Talk about the law of universal gravitation.

Teaching purpose of the law of universal gravitation/200406/CA431316.htm

Discovery and application of the law of universal gravitation

Physics thesis

PB0500082 1 Wu Ruiyang

The discovery of the law of universal gravitation

We all know that Newton discovered the law of universal gravitation. When we were young, we also heard the story that Newton found gravity when he saw an apple fall to the ground. But its discovery is not as simple as seeing an apple fall to the ground.

Gravitational formula: where g is the gravitational constant. In Newton's time, some scientists thought that everything had gravity. Moreover, Newton and Hooke once argued about the right to discover gravity. According to some data, the concept of universal gravitation was first put forward by Hooke, but Hooke's mathematical attainments are far less than Newton's, so he can't explain the elliptical orbit of the planet. Newton not only put forward that gravity is proportional to the square of distance, but also solved the problem of elliptical orbit of the planet satisfactorily, so the priority of discovery of gravity naturally belongs to Newton.

As he said, Newton stood on the shoulders of giants. Kepler's research results made an indelible contribution to the discovery of gravity. Kepler is a German astronomer, and his teacher, Brother Gu, left him a lifetime of astronomical observation data. On this basis, Kepler published the first and second laws of planetary motion in 1609 after 20 years of calculation and arrangement. Later, the third law of planetary motion was published.

According to Newton's memoirs, Newton first studied the movement of the moon. Newton's inverse square law was derived from Kepler's third law of planetary motion. To calculate elliptical orbits, it is obvious that Newton must also have some concepts about calculus and basic mechanical laws. Newton made many discoveries in basic mechanics, and Newton and Leibniz independently discovered calculus. Newton applied calculus to calculate universal gravitation. As for the right to discover the law of gravity, the historical conclusion was discovered by Newton. The expression of gravity is that its establishment is the comprehensive result of Newton's law and Kepler's law, and Newton played a key role in it.

The establishment of the law of universal gravitation

I. Determination of inverse square law

1. Inverse Square Hypothesis of Theoretical Calculation:

For simplicity, the orbit of the planet can be regarded as a circle (when the orbit of the planet is regarded as a circle, it has been proved in textbooks). In this way, according to the law of area, if a planet wants to make a uniform circular motion, only centripetal acceleration a=v2/r, where v is the speed of the planet and r is the radius of the circular orbit.

According to Newton's second law: f=ma

have

Youyou

According to Kepler's third law, k is a solar constant that has nothing to do with planets.

that is

So ... (1)

Newton got the first result: if the gravity of the sun is the cause of planetary motion, this force should be inversely proportional to the square of the distance between the planet and the sun.

2. Verification of inverse square hypothesis:

The story of Newton's "Apple Falling to the Ground" is widely circulated. The main idea of the story is that 1665- 1666, Newton resigned from Cambridge University and stayed at home. One day, while he was thinking about the dynamics of gravity in the garden, he happened to see an apple fall to the ground, which aroused his thinking. At the farthest distance we can climb and the highest mountain top, gravity has not obviously weakened, and this force will inevitably reach a place much farther than usual. That should be as high as the moon. If so, the movement of the moon must be affected by it. Perhaps it is for this reason that the moon stays in its orbit.

Suppose that the moon is at any point in its orbit A (see figure) and O is the center of the earth. If it is not subjected to external force, it will move along a straight line AB, but its orbit is actually an arc AP, and AB is tangent to the orbit of point A, so the distance from the moon to O is BP=y, making the arc length AP=s=2πrt/T,

And cos θ ≈ 1-/2, θ = s/r.

y = r( 1-cosθ)≈S2/2r = 4π2r 2t 2/2r T2 = 2π2r T2/T2

On the ground, the distance a heavy object falls in t time is

y=gt2/2

From this

y/y' =4π2r/gT2

The period of the moon's orbit around the earth is t = 27.3d ≈ 2.36x106s, the ground gravity acceleration is g=9.8 m/s2, and the exact value of radius of the earth r is 6400km. By observing the duration of the total lunar eclipse, the ancient Greek astronomer Iba Valley once estimated that the distance between the earth and the moon was 60 times that of radius of the earth, so R = 60r = 3.84×.

y/y' = 1/3600

R2/r2= 1/3600。

y/y'=a/g=ma/mg=f/mg= R2/r2

So: f=mg R2/r2, that is, the force is inversely proportional to the square of the distance.

