1 Ask a high school physics question about gravitation.

I'm a high school physics teacher, and I'm working on this chapter, but it's getting late, so I'll answer you tomorrow.

1. You can simply think about it, the process from B to A needs to be accelerated (because in order to keep the rocket away from the earth, it is necessary to overcome the earth's gravity to attack), and the result of the acceleration is naturally the conversion of chemical energy into mechanical energy (the mechanical energy at this time is the sum of gravitational potential energy and kinetic energy). Therefore, the mechanical energy increases, and the gravitational potential energy of Chang'e-Chang'e at the tangent point of the two orbits (gravitational potential energy and gravitational potential energy are also energy-related only to position) is the same, so the velocity of ellipse A is greater than that of circle B.

2. Mathematically calculated, the radius of curvature of ellipse a at the tangent point of the two orbits is r2=b*b a (b is the length of the semi-minor axis, a is the length of the semi-major axis), there is a gravitational formula, gmm (r*r)=mv*v r, and the radius of curvature r2 of the circle radius r1 and ellipse a at the tangent point of the two orbits are substituted respectively, and the velocity of the ellipse a is greater than the speed of the circle b.

I'm sorry I read the title wrong, but no matter what the topic is, the analysis is the same, from an elliptical orbit to a near-lunar orbit, the rocket will perform a reverse Spitfire acceleration (that is, a deceleration), so the principle is the same.

First of all, energy is conserved.

At apogee, the potential energy is maximum, the corresponding kinetic energy is small, the launch velocity is small, and at the perigee, the potential energy is converted into kinetic energy and the velocity is maximum.

Figure it out for yourself.

At the tangent point of the two tracks, the velocity of the ellipse a is greater.

The object on the ellipse b moves in a uniform circular motion, and the gravitational force provides the centripetal force gmm r 2 = mv 2 r - v is the velocity of the ellipse b after the spacecraft does centrifugal motion, indicating that the gravitational force is not enough to provide the centripetal force, and the gravitational force f=gmm r 2 [unchanged].

Centripetal force FO = MV1 2 r ; v1 is the velocity of the ellipse a.

In this case, f fo, then v1 > v; [i.e., the velocity of the ellipse a is large].

The gravitational force is the same at the same point.

But the centripetal force is different.

The centripetal force of the circle is the gravitational force.

Can the centripetal force of an ellipse be greater than the gravitational force?

Still, you'd better ask the teacher

The satellite needs to do centripetal motion at the tangent point from orbit A to orbit B, and must brake to decelerate, so VB VA

The motion formulas of the Earth and Mars are listed separately: gravitational force is equal to the centripetal force, gm too r earth 2 = r earth * 4 2 t earth 2

gm too r fire 2 = r fire * 4 2 t fire 2

Compare the two formulas to know: BCD

BCD Let the distance between the Earth and the Sun be r The distance between Mars and the Sun is r by m earth gm too r 2=m earth r4 2 t earth 2 find m too by m fire gm too r 2=m fire r4 2 t fire 2=m fire a fire = m fire v 2 r

You can find the distance between Mars and the Sun, the acceleration of Mars, the speed at which Mars moves in a uniform circle.

Obviously, I basically forgot about high school physics after I hadn't been exposed to it for a year. They should be right in my recollection!

A right. BCD Wrong.

Item a. The potential energy mentioned here can be understood as gravitational potential energy, i.e., gravitational potential energy. It is obvious that the orbital altitude of the Heavenly Palace is higher than that of the Divine Eight. From the formula ep=mgh, the potential energy of the Heavenly Palace is greater than that of the Eight Gods.

Item b. Shen Ba should speed up the docking with the Heavenly Palace.

Item c. The period of Tiangong-1 is greater than that of Shenzhou-8.

Item d is the maximum orbital velocity of the satellite around the earth, because Tiangong-1 and Shenzhou-8 both rotate around the earth, so their velocities are less than.

It is known that the mass of the Moon is 1 81 of the mass of the Earth, and the radius of the Moon is 1 of the radius of the Earth

1) What is the ratio of the maximum altitude of an object thrown vertically upwards at the same initial velocity near the Moon and the Earth's surface?

2) What is the ratio of the horizontal range of an object when it is thrown horizontally at the same altitude from the surface of the Moon and the Earth, respectively, at the same initial velocity?

ad。The gravitational potential energy takes infinity as the 0 potential energy surface, so the potential energy on the orbit is negative, and the larger the orbit, the greater the potential energy.

Refers to the orbital velocity when on the surface of the earth, and neither Shenhachi nor Tianyi is on the surface of the earth.

From gravitational gravity: gmm r = mg g = gm r then g fire g 1 10 m ground (1 2 r ground).

地 r 地 = 地.

When the spacecraft decelerates vertically near the surface of Mars and descends to the surface of Mars, the astronaut is in an overweight state, and the pressure is mg fire + mA

F pressure: KMG = 10mg 10mg = mg fire + MAA = 10g - g fire 10*

Earth's surface.

gmm/r²=mg

The surface of Mars. gm1m/r1²=mg1

and r=2r1

m=10m1

The vehicle decelerates vertically near the surface of Mars and descends to the surface of Mars, fn-mg1=mA

fn=kmg

The maximum acceleration of the aircraft is a=

Gravitational acceleration g=gm r 2, then we can know that the gravitational acceleration on the surface of Mars is times that of the earth, the process of vertical deceleration and landing, the direction of acceleration is vertically upward, there is an acceleration provided by the aircraft to the pressure of the astronaut, then there is, the maximum value of n is k times of its own gravity, k=10, then there is 10*, and a = is obtained

1. Orbit n times around the earth for t

then t=t n

2. According to the law of gravitation.

gmm (r+h) 2=m(r+h)4 2 t 2 An object on the earth's surface, the gravitational force is equal to the gravitational force.

gmm/r^2=mg

The two formulas are combined, and h can be obtained

H = three root number under gr 2t 2 4 2n 2 subtract r3, know h, period t, v how to seek, leave space for the landlord to think, I hope it can help you.

By gravitational force f=mmg r 2 when on the surface of the earth f = mg r 2=g ; m=gr 2 g; m is the mass of the earth, the period of Shenzhou orbiting the earth t=t n, the centripetal force is mw 2r=m(2 t) 2 r = mmg r 2r is the orbital radius of Shenzhou;

r 3 = mgt 2 (4 2) = g(rt 2) 2r = root number 3 (g(rt (2 n)) 2).

The altitude above the ground during the flight of Shenzhou h=r-r=root number 3(g(rt (2 n)) 2)-r

The circumnavigation speed of Shenzhou v=wr=2 r t=root number 3 (2 ngr 2 t).

f=mv*v (r+h)=mg to obtain v*v (r+h)=g v*v=g*(r+h).

Again, 2*r*pi*n=vt, so v=2*r*pi*n t

h=(2*r*pi*n t)(2*r*pi*n t) g-r

It's been a long time since I graduated from high school, and I don't know if it's right, because of terrestrial planets, the density is about the same as the earth, there is a gravitational force formula, and the density formula, and the centripetal force formula, you can get that g is proportional to r, v is proportional to r, a, c,

Habitable for humans, i.e. the planet's g is close to g.

So a b is wrong.

From v=root gr, we can see that v is greater than.

Gravitational acceleration g = gm r 2, we can see that the acceleration cannot be compared with that of the earth, because the mass of the planet is unknown. But according to the title, the acceleration should be approximate to the gravitational acceleration of the Earth.

Orbital velocity v = under the root sign (gr) can be seen and should be greater than.

Therefore, C is chosen