The question of gravitation is related to potential energy

Solution: The mass of the satellite is m

Let the centripetal force of the satellite in a circular motion on the earth's surface be f, f=gmm, r =mg....When the satellite moves in a circular motion in an orbit 2r from the center of the earth, the velocity is v and the centripetal force is f1 = gmm (2r) = mv 2r ....②

Find v= (gr 2) ....

Let the satellite eject an object with a mass of (7-4 3) m in the opposite direction of flight, the satellite has a mass m and a velocity v The flight altitude is r, and the mass of an object with a mass of (7-4 3) m is m and the velocity is v, and the velocity direction of the satellite is positive, which is obtained by the law of conservation of momentum: mv=m v +m (-v).

From the above, v = (gr 2) m = m - (7-4 3)mr = 2 (v) g ....

The change in the flight altitude of the satellite δh = r -2r ....

ΔH is obtained by

Orbit I and Orbit II are two orbits, and it is impossible for a satellite to enter I while II is in operation. You have to know this. It's not just from i to ii, or from ii to i.

They just coincide at point A, and when the satellite is in orbit II, the velocity through point A is smaller, so the centripetal force required for it is smaller, but the gravitational force is just as large. That extra gravitational pull will pull the satellite's orbit lower, thus reaching point b.

When the satellite is in orbit I, the speed through point A is larger, and the centripetal force required by it is relatively large, and the gravitational force acts as the centripetal force, so that the satellite can do a uniform circular motion.

In order to enter orbit I from point A of orbit II, the satellite must accelerate at point A in order to change orbit. Conversely, from I to II, the satellite needs to slow down at point A.

To correct your mistake, A is a fixed point at which there is only a fixed gravitational force, no fixed velocity, no fixed centripetal force, and no fixed kinetic energy.

Also, the gravitational potential energy of orbit I at point A is the same as that of II, but the kinetic energy is greater than that of orbit II.

Halo, which is relative.

First of all, I would like to talk about the problem of small potential energy, which is that the potential energy of people on the earth is smaller than that of people in the air. Because gravity can be analogously regarded as the gravitational force of the earth to people, then according to the gravitational formula, the closer the distance, the greater the gravitational force (gravity), and the person falling from the sky is the work of gravity, so the gravitational potential energy is reduced, so that the gravitational potential energy of the people on the ground is smaller than that of the people in the air.

Similarly, through the conservation of mechanical energy, we can conclude that the kinetic energy of a person on the ground is larger.

Set the mass of the particle to be m, the distance between the two particles represented by x, and g is the gravitational constant. l = 10m;

In fact, this problem is not simple, and it is very difficult to do it in the first year, especially if it requires a lot of money.

I hope you understand.

According to Newton's second law, 1) a particle, m* dv dt = transderivative 2 2;

acceleration a = dv dt = dv dx dt = v dt = v * dv dx;

Try to substitute 2 = gm * dx into equations (1), (2).

v from the integral 0 - v x from x to l integral.

Eq. (2), the integration is obtained by 2 2 V on both sides of 2 = g (1 x-1 l);

sqrt [2gm(1x-1l)](3); The square root of sqrt.

3) the relationship between velocity and displacement, since dx dt = v; dt = dx / v =??dx [sqrt [2gm(1x-1l)]];

T 0-t on the left is an integral right, x long x integral, this integral is not too simple, you can look up the book, the integral is full of points, so x = l 2 time.

Earth's surface (gmm) (r 2)=mg Find m is the mass of the earth Then you can find r is the radius of the spacecraft, r-r is the height of the spacecraft 2(gmm) (r 2)=mv 2 r=mr 2=mr(4 2) t 2 can be used to find the speed of the run.

It's very simple, just look at the formula, and the values come out.

Take a very small arc s on the ring, then the ring is made up of many such arcs.

Each arc is subject to the gravitational pull of the ball, and its components are vertically upward and horizontally toward the center of the ring.

All the horizontal components of the small arc are canceled out, only the upward component.

The mass of the small arc is s*m 2* *r

The upward component of the gravitational force of the small arc is: l (r 2 + l 2) 1 2 * gm s*m 2* *r*(r 2+l 2).

The upward component of the gravitational force of all small arcs is equal to the gravitational force between the ring and the sphere, and s is replaced by 2* *r where in the above equation.

This yields gmml (r*r+l*l) 3 2

According to the symmetry, only the vertical direction is divided, and the horizontal direction cancels out.

When you get to college, you can use calculus to find the gravitational resultant force. The result is the same.

The distance you are talking about is only the distance from the satellite to the ground.

Gravitational force is indeed looking at the distance between two objects. If we think of the Earth as an object with a large volume, then we can think of it as being made up of countless small particles, and the gravitational pull of the whole Earth on the satellite can be thought of as the sum of the gravitational pull of every small part of the Earth on the satellite. Since the volume of the Earth is to be considered, the distance from each small part to the satellite is not the same.

And the average distance they have to the satellite is r (the distance from the center of mass of the Earth to the satellite), not r minus the radius of the Earth.