2. Determination proportional to M and M

① The formula shows that the force is directly proportional to the mass m of the attracted object. According to Newton's third law, the force interacts, F is the action of M on M, F' is the action of M on M, and F is directly proportional to M, so F' must be directly proportional to M and F', and F must be directly proportional to M and M at the same time. (1) type can be written as:

f=GMm/r2，…②

Where g is the gravitational constant.

Three. Determination of gravity constant

Now that we have the expression of gravity, we must measure the gravitational constant. In order to measure the value of the gravitational constant g, it is necessary to measure the gravitational force between two objects with known mass. 1798, Cavendish made the first accurate measurement.

He used a torsion balance device. As shown in the figure, two balls with mass m are fixed at both ends of a polished rod, and a timely filament is hung horizontally on the two polished rods, and a big ball with mass m is placed near each ball. According to the law of universal gravitation, when the big ball is in AA position, because the small ball is attracted, the suspender rotates due to a moment, which distorts the messenger wire. The gravity moment is finally balanced by the elastic recovery moment of the suspended wire. The angle θ of suspension wire torsion can be measured by mirror ruler system. In order to improve the sensitivity of measurement, you can also put the big ball in the position of BB to attract the small ball in the opposite direction. In this way, the correction of the included angle between the two boom balances is doubled. If the mass m, m of big ball and small ball, the distance between them and the correlation coefficient of messenger wire torsion are known, G can be calculated from the measured θ. Cavendish's gravitational constant value is:

g = 6.754× 10- 1 1 m3/kg？ S2. Gravitational constant is the most inaccurate basic physical constant measured at present. Because gravity is too weak to shield its interference, it is difficult to do experiments. It has been nearly 200 years since Cavendish. Many people use the same or different methods to measure the value of g, and constantly improve its accuracy. CODATA) 1986 recommended values are

g = 6.67259(85)× 10- 1 1 m3/kg？ s2，

The uncertainty is 128/ 100000 (i.e. 1.28 per ten thousandth).

Experimental demonstration of gravity

First, the experimental phenomenon:

In some science and technology museums, there are demonstration devices as shown in the figure: a disk similar to a bowl, but the wall of the bowl arches inward. A small ball with ink enters at a low speed from the edge of the disk, rolls around the center on the disk and leaves traces. It can be observed that with the change of time, the speed of the ball is getting faster and faster, and finally it falls into the middle hole. And the radius of the ball changes slowly in the middle (that is, the trajectory of the ball is densest in the middle). The trajectory of the ball is not a perfect circle, but an arc with smaller and smaller radius. (If two balls are put into the dish successively, they will catch up before and after angular displacement. )

Second, the principle explanation:

1, why use it to demonstrate the law of universal gravitation?

It can be inferred from the expression of the law of universal gravitation that the potential energy of the planet is

Gravitational potential energy MGH =-GMM/R is used instead of gravitational potential energy in the experiment.

So as long as h=-GM/gr is satisfied, gravitational potential energy can be used instead of gravitational potential energy.

At the same time, r represents the distance between objects. When the illustrated curve rotates around the H axis, the surface used in the experiment is formed. When the graph is dh/dr=GM/gr2,

f = mgtanθ= mgdh/dr = mgGM/gr2 = GMm/R2

Therefore, this experimental model can simulate and demonstrate gravity from the point of view of energy and force.

2. Why is the ball getting faster and faster?

According to the centrifugal force f=mv2/r=F and the kinetic energy e k =1/2mv2 =1/2gmm/r, it can be seen from the formula that the smaller the r, the greater the kinetic energy and the faster the natural speed.

3. Why is the trajectory from the ball to the middle so dense?

This is a working system with dissipative force. When rotating, friction does work and heat dissipates a part of energy w=, and the magnitude of f is only related to contact pressure and friction coefficient. When the distance is r, the work done by one revolution of friction force: w= =2πrmg? cos[arctan(dh/dr)]=

DE/dr= GMm/r2。 It can be seen that the greater R, the slower the change of potential energy. Therefore, in the outer ring, the change of potential energy is not as great as that in the inner ring after changing a small amount of dr, but the energy consumed in one ring is greater than that in the inner ring. Therefore, when the inner ring rotates, every time the rotation position drops for a short period, the ball rotates with more energy, while the friction consumes less energy every time the ball rotates, so in the same short distance, the ball will rotate several times more than the outer ring.

4. Why is the trajectory of the ball an arc that keeps shrinking inward?

If the disk is smooth enough, that is, there is no friction to do work, what will the trajectory of the ball be like?

