High One Physics, the Problem of Earth Satellites.

Yes, the maximum orbital velocity, the minimum firing velocity.

If a satellite that is moving in a circle around the Earth suddenly accelerates, causing it to travel faster than that. First, it shows that the spacecraft turned out to be right, and at this time it flew on the surface of the earth, and as soon as it exceeded, it was centrifuged, and the radius became larger. 2. If it is not next to the earth, the spacecraft will not be able to overtake it in an instant, and when it accelerates, it will slowly fly away from the earth.

This question involves a change of track, and the questioner can take a look at it.

Yes, it is the maximum speed of orbit around the Earth, the speed of circular orbit is constant, the perigee velocity of elliptical orbit is maximum, and the apogee velocity is minimal.

After surpassing, it will be free from the shackles of the earth and away from the earth.

Why not mine? Is his answer more intelligent? --The fire is big

Answer: BC for a:gm r 2=v 2 r, we get v= (gm r), where m is the mass of the earth and r is the radius of the orbit. It can be seen that the linear velocity decreases with the increase of the rail radius, so a is wrong.

For b:GM r 2=(2 t) 2 r, we get t=2 (r 3 gm). It can be seen that the period increases with the increase of the radius of the orbit, so b is correct.

For c: From b, it can be seen that when the radius of the orbit is smallest (i.e., the radius of the Earth), the period is also the smallest. At this time, the relevant data is consulted and the minimum cycle is calculated to be about 84min, so the running cycle cannot be less than 84 minutes. i.e. c is correct.

Hope it helps.

Question 1: No.

Problem 2: The satellite trajectory goes out.

The satellite will not move to a high orbit for no reason, and there must be an external force to do the work, such as the thrust of the rocket, and the external force is converted into the kinetic energy and potential energy of the satellite. Why should the kinetic energy be increased, because if you don't increase it, you can't get out of the original track. After the speed increases, the original orbit can no longer hold the satellite, and then the satellite moves to a high orbit, and in the process of moving to a high orbit, it will slow down, and when the speed decreases to the point that it can exist in that orbit, the satellite will move in a circle in the high orbit.

It is simple to understand that the increase in potential energy is greater than the decrease in kinetic energy, and for the satellite to fall into its original orbit, it must be faster than the original speed.

In the whole process, because there is an external force doing work (rocket thrust, etc.), the mechanical energy is not conserved.

It can be understood from the following aspects:

1.The high orbit, low speed and large period are suitable for the problem of the same circular orbit.

2.The elliptical orbital velocity is variable, and the velocity may be large or small compared to other orbits.

3.In Kepler's third law, r is the radius of a circular orbit or the semi-major axis of an elliptical orbit.

This kind of problem can only be solved by quantitative analysis. From gmm r 2=mv 2 r=mrw 2=mr(2 t) 2=mg v= gm r so a is wrong, t = (4 r 3 gm) so b is correct, and the radius t brought into the earth is minimum = about minutes, so c is correct.

1,g=m/gr^2

m/g=gr^2

mm/g(r+h)^2=mωb^2(r+h)m/g(r+h)=ω^2

b^2=gr^2/(r+h)

t=2π/ωb

2, the period of the synchronous satellite is the same as the earth, it is 0, and asking when it is closest again is to ask when the two satellites are operating at a different angle of 2, and the difference of 2 is that the two satellites are 360 degrees different from the original angle of difference, a circle, that is, back to the original angle again. At the same time, A has traveled 0 t degrees, B has traveled B t degrees, and there is a difference of 2, because B is in the same direction as the Earth's rotation, so B>= 0 is B t- 0 t=2

The time required is 2 ( b- 0).

Yes, you yourself are right!

1。Artificial geostationary satellites, with a period of 24 hours, are stationary relative to the ground, at a fixed altitude at the equator.

2。The center of the orbit of the satellite must pass through the center of the Earth's sphere.

a.In a circular orbit at any height above the ground on the Earth's equatorial plane, and is always stationary relative to the Earth.

Mistake. Relative to the ground, it is always stationary, that is, a geosynchronous satellite, and must be fixed at an altitude.

b.It moves in a uniform circular motion in a circular orbit coplanar to the Earth's equator, but is not necessarily stationary relative to the ground.

b Correct. c.It is possible to make a uniform circular motion around the earth in a plane determined by any latitude of the earthc error The plane ratio of the latitude must be past the center of the sphere.

d。I can't guarantee a ball center, a mistake.

Select Answer B