The law of gravitation is the distance between two objects in space, if the earth is regarded as a point with a large mass, then this point is the center of mass of the earth, because there is no center of mass, just like the center of gravity, it is an artificial abstract concept; Since the earth is abstracted as a point, the distance between him and the satellite is, of course, the distance from the center of mass to the satellite.

You are right, the law of gravitation is the distance between two objects, and the orbital radius r is the sum of the radius of the earth (r) and the distance of the satellite from the ground (h) i.e. r=r+h gravitational force f=gmm r square only when the satellite is orbiting on the earth's surface r=r f=gmm r square.

The orbital radius is commonly represented by r, and the radius of the planet is often represented by r, and the meaning of their representation is different.

If you don't understand anything in other chapters, you can also ask me at any time.

When the distance is long, the radius of the earth is negligible.

Solution: From the problem, we can know that f=gmm (2r)*2

Density p=m4 3 (r)*3

When two large solid spheres with a radius twice the size of the sphere are close together, the law of gravitation shows that f*=gmm (4r)*2

where m=vp=4 3 (2r)*3 x m 4 3 (r)*3 substituting the solution to obtain f*=16f

Then the magnitude of the gravitational force between them is 16f

"If two large solid balls with a radius twice the size of the ball are close together"I don't understand this sentence, your previous question** reflects the big ball, the small ball?

The gravitational force between two objects that can be regarded as particles can be used in order to use the formula, your topic is the Orsay?

First of all, it should be stated that the first question can be solved with flat throwing knowledge. The second question can actually be solved with knowledge of kinematics. Since physics is a discipline that simplifies problems, I will write two methods for your reference.

1.According to the knowledge of free fall motion: object motion time t = (2h) g = 2 5 horizontal motion distance s = vt = 1800 5

Apply kinematics to solve.

Change unit: deceleration: a minus = ff m =

According to the formula, so s=(vt 2-v0 2) 2a=5m is solved using the kinetic energy theorem.

Swap unit: The kinetic energy of the truck is w= j

Because the resistance is ff=

So there is l=w ff=5m

Question 1 The time it takes for an object to fly is 100=solution t=4 5 seconds.

So the horizontal distance is 4 5*900=3600 5 meters.

The velocity in the vertical direction when landing is v=gt=4 5*10=40 5 meters per second.

So the sum speed is 900 2 + (40 5) 2 = 100 meters per second.

Question M s The kinetic energy of a truck is w = j

Because the resistance level is to the 4th power n

So there is l=w f=5 meters.

Answer: Whether it is the Earth or Mars, the circumferential motion around the Sun is:

Gravitational force = centripetal force required for circular motion, i.e.:

To the earth there are: gmm r = m r = 4 mr t , t = 4 r gm

Where: g = gravitational constant; m = solar mass; m = mass of the earth;

the angular velocity of the Earth; r = solar-terrestrial distance); t = Earth's orbital period.

If m, t, r are known, then the mass of the sun m can be calculated.

The orbital radius of Mars can be calculated when m and t (the orbital period of Mars around the Sun) are known

r = (gmt²/4π²)1/3)

Mars orbital velocity = 2 r t, substituting the above equation, it can be calculated.

The acceleration of Mars = r, substituting the above equation can be calculated.

Based on the above conditions, the mass of Mars cannot be calculated.

The centripetal acceleration of Mars' orbital motion calculated above, not the gravitational acceleration on Mars.

Gravitational acceleration on Mars (gravitational acceleration on Mars) = gm r', r' = Martian radius.

Because Mars lacks its own radius, it is impossible to calculate the gravitational acceleration on Mars.

Based on the above conditions, the orbital centripetal acceleration of Mars can be calculated.

Earth-sun spacing r

The mass of the earth m the period of motion of the earth around the sun t

Mars orbits the Sun for a period t

Find: Mars circular velocity v Mars mass m is the same according to the gravitational pull of the Sun on Earth and Mars:

1) mr t2 (squared) = mr t2 (squared);

v=2 r t ; Relationships are listed by period ratios;

2) Available: r (Mars-Sun distance) m v

The mass of Mars cannot be calculated under these conditions, and the mass of Mars can be calculated according to the radius of Mars and the period of the near-Earth satellite.

The orbital radius of Mars is missing.

1 All assume that an object m motions in a circle around the planet attached to the surface of the earth, then.

1/6 mg=mmg/r^2

mmg r 2=mv1 2 r Find the first cosmic velocity of this planet v1=√1/6gr

The second cosmic velocity is the first cosmic velocity. 2 times, so v2=.√2v1=.1 3gr option c

Solution: According to the meaning of the topic: Suppose that an object m on the surface of the earth moves in a circle around the planet, then 1 6 mg = mmg r 2

mmg r 2=mv1 2 r Find the first cosmic velocity of this planet v1=√1/6gr

The second cosmic velocity is the first cosmic velocity. 2 times, so v2=.√2v1=.1 3gr option c