High School Physical Earth Satellite Position Adjustment Problem

A geostationary satellite is 10,000 meters above the equator.

From this, it is known that this satellite is a geosynchronous satellite, that is to say, it is always at 103 degrees east longitude, and now it has to be adjusted to 104 degrees east longitude, that is, to advance by 1 degree, the angular velocity of the satellite must increase, which is greater than the angular velocity of the earth's rotation. Because the lower the satellite, the greater the angular velocity ( = (gm r ) so that the satellite is adjusted to reduce its altitude and increase the angular velocity, and when the satellite reaches 104 degrees east longitude, it is adjusted back to its original altitude and becomes a geostationary satellite.

Explanation: Of course, you can also increase the altitude of the satellite first, the angular velocity will be slower, and the earth's rotation will be faster than the satellite, and the earth will rotate 359 ° more than the satellite, and the satellite will reach 104 ° east longitude. But this kind of question should be based on the nearest route.

East longitude 104 is east of 103 geostationary satellites, like the Earth, flying from west to east.

In other words, it is equivalent to speeding up for a while.

If you lower the altitude, the speed will become faster (you can push this yourself with a formula), and then you will increase the altitude and return to the original speed, and you will naturally fly eastward.

Ask Yang Liwei else, in fact, Yuli. Gagarin knew better.

Zhen Khan: There is still such a question.

According to the law of gravitation and centripetal motion, gmm (r 2) = m * w 2 * r gives w = [g (r 3)] 1 2), we can know that as the orbital altitude of the satellite increases, its distance r from the center of the earth increases, resulting in a decrease in the angular velocity w of the satellite's motion around the earth. Now to move the satellite from 103 degrees east longitude to 104 degrees, the angular velocity needs to increase (the satellite orbits the earth from west to east), so it is necessary to first reduce the altitude to increase the angular velocity, and then rise back to the original orbital altitude.

The Earth's geostationary satellites are located in orbits 10,000 kilometers above the equator, and their orbit around the Earth (from west to east) is 24 hours, so their orbit rate is equal to the rate of the Earth's rotation (from west to east).

Idea: If the satellite wants to move from 103 degrees east longitude to 104 degrees east longitude, it must first make the velocity greater than the rate of the earth's rotation, and then make the velocity equal to the earth's rotation rate after reaching the position.

Analysis: According to the law of gravitation, f=gmm(r+h) squared = m(v, squared) (r+h), "v=rooted(gm (r+h))", so the velocity of the satellite is inversely proportional to its altitude above the earth.

Result: The satellite altitude was lowered to a rate greater than the rate of the Earth's rotation, and it moved eastward to 104 degrees east longitude.

Then raise the altitude of the satellite so that its rate is equal to the rate of Earth disposal, and the fixed point is achieved.

High school relies on "knowledge points", and you will never learn well. High school is no longer like junior high school, where you can apply the formulas and theorems of the conclusion as long as you memorize them. That way seems simple and efficient, but in fact you can't learn anything at all, and you don't have the ability to analyze.

This is not playable in the case of high school physics being flexible. What you need is not someone else to sort out ready-made knowledge points for you. It's about understanding all the basics (what it means, what it's for, how to use it), and then learning to analyze the problem and use it flexibly.

The difference is that, first, the orbital radius is different, for low-Earth satellites, the radius is understood as the radius of the earth (the height of the satellite is much smaller than the radius of the earth), and the orbital radius of the geostationary satellite can be calculated to be about a multiple of the radius of the earth (the height of the satellite is about 36,000km).

Second, the period is different, the period of the near-Earth satellite is the minimum period of the satellite, about 5000s (84min); The period of the geostationary satellite is, of course, the period of the Earth's rotation, which is 86400s

The common denominator is that the resultant force of the two is the gravitational force of the earth, so to use Newton's second law, the gravitational force of the earth is equal to the centripetal force required for the satellite to move in a circular motion.

Earth's gravitational force = satellite mass and satellite acceleration.

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!

The satellite moves in II, with a large period.

At point p: the linear velocity i is small.

The angular velocity i is small, the centripetal acceleration is as great, the gravitational force is as great, and the centripetal force is as great.

Kinetic energy is small. In motion i point p and point q the gravitational force is equal to the centripetal force.

1. According to Kepler's law, the square of the period is proportional to the cube of the semi-major axis of the orbit, and it is obvious that the semi-major axis of the elliptical orbit is small and the period is small.

2, this is a lot, one by one.

Compared with orbit 1 and orbit 2 at point p, orbit 1 is obviously smaller than orbit 2, and the reason is also very simple, assuming that a satellite is flat at point p, without a formula, the conclusion is also obvious.

The kinetic energy is obviously also small, and the angular velocity is also small because the radius r of the circumambulation at this moment is equal;

The gravitational force is definitely equal, orbit 2 all provides centripetal acceleration, orbit 1 only partially provides centripetal force and part provides falling acceleration, so the centripetal acceleration is also small and the centripetal force is also small.

3. The gravitational force is not equal to the centripetal force, the gravitational force is large at point p, and the gravitational force at point q is insufficient.

r:r is the velocity ratio of the geostationary satellite and the object on the ground, and the question asks the velocity of the geostationary satellite and the velocity ratio of the first universe.

Because the formula gm r 2=v 2 r

So for synchronous satellites and the first cosmic velocity ratio v1:v2= r:r

First, the cosmic velocity is faster than the rotation of the Earth, and the equation is mv2 r=mmg r2

r is the distance from the earth, and you actually have to consider the distance from the center of the earth.

Of course, the velocity is high when passing point P on the A track. From an elliptical orbit to a circular orbit, it is necessary to go through a deceleration process. On the other hand, if the satellite returns to the earth from the moon, it first makes a circular motion in a circular orbit B, and then accelerates and increases its kinetic energy before it can change its orbit to an elliptical A orbit and finally fly away from the moon.

From the perspective of common sense, it is said in the news that only by accelerating the orbit change can we go around a larger circle.

Next, look at this, a=v 2 2 This is the formula for uniform circular motion, where a is the centripetal acceleration required for uniform circular motion. When V becomes larger and acceleration A remains unchanged, the uniform circular motion state is broken, and because the velocity is large, the inward centripetal force is insufficient, so it will float outward, and it will change from B to A. If you don't understand, continue to ask.

The elliptical orbit can find a focal point as the radius, which can be easily compared with MV 2 R, and if R is larger, V must be larger.