How long does it take for a reconnaissance satellite to orbit 150 km at 7 9 km per second, once arou

The calculation is as follows:

Average radius of the Earth = kilometers.

Reconnaissance satellite flight radius = km.

time=2 r v=2 hours).

Extended information: flight time = flight distance flight speed.

According to the calculations, the flight distance = kilometers, the known speed = kilometers in seconds, and the calculation time is about hours.

The Earth is not a regular object. First of all, it is not a positive sphere, but an ellipsoid, to be precise, an oblate sphere with slightly flattened poles and slightly bulging equators;

Secondly, the north pole and north pole of the earth are also asymmetrical, in terms of sea level, the north pole is slightly convex, and the south pole is slightly concave; Third, the Earth's external topography is undulating (which has an impact on measuring the Earth's radius). The average is about 3959 miles (km).

Polar radius of the Earth:

The distance from the center of the earth to the North or South Pole is about 3,950 miles (kilometers) (the difference between the poles is negligible).

Earth's equatorial radius:

It is the distance from the center of the earth to the equator, about 3,963 miles (kilometers).

The average radius of the Earth:

Approximately 3959 miles (km).

This number is the average of the distance from the center of the earth to all points on the earth's surface.

Hello, I am glad to send you a code to answer the question of how many minutes is the period of reconnaissance satellite with an orbital altitude of 600 kilometers around the earth: The orbit of an artificial satellite depends on the mission requirements of the satellite, which is divided into low orbit, medium and high orbit, geosynchronous orbit, geostationary orbit, sun-synchronous orbit, large elliptical orbit and polar orbit. At present, the shortest time to orbit the earth is about 84 minutes, and the slowest is a geostationary satellite.

Extended information: The orbital altitude of near-Earth satellites is generally 300 to 1000 km, and the orbital period of the satellite is generally between 90 minutes and more than 100 minutes in this altitude range.

Analysis] analysis can find that the satellite passes over the equator twice in a cycle, one of which is located in the sunshine region and the other is located in the night zone. Therefore, the satellite passes through the sunshine zone once once it makes one revolution. …1）

Compared with geocentric observation, satellite photography over the equator is equivalent to "fixed-point" photography, and each point on the equator through the Earth's self-transfer may be photographed by a satellite once in 24 hours.

If we assume that the orbital period of a satellite is t, its relation to the Earth's rotation period t is "t=nt". This means that the Earth will rotate N times in one rotation (t time). Therefore, from the analysis of (1), it is known that the satellite passes over the equator in the sunshine region n times in t and takes n photos.

The problem requires that in one day (t), the total arc length of these n shots should be "equatorial circumference l", and the arc length of each imitation shot should be l=l n.

It is suggested that the landlord should use the known conditions in the question to calculate according to this idea, and it should not be difficult to get the result.

If you still have any questions, you can send me a message.

Answer: Arc length l=(4 2)*[r+h) (3 2)] t*g (1 2)]

Warmly remind the downstairs that the number of times it passes over the equator every day is indeed 2t t0, but the number of times it passes over the equatorial "sunshine zone" every day is half, that is, t t0(The question is about taking pictures of the sunshine area).

First, calculate the period t2 of the satellite's orbit around the Earth. I know h, g, r, I'm sure you can figure it out.

Then the motion of the satellite is not affected by the rotation of the Earth. So we can simplify this problem to drawing lines from top to bottom on a world map while the world map moves. Ask how wide the equator should be in the sunshine range in one day, and how wide the line should be.

Then let's take the time when the satellite first crosses the equator to 0, and take this point as the coordinate 0 point. Then at t=(t2 2)*n (n=1,2,3......), the satellite passes through the equator. Then the coordinates of each satellite passing through the equator are x=(-2*pi t)*t, because the earth is from west to east, that is, from left to right on the coordinate axis.

2*pi t is the speed of the earth, which is the speed of the world map.

So is it daytime at this time?

Let's assume that the sunshine region is [-b,-b+pi] at t=0. The length of the world map is considered to be 2*piThen the sunshine region at the moment of t=t is [-b+x,-b+x+pi].

In fact, you can understand it this way, if the satellite is a pen to draw lines, the sun is a flashlight. The pen is always drawn vertically, and the flashlight is not moving. Only the world map is pulled.

Therefore, as long as you cross the equator for the first time in sunshine conditions, you must also be in sunshine conditions when you cross the equator in the future.

Let's assume a 2W radian per shot. Then the area of each shot is [x-w, x+w]. If you want to be as short as possible, then you want to have no overlap in each shot, right?

