An artificial earth satellite moves in a uniform circular motion around the earth, and its orbital r

Solution: Let the mass of the earth be m, the mass of the satellite be m, and the centripetal force of the satellite in a circular motion near the ground is f, f=mg=gmm r....

When the satellite moves in a circular motion with an orbital radius of 2 times the radius of the Earth r, the velocity is V, the centripetal force is F1, F1 = GMM (2R) = MV 2R ....②

From the formula v= (gr 2) f1=mgr 4r =mg 4 centripetal acceleration a=f1 m=g 4

Period t = 2 2r v = 4 r ( (gr 2)) = 4 2r g) = (32r g).

The acceleration due to gravity is inversely proportional to the square of the radius: a g = (r (2r)) 2 centrifugal acceleration is equal to the acceleration due to gravity: w 2*2r = a period: t = 2 w

Solution: a=g 4, t=4 2r g).

Let the Earth's mass be m, the satellite mass be m, the centripetal acceleration be a, and the period t The Earth's gravitational pull provides the centripetal force for the satellite's circular motion.

gmm (2r) 2=ma=m(2 t) 2(2r) near the ground: mg=gmm r2 =>

gm/r^2=g

Solve it yourself.

Select C, the satellite begins to do a uniform circular motion around the earth, it is not only constrained by the law of circular motion, but also satisfies the law of gravitation, that is: gmm r2=m(4 2 t2), get t=(4 r 1 3) gm, that is, under this motion model, the change of the satellite period is not directly related to the velocity, t is determined by the radius of motion. Your mistake is to focus only on its kinematic constraints and ignore its dynamics, and you have to be careful and think more about these kinds of problems that are easy to solve.

For a, if only the linear velocity is reduced to half of the original, then gmm r squared = mv squared r, gm = rv squared, then the radius will become a quarter of the original, so a is wrong; In the same way for b, if the radius becomes twice as long, then the linear velocity will also be twice as long as the original root number, so the linear velocity cannot be constant. Hence ab wrong. Have a nice day!

Analysis: The artificial earth satellite moves in a uniform circular motion around the earth, and the centripetal force is provided by the gravitational force of the earth, Newton's second law deduces the relationship between the period t and the radius, and selects the possible methods If the radius r does not change, so that the linear velocity of the satellite decreases, the satellite will do a near-centric motion, and the period decreases If v does not change, the satellite can only move in the original orbit, and the period does not change Answer: Solution:

a. If the radius r is unchanged, the linear velocity of the satellite decreases, and the satellite will do a pericentric motion, and the period decreases, so a is wrong

b. If v is unchanged, the satellite can only move in the original orbit, the radius is unchanged, and the period is also unchanged

c. Let the mass of the earth be m and the mass of the satellite be m obtained from Newton's second law: gmmr2 m(2 t)2r, and t 2 r3gm can be obtained according to mathematical knowledge, when the orbital radius r becomes 34r, the period of the satellite becomes 2t

d. According to the previous analysis, the speed and orbit radius are doubled, which cannot be realized, so D is wrong

Therefore, choose C Comment: This question examines the problem of orbit change of satellites, when the speed of the satellite increases, centrifugal motion is done, and the radius increases; When the velocity of the satellite decreases, the radius decreases due to the proximity motion

Because gmm (r 2) = mv 2 r, so v = gm r, so the speed of the artificial earth satellite doing a uniform circular motion around the earth is determined by r, when r is constant, it is certain.

And because gmm (r 2) = mr4 2 (t 2), the period of the satellite becomes 2 t

The radius of the track becomes (4 times the 3rd root number) times r

The radius of the track becomes 4r under the cubic root number

On the surface of the earth, gravity and gravitational force are equal, so GMR, liquid imitation branches, mg, can obtain GM=G

rThe gravitational force that provides a circular motion towards the large wheel of gravity has:

gmmr+m(r+

t can obtain the sensitivity period of the satellite t=

r+rrs＝2×1s

Answer: The period of the satellite is 2 104s

Let the mass of the Earth be m, and the mass of the artificial earth satellite be m, and the law of gravitation has gmm(3r) mA

Near the ground: mg gmm

The centripetal acceleration of the artificial earth satellite is a 1g

It is determined by Equation A (2 t

The period of the artificial earth satellite is t 6 3rgAnswer: The centripetal acceleration of this artificial earth satellite is: a 1g The period is t 6 3rg

First, cosmic velocity is the velocity of an object as it moves around the earth's surface.

From the gravitational force to determine the ascending law, the old force of the gravitational force = gmm r = centripetal force = mv r is solved, in the case of g and m are constant, the larger r is, the smaller v is.

is the velocity when r = radius of the earth.

Then, when r = 2 times the Earth's semi-violent diameter, the velocity should = thus choose b

Let the Earth's mass be m, the satellite mass be m, the centripetal acceleration be a, and the period t The Earth's gravitational pull provides the centripetal force for the satellite's circular motion.

gmm (3r) 2=ma=m(2 t) 2(3r) near the ground: mg=gmm r 2 => gm r 2=ga=gm 9r 2=g 9

t=6√3π√r/√g

The height of the satellite from the ground is equal to the radius of the Earth, and the radius of motion is 2 times the half-longitude of the Earth, set to 2r

gmm/(2r)²=ma

Replace gm=gr (gmm r = mg) from the object m on the earth's surface

a=g/4

gmm r 2 = mv 2 r mv 2 = gmm r kinetic energy is reduced to the original 1 4The orbital radius r=4r d is wrong.

gmm r 2 = ma a1 a2 = 16 1 a wrong.

gmm r 2=m 2r 1 yard high eggplant 2=8 1 b wrong.

t=2 c mindfulness.

Why is the kinetic energy delayed as the original 1 4, and the linear velocity is the original 1 2ek=