If the two satellites collide with each other, the velocity is equal

The title is very clear, [if the velocity of the two satellites before the collision is equal] can collide, it means that their orbits have a junction, the gravitational force of the satellite and the planet provides centripetal force, the centripetal force f=m*m r 2=m*v 2 r can get m r=v 2, m is the mass of the planet, we can see that as long as the value of r is fixed, the linear velocity v is correspondingly fixed! In other words, according to [if the velocity of the two satellites is equal before they collide], we can launch their orbital radius to be the same! The period t=2 (r3 gm)1 2=2 r v, the linear velocity is the same, the radius is the same, and the period is the same!

In fact, the orbits of these two satellites are exactly the same, and the reason why they can collide is because their orbits are not on the same plane, much like the previous ** TV station logo! If you imagine what that track looks like, it's not hard to understand!

Whether the cycle is certain or not is discussed has to be discussed! I haven't rolled it out yet.

But the judgment of the orbit is wrong, imagine that the two satellites are launched from the collision point, it is obvious that the launch direction is different, and the flattening rate of their elliptical orbits must be different.

When orbiting the earth, the gravitational force of the earth on the satellite provides the centripetal force of the satellite to make a circular motion, and it can be calculated that there is only one optimal velocity for each orbit, and if it is larger than this speed, the satellite will be "thrown out"; Less than this, the satellite will "fall". Therefore, if the two satellites have the same velocity, it is certain that they must be moving in the same orbit.

Personally, I think this velocity should be tangential velocity, because the tangential velocity v=gmm r 2, no matter how the satellite moves, the force is always only gravitational.

The motion of the satellite around the same celestial body is, agreeing that the velocity of his radius is the same, and the period which is only related to the radius.

The larger the radius, the angular velocity.

The smaller, the gravitational pull.

Deduction. Gravitational force: pre-combustion f gmm r 2

Assuming that the mass of each planet is constant, r, f, f=m·r(w 2), angular velocity w, an=r(w 2), an

It can also be understood in this way, linear velocity.

v = angular velocity w radius r (if you are high school this formula is enough, because there is no tangential thought plexus acceleration, etc.) linear velocity is constant, angular velocity w is inversely proportional to radius r.

Relationship. Centripetal acceleration an = angular velocity w radius r

So the moon is like Mars, with a smaller angular velocity than the Earth and a slower rotation than the Earth. For a good understanding, you can compare the Earth to the minute hand, Mars (that is, a certain moon) to the hour hand (strictly speaking, the hour hand here is longer than the minute hand is more consistent), do you think that after the time hand and the minute hand meet, and then n minutes later, they meet for the second time, is the minute hand more than the hour hand? It is equivalent to the hour hand not moving the minute hand to chase the hour hand.

Solar system. Planets, radius r, linear velocity v, period t, angular velocity w, centripetal acceleration an

The larger the radius of the satellite operating in orbit, the smaller the linear velocity, the smaller the angular velocity, the smaller the centripetal acceleration, and the greater the period.

But if it's a geostationary satellite.

The above conclusion is not valid. The larger the radius, the smaller the angular velocity, provided that the linear velocity is certain, and the angular velocity of the satellite must be the same as that of the Earth, and the linear velocity must be larger than that of the Earth.

GM R 2=A=[(2 factions t) 2]r, so the disadvantages are r=[(gmt 2) brother Buji (4 factions squared)] 1 3,a=4 factions squared R t 2=(16gm faction 4 T 4) 1 3So a1 a2=(t2 t1) 4 envy3

The geostationary satellite is the same height from the ground and the same angular velocity of rotation as the Earth, so there are the same: angular velocity, linear velocity, altitude above the ground and centripetal acceleration.

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Question 2: Both of them move around the earth, so the centripetal acceleration is the same, let the radius of the earth be r, the radius of motion of A is the ground, and the radius of motion of B is 6r.

From the formula a=r 2, where a is the centripetal acceleration, r is the radius of motion, and is the angular velocity, we get:

r A A 2=r B B 2, when the angle of the two motions is different by one radian (i.e., 180 degrees), the distance between the two satellites is the largest, so the elapsed time t = A - B).

In the period of converting the solved t to b: t (2 B posture comics) <>

If the world map is the only one that is high.

Suppose an elliptical orbit satellite a (orbital parameters a, c, perigee velocity v1, apogee velocity v2), then there is:

A at perigee mechanical energy is equal to a apogee mechanical energy: e1 = 1 2m v1 -gmm (a-c) = 1 2m v2 -g(mm) (a+c).

a is conserved at the far and perigee angular momentum or using Kepler's second law: v1(a-c) = v2(a+c).

Solution: e1=-gmm (2a).

Let the circular orbit satellite b (orbital radius r, velocity v), then there is:

The velocity of b is related to the radius of the orbit r: v r = gm (r ).

The mechanical energy of b: E2=1 2mv -g(mm) r

Solution: e2=-gmm (2r).

It may be useful to assume that the two satellites have equal mass (the orbit is fixed, and the mass does not affect the period). Since a and b are equal in magnitude (same kinetic energy) velocity (same kinetic energy) at the intersection of orbits, the mechanical energy of both is equal, so that there is:

gmm/(2a)=-gmm/(2r)

Get: a=r

There should be a premise for saying that "if the velocity of two satellites is equal before they collide, then the period of motion must be equal", that is, satellites in circular orbit. If the velocity of a satellite in an elliptical orbit is equal to that of a satellite in another circular orbit, the periods are not necessarily equal.

Physical questions are all relative, whether you haven't finished this question, this satellite cycle problem, is to have a certain central celestial body. Even if the period or velocity of the moons opposite different central bodies is the same, the other quantities are not necessarily the same.

Answer: (1) "If two artificial satellites of the earth collide, are the two satellites in the same orbit?" - Not necessarily equal, two satellites can be in different orbits.

On the road, as long as there is an intersection between two tracks, and they reach that intersection at the same time, they will collide.

2)"Are the magnitudes of the period and centripetal acceleration of both necessarily equal? "— Only when the radius of the two orbits is equal, do a circular motion.

The period and centripetal acceleration of the satellite are equal; The orbital radius is different, and the corresponding period and centripetal acceleration are not equal.

When two satellites collide, the satellites are generally not in the same orbit because the satellites in the same orbit have the same speed and path, and generally do not collide unless they are operating in reverse. Most of the colliding satellites are satellites with intersecting orbits but not the same.

When the orbits are the same, the period and centripetal velocity of the two must be equal.

When the orbits are not the same, the period and centripetal velocity of the two are generally not equal. However, two satellites with different orbits but the same velocity before the collision should still have the same period.

Since they collided, they must be at the same altitude, so the potential energy of the two with respect to the earth must be the same. Ignoring friction (there is some very thin air in low Earth orbit), the next time you reach this point, the two must still be the same.

If the Earth is regarded as an ideal mass point (in fact, the Earth is an irregular sphere, with slightly flattened poles and slightly bulging equators, and not the gravitational acceleration at every point is the same), the height and velocity must be the same at the moment before the collision (because the potential energy is the same, the height must be the same, and the velocity must be the same), and the same is true when it is pushed forward, so that it is the same until the moment of the last collision.

Therefore, the period, radius, and long axis are all the same.

The orbit of the satellite is fully determined by the position vector and the velocity vector. If the velocity of the two satellites is equal before the collision, the position vectors of the two satellites at the collision position are exactly the same, and the velocity vectors are opposite, so it can be determined that the orbits of the two satellites overlap, but the normal direction of the orbits is opposite. So with the same long and short axes and periods.