Then you need to take x1-w=x2+w, bring in the above formula, and bring n. then 2w=(pi*t2 t)*(n1-n2). N1-N2 is 1 at the minimum.

It seems that the answer is on the horizon. However, you may have doubts that the shooting area seems to be smaller. You don't need to photograph the area between two adjacent equator crossings every time.

Instead, wait for the next time you pass by to make up for it. But think of our model and wait for the next time you pass by this place. The Earth has been rotating for a week, that is, 1 day has passed.

In summary, the minimum arc length is pi*t2 t

Multiply 10 by the cubic meter of the second * 2 times 10 by the quadratic second = 5 to the power of each second = multiply 10 by the 6th power.

If the radius of the earth is r, the mass is m, the mass of the satellite is m, and the motion of the satellite is uniform circular motion, the velocity is v.

The gravitational force experienced by the satellite provides the centripetal force around its uniform circumference, then:

gmm/(r+h)^2=m v^2/(r+h)v=√gm/(r+h)

Whereas, the orbital circumference of the satellite l=2 (r+h).

The time for a satellite to make one revolution t=l v

The time of a day is t, then the number of revolutions that the satellite can circle in a day n=t t t and the satellite passes through the equator twice around a circle, so the number of times the satellite passes through the equator in a day is 2n, so it is natural to shoot the equator, and the equator is divided into 2n parts on average, and the arc length l=l 2n. After that, just bring in the above equation.

The calculation here assumes that the Earth is a sphere, and that the satellites are moving in a uniform circular motion. However, in reality, the orbit of the satellite is elliptical, and the Earth is not a sphere. This can only be calculated by giving the orbit of the satellite.

v1= km/s。

First, cosmic velocity is divided into the minimum launch velocity of the spacecraft and the maximum operating velocity of the spacecraft. In some cases, it is said that when a spacecraft is moving at the first cosmic velocity, it means that the spacecraft is moving along the Earth's surface. According to the theory of mechanics, v1= km s can be calculated.

When a spacecraft is moving at the first cosmic velocity, it means that the spacecraft is moving along the Earth's surface. According to the theory of mechanics, it can be calculated that v1 = kilometer second. In fact, there is a dense atmosphere on the surface of the earth, and it is impossible for a spacecraft to move in a circle close to the surface of the earth, and it must fly at an altitude of 150 kilometers to make a circular motion around the earth.

Classmate, this one also asks. The first cosmic velocity is v1 = root number (gr) =. Second, the cosmic velocity v2 = v1 = s. of 2 times the root number

What it means: You want to launch a satellite, but the speed must be greater than v1 at the time of launchBut if you want him to orbit around a circle, it must be less than v1

And if you want to orbit around an elliptical, the velocity must be greater than v1But it has to be smaller than v2. Push down method, and the resultant force provides centripetal force.

You try!

Reconnaissance satellites are not too high, so satellites are faster than Earth.

n (t t detection) 2, where (t t detection) is the ratio of the earth's rotation period to the reconnaissance satellite's orbit period, which indicates the number of times the satellite orbits the earth in one day. And one revolution, that is, one revolution around the earth, will pass the equator twice, so it is also multiplied by 2 to pass the equator in one day.

The short-haul principle of the photovoltaic industry.

Satellites are divided into low-orbit satellites, medium-orbit satellites, and high-orbit satellites. Among them, the orbital altitude of high-orbit satellites is about 36,000 kilometers from the earth's surface, and it is relatively stationary with the earth, so the average speed around the earth is theoretically 0. Because of the gravitational force, the lower the altitude, the greater the gravitational force, the greater the orbital flight speed, so the speed of low-orbit satellites and medium-orbit satellites around the earth is not 0.

Due to the fact that the spacecraft is flying around the Earth at a high speed, reaching a speed of kilometers per second. It takes about 90 minutes to go around the globe.

Earth radius: 6370 km; Altitude of spacecraft above the ground: 343 km; One minute: 60 seconds.

It's a very short time

Instantaneous velocity is the average velocity over a "very short period of time". Have you ever seen a satellite that was able to circle around it in a "very short time".

What is a very short time, the so-called "instant", 1 second? A ten-thousandth of a second? Or a billionth of a second, which can be called a very short time.

That's right. The average velocity is the displacement time, and the displacement is 0 after a week of flight, so the average velocity is 0

The spacecraft cannot be stopped in flight, so its instantaneous velocity at every moment is not zero

Note that in the first sentence, the average speed is not the average rate, and the average rate is the distance time, which is not 0, so be aware of the difference between the two.

Set time is x minutes.

Then 270:3=x:45

Solution x=4050

It takes 4050 minutes.

m) So the answer is:

m 474 90 = 42,669 km.