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A and B are two solid balls. The mass of a ball is 1 / 2 of that of B ball. The volume of a ball is twice that of B ball. Their density ratio is A 1;4 B 1；1 C 2;1 D 4;1
Density is the ratio of mass to volume
Therefore, the density ratio of a and B is 1 / 2 / 2 = 1:4
Choose a
Why does the formula of universal gravitation only calculate the central celestial body,
Relative in motion
Now I assume that the earth does not move, take the earth as the reference system, then the sun revolves around the earth, t = 365,; let m be the mass of the earth, m be the mass of the sun, because the sun revolves around the earth, its centripetal force is provided by the earth's gravity on the sun, then GMM / R3 = 4 π 2MR / T2, eliminate the mass of the sun, then the mass of the earth M = 4 π 2r3 / GT2 can be obtained
Because Newton's formula of universal gravitation is calculated according to the third law of Kepler, then the ratio K is a constant for the fixed celestial body around which all planets revolve. Its different periods and semi major axes are only applicable to the corresponding planets around. For example, the earth revolves around the sun about 365 days a cycle. If we take the earth as the reference system, the ratio K is a constant, So the sun around the earth a cycle is a day, so different
The formula of gravitation between celestial bodies in high school physics?
(GmM)/(r^2)=mv^2/r=mrω^2=mr（4π^2）/T^2
G is the gravitational constant
In general, we just need to remember that (GMM) / (R ^ 2) = MV ^ 2 / R. the last two formulas can be deduced from v = R ω; and ω = 2 π / T. m is the mass of the central body, and M is the mass of other stars around other stars
In addition, we often use (GMM) / (R ^ 2) = mg, which can be used as other stars running on the surface of the central celestial body. Gravity is equal to gravity
Universal gravitation (mass change of celestial bodies)
Suppose the earth moves around the sun at a uniform speed, and the sun shines due to the nuclear reaction inside the sun. In this process, the mass of the sun decreases continuously,
The acceleration decreases
Gravity decreases
Frequency reduction
The angular velocity decreases
The linear velocity decreases
At this time, the average density of the earth will decrease, and the density of the sun will decrease in later years
How to calculate the mass, volume and distance of celestial bodies?
Using the law of universal gravitation can be easily calculated, then you know something about it. The method of calculating mass: calculate the centripetal acceleration of celestial body according to the motion condition, and the centripetal force is provided by universal gravitation, so the equation can be solved by listing. Let m 'be the mass of the sun, m be the mass of a planet, R be the distance, t be the revolution period, then
How is the mass of a celestial body measured?
How is the mass of the earth, the mass of the sun, and even the mass of Pluto calculated?
This involves a lot of astronomical knowledge, celestial observation is generally directly to the track of the stars, position, so that we can analyze the possible period, stars can also observe the spectrum, we can know the composition of the star's material combination, comprehensive calculation can get the star's mass density and other physical quantities, planetary
How to calculate the mass of celestial bodies?
1、 The mass of celestial body is estimated by the law of universal gravitation and Newton's law of motion. In the motion of celestial body, it is approximately considered that the motion of celestial body is uniform circular motion. The decisive factor in the motion process is universal gravitation, that is, universal gravitation provides the centripetal force for celestial body to do uniform circular motion, G (mm / r2) = m × (2 π / T) 2 ×
What is the mass of celestial bodies in the universe
How to estimate the mass of celestial bodies?
It can pass the period of its satellite operation: T ^ 2 / R ^ 3 = k = GM / 4 (PI) ^ 2 (Kepler's third law + law of gravitation), R is the radius of the satellite, G is the constant of gravitation (6.67 * 10 ^ (- 11) n * m ^ 2 / kg ^ 2), M is the mass of the celestial body
Assuming that the density of celestial bodies in the solar system remains unchanged, the diameter of celestial bodies and the distance between celestial bodies are reduced to half of the original,
Then the correct change of the following physical quantity is ()
A the centripetal force of the earth is half of what it was before
B the centripetal force of the earth is 1 / 16 of that before it shrinks
C the earth's cycle around the sun is the same as before
D the period of the earth's revolution around the sun becomes half of the previous period
B
The mass becomes 1 / 8 of the original, so the new M1 * M2 = 1 / 8m1 * 1 / 8m2 = 1 / 64m1 * m2
And the distance becomes half of the original
Substituting f = gm1m2 / R ^ 2, we can get a new F = 1 / 16F
Assuming that the density of celestial bodies in the solar system remains unchanged, the diameter of celestial bodies and the distance between celestial bodies are reduced to one third of the original, and the earth's revolution around the sun is approximately circular,
As a result, the centripetal force of the earth becomes m times of that before the reduction, the earth's cycle around the sun becomes n times of that before the reduction, and the earth's linear speed around the sun is Q times?
I think that all masses become 1 / 27, all lengths become 1 / 3, M = 1 / 81, n = 1, q = 1 / 9
The answer is 1 / 243
Ask for advice
The mass of the sun, R1 is the radius of the sun: M = 4 / 3 π × R1 & # 179; the mass of the planet, R2 is the radius of the planet: M = 4 / 3 π × R2 & # 179; the gravitational acceleration of the planet is equal to the centripetal acceleration, R is the radius of the planet's orbit: GM / R & # 178; = V & # 178 / / r the gravity of the planet: F = MV & # 178 / / r the cycle of the planet: T = 2 π R
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