Compare the acceleration of geostationary satellites, near Earth satellites, and objects on the grou

The angular velocity of geostationary satellites and ground objects is the same, and the linear velocity of acceleration of geostationary satellites is greater than that of ground objects (v=wr a=v 2 r), and the acceleration linear velocity angular velocity of near-earth satellites is the largest (gravitational and centripetal force formulas).

To compare the acceleration, linear velocity, and angular velocity of geostationary satellites, near-Earth satellites, and objects on the ground (equator), we can use the following formula:

Centripetal acceleration a = r w 2, linear velocity v = w r, angular velocity w = v r

where r denotes the distance of the object to the center of the earth, w denotes angular velocity, v denotes linear velocity, and a denotes centripetal acceleration.

Compare Acceleration:

Centripetal acceleration of a geostationary satellite: $6400000 times

Centripetal acceleration of a near-Earth satellite: $320000 times

Centripetal acceleration of an object on the ground: $3200000 times

Since > the centripetal acceleration of geostationary satellites is the largest, followed by near-Earth satellites, and objects on the ground are the smallest.

Compare Linear Speeds:

Linear velocity of geostationary satellites: $6400000 times

Linear velocity of near-Earth satellites: $320000 times

Linear velocity of objects on the ground: $3200000 times

Since > the geostationary satellite has the highest linear velocity, followed by the near-Earth satellite, and the object on the ground is the smallest.

Compare angular velocities:

Angular velocity of a geostationary satellite:

Angular velocity of a near-Earth satellite:

Angular velocity of an object on the ground:

Since = > the angular velocity of the geostationary satellite and the object on the ground is equal and maximum, the angular velocity of the near-Earth satellite is the smallest.

The high speed and low orbit have a long period, and v = gm r is obtained from gmm r = mv r, so the larger r is, the smaller v is.

From gmm r =m r =gm r So, the larger the r, the smaller it is.

From gmm r = 4m t t t to get t = 4 r gm so the larger r is, the greater t is.

Remember the mantra: the higher the slower, the closer the faster (the object is not on Earth).

Period: Near-Earth "Synchronization = Equator, Speed: Equator" Synchronization "Near-Earth, Acceleration: Synchronization "Near-Earth = Equator.

An object on the equator of the type finger beam: the period is 24 hours, and the acceleration due to gravity is about g.

Near-Earth satellite: The acceleration is about g.

Geostationary satellites: The period is 24 hours.

The centripetal acceleration of an object at the equator is not equal to the acceleration due to gravity and should be distinguished from the other two.

Near-Earth satellites are lower than geostationary satellites and therefore have shorter periods than geostationary satellites.

Geostationary satellites are far from the ground, the gravitational acceleration is less than that of near-Earth satellites, and the kinetic energy and velocity are lower than those of near-Earth satellites.

The angular velocity of a geostationary satellite is equal to that of an equatorial object, but the orbital radius is much larger, so the velocity is greater than that of an equatorial object.

Geostationary satellites are synchronized with the Earth's rotation period, i.e. the period is 24 hours.

Objects on the equator are not satellites, they are placed on the ground, for example, a stone is placed on the ground, and a person stands on the ground, and the period is synchronized with the Earth, which is 24 hours. Note that an object at the equator is a little different from a satellite, the satellite is the gravitational force that completely provides the centripetal force, while the object on the equator has a small centripetal acceleration and the difference between the gravitational and support forces provides the centripetal force.

Near-Earth satellites are satellites orbiting the air near the ground and operating at the first cosmic velocity.

Comparison of the three:

In terms of period, objects on the equator are the same as geostationary satellites, both are larger than near-Earth satellites, and the period calculation of near-Earth satellites.

gmm r 2 = m(2 t) 2 r, t can be calculated or the equatorial circumference divided by the first cosmic velocity, which is roughly calculated to be between 5000 and 5100 seconds, much less than 24 hours.

In terms of velocity, you can use gmm r 2 = mv 2 r, and we can get that v 2r = gm is a constant, that is, the larger the orbital radius of the satellite, the smaller the linear velocity. The linear velocity of the object on the surface of the equator is brought by the rotation of the earth, and it is not in the same system as the satellite, which is about 463 meters and seconds, so the linear velocity of the satellite is not necessarily related to the autobiographical speed of the earth. But it is very easy to compare geostationary satellites with the rotation of the earth, they have the same angular velocity, and the radius of geostationary satellites is larger than that of the earth, so the linear velocity is large.

The acceleration, which is the largest for near-Earth satellites, is approximately equal to g, and gradually decreases as the radius increases, calculated using gm r 2.

