object, m is surrounded by the mass of the object, r is surrounded by radius, v velocity. It is concluded that v2 = g*m r, the radius of the moon is about 1738 km, which is the earth's tomb 3 11. The mass is about 735 billion tons, which is equivalent to 1 81 of the mass of the Earth.

The first cosmic velocity of the Moon is approximately in the semi-long diameter of the orbit of an artificial celestial body according to: V 2 gm (2 R-1 A) a. a, get the second cosmic velocity v2=.

1. Second Cosmic Speed When an object (spacecraft) flies at a speed of kilometers and seconds, it can get rid of the shackles of the earth's gravity, fly away from the earth and enter an orbit around the sun, and no longer orbit the earth. This minimum velocity out of the Earth's gravitational pull is the second cosmic velocity. The starting flight speed of various planetary or satellite probes is higher than the second cosmic velocity.

2. The second cosmic velocity is the escape velocity, the velocity of an object when its kinetic energy is equal to the gravitational potential energy of the object. Escape velocity is generally described as the minimum velocity required to escape the gravitational constraints of a gravitational field and fly away from that gravitational field.

Derivation of the second cosmic velocity:

An object with mass m has velocity v, then it has kinetic energy mv 2 2. This assumption is plausible because the gravitational potential energy at infinity is zero (it should be that 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 a negative sign indicates that the potential energy of the object is less than the potential energy of the infinite point). And because the gravitational pull of the earth on an object can be regarded as the weight of an object, there are:

gmm r2 = ma, i.e. a = (gm) r2.

Therefore, 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, it can be seen that 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 the earth's surface, i.e., MV2 2+(-GMM R)=MV2 2+(-GMM R).

When an object gets rid of the gravitational pull of the earth, r can be seen as infinity, and the gravitational potential energy is zero, then the above equation becomes: mv2 2-gmm r is equal to mv2 2.

Obviously, when v is equal to zero, the required detachment velocity v is the smallest, i.e., v=2gm r open root number, and because gmm r2=mg, v is equal to 2gr open root number, in addition, it can be seen from the above equation that the separation velocity (second cosmic velocity) is exactly 2 times the root number of the first cosmic velocity.

Second, the characteristics of the cosmic velocity

The escape velocity, depending 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 large. 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, and the closer the distance, the greater the escape velocity.

If the mass and surface gravity of a celestial body are so great that the escape velocity reaches or exceeds the speed of light, the object is a black hole. Black holes can escape at speeds of up to 300,000 kilometers per second.

According to the kinetic energy formula, the kinetic energy of an object with mass m and velocity v is e1 = according to the gravitational potential energy formula, when the object is about equal to the planetary radius r and the gravitational acceleration is g, its gravitational potential energy e2 is: e2 = mgr and mg=gmm r 2 can be obtained: e2=gmm r When e1-e2=0, the aircraft is a flying vehicle.