A, when the ball enters the edge of the disk, it just gets enough kinetic energy to move on the edge of the disk;

Mgh= 1/2 mv2, that is, F=mv2/r, the ball will move in a circular trajectory along the edge of the disk.

B, when the ball enters the disk edge, it does not reach the kinetic energy required for the disk edge movement;

Mgh÷ 1/2mv 2, F & gtMv2/r, the ball will make a circular motion, and at the same time there will be a radial deflection motion (reducing the radius will convert potential energy into kinetic energy until it meets the balance). When it reaches a certain radius, it will reach F=mv2/r, where it will do a circular motion.

C, the kinetic energy of the ball when it enters the edge of the disk exceeds the kinetic energy needed to move at the edge of the disk:

Mgh< 1/2mv2, F<'s mechanical analysis of mv2/r on the disk shows that the ball will fly along the edge.

Only case b was simulated in the experiment.

Let's consider friction:

Because friction does negative work, the kinetic energy of the ball is constantly lost in the process of motion, which requires the ball to continuously shrink its radius to find a new balance until the second half passes through a small hole in the middle due to lack of energy.

Therefore, there is: the trajectory of the ball is a circle that shrinks inward.

5. Why do two balls have an angle to catch up with each other?

According to the formula of universal gravitation, the angular velocity ω =, so the smaller the radius, the greater the angular velocity. The angular velocity of the two balls that enter successively is always greater than that of the latter, so for a period of time, the former ball always rotates much more than the latter ball, so the angle difference is increasing, and the ball that enters first will exceed the ball that enters later by many times, so it seems that the two balls are always chasing each other.

Application of gravity

As one of the most basic laws of nature, the law of universal gravitation has been widely used in theoretical research and engineering design. For example, in aerospace, the gravity of spacecraft approaching celestial bodies can be used as an effective acceleration method (slingshot effect); In cosmic physics, the position and mass of celestial bodies are often determined by measuring the action of gravity. In areas where electromagnetic detection is limited, the density of underground materials can be detected by gravity measurement and calculation, so as to determine the distribution of underground mineral deposits or the scale and location of underground tombs; In other fields, such as the manufacture of super-circular rolling spheres in precision industry, we can choose to produce them in space because there is an ideal stress environment (because objects are in weightlessness on spacecraft, the gravity of other stars can be ignored because of the uniformity of large-scale distribution in the universe); The project aimed at studying the weightlessness of organisms in space (that is, the cancellation of centrifugal force balance in gravity language) has developed into a high-tech frontier technology. If the seeds of vegetables and fruits are taken into space, under the influence of weightlessness and cosmic rays, the quality of some mutant varieties is greatly improved than that of varieties on earth.

In fact, the law of universal gravitation is often one of the most basic and commonly used formulas in theoretical research. Here is a practical example to illustrate this point.

The launching process of artificial satellite;

1, when we want to launch an earth satellite, we only need to launch the satellite into space at a certain angle and at a certain initial speed. The theoretical value of this speed can be inferred from the law of universal gravitation: the law of universal gravitation of 7.9 km/s determines the boundary conditions from the satellite to the sky for us. Of course, the drag problem should be considered in the actual launch, which is not an instantaneous acceleration to this value, but a gradual acceleration process, which is more complicated.

2. When we require the satellite to be the sun's satellite, our theoretical launch speed will be as high as 1 1.2km/s ... Similarly, in the actual process, the speed will not reach this value, but will gradually accelerate.

3. When we ask the satellite to become an extrasolar body, our theoretical launch speed will be as high as 16.7km/s/s ... In practice, the speed will not reach this value. In fact, we deliberately launch the aircraft near the celestial bodies in the solar system, and use the slingshot effect to change the speed and direction of the aircraft by using the gravity between the aircraft and the celestial bodies.

From the above application, we can see the important position of gravity. As soon as the concept of gravity was put forward, it caused a "scientific revolution". The subsequent era triggered a wave of research and exploration of the universe, resulting in many new disciplines and projects and many new discoveries. These research results are still closely related to our lives. The discovery of the law of gravity has promoted the whole process of human civilization, which is a big step for human beings to understand the universe and an extremely important step!

We have more reason to believe that gravity will still play a very important role in future scientific exploration and research.

Review on the Law of Universal Gravitation (I)

The law of universal gravitation and its application in astronomy

Key points of knowledge

(A) the law of universal gravitation

1. Content:.

2. Formula:. G is the gravitational constant, usually taken as g = 6.67×10-11.

3. Applicable conditions:

( 1)

(2)

(3)

4. Note:

(1) The mutual attraction between two objects is a pair of force and reaction, which are always equal in size and opposite in direction, and obey Newton's third law.