Objects at the equator, acceleration and satellites are also not part of the same system, but are a constant of the angular velocity of the Earth's rotation * Earth's radius.

4 2 r 86400 2.

Period: Geostationary satellites have the same as objects at the equator.

Periodic velocity: The velocity of a geostationary satellite is greater than that of an object at the equator.

Acceleration: Near-Earth satellites are larger than geostationary satellites and larger than objects at the equator.

First of all, the central bodies of all three are the same, and Kepler's third law can be used, which is r 3 t 2 = k (k is a constant).

The radius of a near-Earth satellite is approximately equal to the radius of the Earth.

The velocity is obtained by gmm r 2 = mv 2 r, and the closer you are to the central object, the faster the velocity.

Since the object at the equator has the support force of the ground, i.e., the centripetal acceleration is particularly small.

The other two are found with a=gm r 2.

Because it's 100 points, you need a method instead of an answer, right? If you need an answer, if you don't understand, you are welcome to ask! ^_

The geostationary satellite, obviously, is synchronous with the earth, and the period of the earth is the same, it is 24 hours, and the problem on the equator is also synchronous with the earth, and it is also 24 hours, so it is equal to a synchronous satellite. The cycle of a near-Earth satellite is very fast, with the fastest period being about 85 minutes. So synchronization = equator is greater than near-Earth satellites.

Cycle; Equator = Synchronous "Near-Earth, Velocity: Equator" Synchronous "Near-Earth, Acceleration: Synchronous "Near-Earth = Equator.

Periodically synchronous satellites = objects at the equator" near-Earth satellites.

Velocity Near-Earth Satellites (GNOS) Objects on the equator.

Acceleration, both are 0, so they are all equal.

Geostationary satellites, near-Earth satellites, and objects at the equator are all celestial bodies that are connected to the Earth, but they differ in periods, velocities, and accelerations.

Geostationary satellites: The orbital period is the same as the Earth's rotation period, about 23 hours and 56 minutes (exactly one day), so they always stay motionless over a certain point of the Earth, with a velocity of kilometers per second and an acceleration of about meters and seconds.

Near-Earth satellite: the orbit altitude is low, the orbit period is short, the speed is fast, the orbit is generally completed in about 90 minutes, the speed is simple kilometers per second, and the acceleration is about meters and seconds.

Objects at the equator: The Earth's rotation cycle is 24 hours, so objects at the equator move on their own on the Earth at a speed of about meters per second (eastward longitude) with zero acceleration.

In general, the speed and acceleration of geostationary and near-Earth satellites are not as high as those of objects at the equator, but they also have shorter periods. For example, geostationary satellites are widely used in communications, meteorology, navigation and other fields, near-Earth satellites are mainly used for observation and detection, detection and scientific research, while objects on the equator are mainly less affected by gravity and are suitable for measuring geophysical properties and astronomical observations.

The period of the geostationary satellite is 1 day, that is, when the 24-hour line is broken. Since it needs to be synchronized with the Earth's rotation at a constant speed, its speed is the Earth's rotation speed, which is about 1670 kilometers per hour, and the acceleration is zero.

Near-Earth satellites orbit the Earth quickly, and their periods are shorter than those of geostationary satellites, typically a few hours to a few days. Due to its high rotation, it is faster than geostationary satellites, but it is also faster than upshifting.

The remnant objects at the equator do not follow the Earth's motion around the sun, so their cycle is one day. It is a bit faster than a geostationary satellite, but it is still slower than a near-Earth satellite, and the acceleration is also less.

In general, the cycles, velocities, and accelerations of these three are different, are affected by different factors, and have their own characteristics and application scenarios.

The velocity of a geostationary satellite is greater than that of an object at the equator because the orbit of the geostationary satellite is a circle on the surface of the Earth's equator, and the time required for the Earth to rotate around is 24 hours. Therefore, in order to keep the satellite and the Earth in sync, the satellite must operate at a speed equal to the speed of the Earth's rotation, i.e. about 1,670 kilometers per hour.

However, objects at the equator have an eastward linear velocity due to the rotation of the Earth, and its magnitude is related to its latitude, and the closer to the equator, the greater the linear velocity, with a maximum of about 1670 kilometers per hour. Therefore, geostationary satellites must be faster than objects at the equator in order to remain synchronized with the Earth.

In addition, the speed of geostationary satellites is also affected by factors such as the Earth's gravity and air resistance. Earth's gravitational pull slows down the satellite, so constant propulsion is required to maintain its speed. Satellites flying in the atmosphere are also affected by air resistance and gradually slow down.

