High School Physics Gravitation is urgent

Kepler's third law a 3 t 2 = k

That is, the cube of the semi-major axis of the elliptical orbit of a satellite orbiting the same celestial body is constant compared to the square of the previous period, and the length of the semi-major axis can be regarded as the radius of its circular orbit.

t = hours.

The centripetal force train equation can be obtained by using gravitational force.

r+h) 3 t 2 = constant.

That is, (r+h1) 3 t1 2=(r+h2) 3 t2 2 is obtained by substituting the data.

Solvable. t = hours. Pick B

Pick b Kepler's third law a 3 t 2 = k

The centripetal force train equation can be obtained by using gravitational force.

r+h) 3 t 2 = constant.

That is, (r+h1) 3 t1 2=(r+h2) 3 t2 2 is obtained by substituting the data.

Solvable. t = hours.

Option b is obtained by gmm r = m r : 1 2= (r2 r1) 3 2 ,a is false.

From gmm r =ma : a1 a2=r2 2 r1 2 , b pair.

After the orbit change, the orbit radius of Tiangong-1 is larger, the speed is reduced, and the kinetic energy is reduced.

In the normal operation of the "Tiangong-1", the spring dynamometer can still be used, d is wrong.

The left exchange of the two equal signs of C and AB is correct, and the spring dynamometer in D can be used.

Wrong. Remember, the centripetal force, is not a force, it is a force that acts as a centripetal force.

In fact, the force of the hot air balloon near the ground is not the same as that of the satellite above the equator, but the gravitational force they are subjected to is the same, the hot air balloon has more buoyancy, and the hot air balloon cannot be regarded as moving in a circle around the earth, so for the hot air balloon, there is no centripetal force.

As for satellites, the essence of gravitation and gravity is actually the same, but when it is near the ground, a g = is quoted, while in celestial bodies, the gravitational formula is directly used. In fact, if you go to calculate and extrapolate, you will know that g = is a convenient Changshu that is calculated according to the gravitational formula combined with the earth's near the ground. And then, still, to be precise, the satellites at high altitudes, where gravity acts as a centripetal force

This is correct, the hot air balloon rotates with the earth, which can be regarded as relatively stationary with the earth. The centripetal force is generated with the rotation of the earth, of course, this force is very small, but it cannot be ignored, if the earth does not rotate, there is no centripetal force. Satellites are only subject to gravitational attraction and don't explain it!

The book says it very clearly.

Hot air balloons have more air buoyancy in the atmosphere than satellites, and if they move in a circle around the earth, the centripetal force is provided by the combined force of gravitational force, air buoyancy, and drag, but the speed is too small to be ignored. The decomposition of gravitational force into gravity and centripetal force is the situation when the object on the ground is not on the equator, and this centripetal force is caused by the rotation of the earth to drive the object to do a circular motion.

The gravitational force of a satellite at high altitude is equal to gravity equals centripetal force, and the gravitational force of a hot air balloon near the ground is equal to gravity equals centripetal force,

1: Let the velocity of an artificial satellite be v1 and the radius of motion be r1, and the velocity will be v2 and the radius of motion will be r2 after the speed is doubled. Then gmm r1 2=mv1 r1, gmm r2 2=mv2 2 r2 divide the two formulas to obtain r2 r1=v1 2 v2 2=1 4.

Let the original period be T1 and then T2, then GMM R1 2=MR1(2 T1) 2, and GMM R2 2=MR2(2 T2) 2. Divide and combine r2 r1=v1 2 v2 2=1 4 to get the period reduced to the original 1 8 2: because gmm r 2=mv 2 r gives v= gm r when r is the radius of the earth v = , the larger the radius the smaller the velocity, while the artificial satellite r is greater than the radius of the earth, so the velocity is less than.

First of all, the answer to the first question should be 1 2, which is obtained by the formula gmm r = mv r = mr(2 t), t = (4 r v ).

From this equation, we know that t is inversely proportional to v, i.e., t(2v) t(v) = v (2v) )1 2

As for the second question, it is the minimum launch velocity and the maximum orbital velocity, if it is in between, it is an elliptical motion.

The velocity out of the gravitational pull of the Sun is the third cosmic velocity of 16 7 km s, which is out of the gravitational pull of the Earth.

Reasons for choosing B and C, A, because the spacecraft is ignited, the internal energy of the fuel is converted into the mechanical energy of the spacecraft, so the mechanical energy of the spacecraft is not conserved.

b. This is the principle of the operation of the spacecraft.

C, W = root number GM R cubic, the larger the radius, the smaller W, and the radius of the geostationary satellite is greater than 343 km.

d, ma=gmm r2 r does not change, so a does not change.

B ca, ignition accelerates, so there is energy conversion, mechanical energy increases.

b. Gravity is equal to gravity, circular motion, centrifugal force, so weightless.

c. The angular velocity is inversely proportional to the period, and the period of a synchronous satellite is 24 hours, so the angular velocity of the spacecraft is large.

d. The centripetal acceleration is related to the distance of the spacecraft to the center of the earth, that is, gmm r 2=ma, the distance to the center of the earth is the same, so the acceleration is the same.

Answer: b, c.

In item a, in addition to gravity, the spacecraft accelerates with external force to do work, so the mechanical energy is not conserved.

In terms c, the period of the synchronous satellite is larger than that of the spacecraft, so it can be seen from w=2 t that c is correct.

D, there is a gravitational force to know f=gmm r 2=ma, r is constant, so the acceleration is equal.