(2) The gravity of an object on the earth's surface is approximately equal to the gravity of the earth on the object. GM = GR2 is known, which is a common transformation.

(3) The higher the height above the ground, the smaller the acceleration of gravity of the object, and the relationship with the height is: r is radius of the earth, and h is the height above the ground.

(2) Application in astronomy

1. Basic method: The motion of celestial bodies (or satellites) is regarded as uniform circular motion, and the required centripetal force is provided by gravity.

When solving problems, you can choose formulas to analyze and calculate according to the situation.

2. The method of finding the mass and density of celestial bodies

(1) By observing the period t and radius r of the satellite whose celestial body moves in a uniform circle, gravity equals centripetal force.

Force, celestial mass m =

(2) If the radius r of a celestial body is known, then the density of this celestial body

Typical example

As we all know, the radius of Mars is half that of radius of the earth, and the mass of Mars is the mass of the earth. If a person with a mass of 60kg on earth goes to Mars, what is the gravity on the surface of Mars?

Example 2: The distance from the earth to the center of the moon is about 4× 108m, and the mass of the earth is estimated. (Result: Keep one significant digit. )

Example 3: The mass of a planet is nine times that of the earth, and its radius is about half that of the earth. How long does it take to throw an object vertically upward from a certain height at the initial speed of 10m/s from throwing to falling back to its original position? (g = 10m/s2)

Example 4: The gravity of an object on the ground is 160N. If it is placed in a spaceship, when the spaceship rises at an acceleration of a=g/2, at a certain height, the force between the object and the bracket in the spaceship is 90N. How far is the satellite from the center of the earth at this time? (R = 6.4× 103km, G = 10m/S2 is known).

Consolidation exercise

1. Stars and stars are found in orbits calculated by people according to the law of universal gravitation. The reason why the orbit of a star deviates from the orbit calculated by the law of universal gravitation is because the star is influenced by the gravity of other planets outside the orbit.

2. Let Saturn make a uniform circular motion around the sun. If the distance from Saturn to the sun is measured as r, the period of Saturn around the sun is t, and the known gravitational constant is g, according to these data, the quantity that can be obtained is ().

A. The magnitude of Saturn's linear velocity B. The magnitude of Saturn's acceleration

C. The mass of Saturn D. The mass of the sun

3. 1789, the British physicist skillfully used this device to accurately measure the gravitational constant in the laboratory for the first time. The measurement of gravitational constant is of great significance: (1)

(2) .

4. If a ball is thrown vertically on an asteroid with initial velocity of v0 (there is no air on the surface of the planet) and the maximum height that the ball can rise is measured to be H, then () can be calculated from it.

A. the mass of the planet

B. the minimum period of a satellite's uniform circular motion around the planet

C. the speed of the first universe on earth

D. the maximum acceleration of a satellite that makes a uniform circular motion around the planet.

A spaceship flies around the surface of an unknown planet. To measure the density of the planet, just ().

A, measuring operation period B, measuring peripheral radius

C, measure the volume of the planet d, and measure the running speed.

6. Which set of data is known below, you can calculate the mass m () of the earth.

A. the period t of the earth's revolution around the sun and the distance r from the earth to the center of the sun, and the distance from the earth to the sun.

B the period of the moon's orbit around the earth is t months, and the distance from the moon to the center of the earth is r.

C. the speed v and operating period t of the artificial earth satellite when it orbits near the earth's surface.

D If the rotation of the earth is not considered, the radius and gravitational acceleration of the earth are known.

7. The masses of the two planets are M 1 and M2, respectively, and the orbital radius ratio around the sun is R 1: R2, so their periodic ratios around the sun.

8. The orbit radius of the earth around the sun is R and the period is T. The orbit radius of the moon around the earth is R and the period is T. What is the mass ratio between the sun and the earth?

9. The period of the moon's orbit around the earth is about114 of that of the earth's orbit around the sun in period of revolution, and the distance between the sun and the earth is about 400 times that of the moon and the earth. What is the mass of the sun? (Take one significant digit)

10. It is proved that if the satellite of the celestial body moves around the surface of the celestial body, the density of the celestial body is

1 1. The time required to measure the height h of an object falling freely on the ground is t, and the time required to measure the same height of an object falling freely at the top of a mountain is increased by Δ t. It is known that radius of the earth is R0. Try to find the height h of the mountain.