Therefore, geostationary satellites need to constantly make attitude adjustments to ensure that their speed is equal to the speed of the Earth's rotation.

In conclusion, geostationary satellites are faster than objects at the equator in order to maintain synchronization with the Earth and overcome the effects of factors such as the Earth's gravitational pull and the drag of the air.

Synchronization: The satellite is always located at the same place and directly above the width of the equator on the ground, and the earth rotates once and the satellite also rotates once with equal periods.

Angular velocity mould = 2 cycles.

So the angular velocity is equal.

The gravitational pull of the Earth on an equatorial object and a near-Earth satellite is the same, but the difference is that the gravitational pull of the Earth on an object at the equator is not all used for centripetal motion like a near-Earth satellite. The gravitational force experienced by an object can be decomposed into two forces: gravity and centripetal force; Since the speed of the earth's rotation is relatively small, only a small part of the earth's gravitational force on the object is used to provide centripetal force for the object to move in a circle around the center of the earth, so the gravitational force is large, so in general we can consider gravity to be equal to the gravitational force of the earth on the object, but strictly speaking, gravity is the smallest at the equator and the maximum at both levels (this is because the lower the latitude, the greater the linear velocity of the object on the ground, the greater the centripetal force required, and thus the smaller the gravitational force).

In addition, the gravitational force of the near-Earth satellite is all used for centripetal motion, so the gravitational force of the gravitational force is 0, that is, it is in a state of complete weightlessness.

Songjz1971 below your answer has given a very comprehensive answer, so I won't be verbose.

Near-Earth satellite A2 = F M = GM R 2

Geostationary satellite A3 = F M = GM R 2

r>r a2>a3

The period of the equatorial object is the same as that of the geostationary satellitea1=4 2r t 2, and the geostationary satellite a3=4 2r t 2,r>r,a3>a1

a2>a3>a1

a1=a2 a1 has gravitational acceleration, g a2 is centripetal acceleration, and gravitational acceleration all acts as centripetal acceleration.

A3 is the smallest because of the large radius!

gm/r^3=w^2

According to the angular velocity of the geostationary satellite is less than that of the near-Earth satellite, the geostationary satellite can also be launched with a large radius gm r 2=a

Because the geostationary satellite has the largest radius, the acceleration is minimal.

The key is to find out the same amount between them.

For both equatorial objects and geostationary satellites, the period is equal to the rotation of the Earth, so their t is the same.

Substituting the expression of a, a=4 2r t 2, we get a1 "a2.".

Near-Earth satellites and geostationary satellites are based on a=f m=gm r 2 to get a2>a3

: The basic principle is that gravitational force is equal to centripetal force.

Find the linear velocity v= from gmm r = mv r

Find the angular velocity =by gmm r =mr

Find the period t= by gmm r =mr(2 t).

Sometimes it is necessary to use the gravitational force near the ground equal to the gravitational force, that is, mg=gmm r, that is, the so-called **substitution relation gm=gr (where g is the gravitational acceleration near the ground, and r is the radius of the earth. On the ground is the person who is except subjected.

approximately equal to gravity), also supported by the ground, its net force f is very small;

A satellite close to the ground is only subject to the gravitational pull of the earth, and its net force is the gravitational force of the earth f>>f;

According to the formula f=mv2 r, the speed of the easy-to-know person is much less than.

Follow-up question: Is there a difference between an airplane that is not supported by the ground and a satellite that is close to the ground? What does f>>f mean?

Answer: The airplane is still supported by the upward force of the air on it; The satellite close to the ground has at least flown out of the atmosphere, and it is only subject to the gravitational pull of the earth (when the satellite is 100km from the ground, 100km is far less than the radius of the earth 6400km, which can be regarded as close to the ground, do not think that the near-earth satellite is 10m above the ground).

f is the resultant force of gravity f and the supporting force, and the magnitude of these two forces is about the same (our teachers often say that gravity is about equal to gravity, and gravity and support force are balanced) so f is very small, f is much smaller than the earth's gravitational force f. If the latter is used, the linear velocity of the geostationary satellite is smaller than that of a person standing on the earth. How do you use these formulas?

Only the net force of the satellite is equal to the gravitational force, there is gmm r2 = mv2 r, and there is v = root number (gm r). (The net force of man is not equal to gravity). Oh, is there a relationship between the speed of the Earth's rotation and the speed of the first universe?

The rotation velocity of an object at the Earth's equator v = rotation angular velocity xr.

You see, they're all about R.

If a person is standing on the ground, which is stationary relative to the ground, the speed of the person is the same as the rotation speed of the person's foothold (or ground).