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If the orbit radius of the earth around the sun is R1 and the revolution period is T1, and the orbit radius of the moon around the earth is R2 and the revolution period is T2, then the mass ratio of the sun to the earth is 0___ ．
According to the centripetal force provided by universal gravitation: for the earth, there is: GM earth m day R21 = m earth 4 π 2t21r1. For the moon, there is: GM earth m month R22 = m month 4 π 2t22r2 simultaneous solution, m day m earth = t22r31t21r32, so the answer is: t22r31t21r32
The period of the earth's revolution around the sun is T1, the orbit radius is R1, and the period of the moon's revolution around the earth is T2, the orbit radius is R2. How many times of the mass of the sun is that of the earth?
According to gmmr2 = MR4 π 2t2, the mass of the central celestial body is m = 4 π 2r3gt2. Because the ratio of the orbital radius of the earth to the orbital radius of the moon is r1:r2, and the ratio of the period is t1:t2, the mass ratio of the sun to the earth is m = r13t22r23t12
If the period of the moon around the earth is T1 and the radius is R1, and the period of the earth around the sun is T2 and the radius is R2, they will move
If the period of the moon's orbit around the earth is T1 and the radius is R1, and the period of the earth's orbit around the sun is T2 and the radius is R2, what is the relationship between the ratio of the orbit, the third power of the radius and the second power of the period?
God! It's so complicated. I study literature. Although I'm good at geography, I don't think it can be solved only by physics!
The masses of the two planets around the sun are M1 and M2 respectively, and the orbital radii are R1 and R2 respectively. If they are only affected by the sun's gravity, then the two planets will have the same row
（1） (2)
Analysis:
(1) Let the mass of the sun be m. according to the law of gravitation, the ratio of gravitation between two planets and the sun is
==.
(2) The motion of two planets around the sun is regarded as uniform circular motion, and the centripetal force is provided by universal gravitation, then = m () 2R
RELATED INFORMATIONS
- 1. If the mass ratio of the two planets is M1: M2 = P and the orbit radius ratio is R1: R2 = q, then the ratio of the two planets to the sun's gravity is F1: F2=______ ．
- 2. The radius of a planet is R1, the autobiographical period is T1, it has a satellite, the orbit radius is R2, the cycle around the planet is T2, if the gravitational constant is g, find the average density of the planet
- 3. 7. There are two forces F1 and F2 acting on the same object. When the directions of F1 and F2 are the same, the resultant force is 100N; when the directions of F1 and F2 are opposite, the resultant force is 100N The size is 50N, and the direction is the same as that of F1
- 4. The two forces acting on the same point are F1 = 5N and F2 = 4N. If the angle between the resultant force F and F1 is θ, then θ may be () A. 45°B. 60°C. 75°D. 90°
- 5. When the horizontal thrust f acts on a, the two objects move together, and the pressure between AB is F1 When the horizontal thrust f acts on a, the two objects move together, and the pressure between AB is F1; when the thrust f acts on B, the two objects move together, and the pressure between AB is F2, then () The acceleration of two motions of a is equal BF1 + F2
- 6. On the horizontal plane, the masses of a and B are M1 and M2 respectively, and the dynamic friction coefficients between them are the same, Slide right together. Find the interaction between a and B
- 7. When the angle between the two forces is 90 degrees, what is the resultant force?
- 8. The masses of the three objects are M1 = 2kg, M2 = 4kg and M3 = 6kg, respectively, moving along a smooth horizontal plane at a certain speed, Now they brake under the same constant resistance, and finally they all stop (1) If the initial velocity is the same, what is the displacement ratio S1: S2: S3? (2) If it takes the same amount of time for the resistance to start and stop, what is the velocity V1: V2: V3 of their uniform motion?
- 9. A body is forced by F 1 = 6N to produce an acceleration of a 1 = 3m / S 2. How much force should be applied to make a 2 = 9m / S 2 acceleration
- 10. The product of the lengths of the real and imaginary semiaxes of the hyperbola is root sign 3, F1 and F2 are the left and right focal points, the straight line L passes through point F2, and the included angle with the straight line F1F2 is θ, Tan θ = root sign 21 / 2, the intersection of the vertical bisector of the straight line L and F1F2 with point P, the intersection of the line PF2 and the hyperbola with Q, and | PQ |: | QF2 | = 2:1, the hyperbolic equation is solved
- 11. In the test of verifying Newton's second law, why do we sometimes make the weight mass m far less than the trolley mass m? Sometimes we don't need to?
- 12. In the test of verifying Newton's second law, why is the mass of the heavy object much less than that of the car?
- 13. Under the action of traction F1 and resistance F2 in the horizontal direction, when F1 = F2, F1 is less than F2 and F1 is greater than F2, the car will be ······························································································································?
- 14. Two forces F1 and F2 acting on the same object, on the same line, their resultant force is 20n. It is known that F2 = 50N, and the direction of F2 is the same as that of resultant force
- 15. In the triangle ABC, the angle ABC = 45 degrees, ad vertical BC. The perpendicular foot is d. be vertical AC. the perpendicular foot is e. ad intersects be with F, connecting cf. if the angle BAC is an acute angle In the triangle ABC, the angle ABC = 45 degrees, ad is vertical BC. The perpendicular foot is d. be vertical AC. the perpendicular foot is e. ad intersects be with F and connects cf. if the angle BAC is an acute angle, it is proved that the triangle CDF is an isosceles right triangle
- 16. As shown in the figure, in △ ABC, ab = AC, AE ⊥ AB in a, ∠ BAC = 120 ° AE = 3cm
- 17. As shown in the figure, △ ABC, ab = AC, D is on BC (D is not at the midpoint of BC), de ⊥ AB is on e, DF ⊥ AC is on F, BG ⊥ AC is on G, proving: de + DF = BG
- 18. As shown in the figure, it is known that ab ‖ CD, BD bisection ∠ ABC intersects AC in O, CE bisection ∠ DCG. If ∠ ace = 90 °, please judge the position relationship between BD and AC and explain the reason
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- 20. As shown in the figure, in RT △ ABC, ∠ C = 90 °, a = 30 °, the bisector of ∠ C intersects with the bisector of the outer corner of ∠ B at point E, connecting AE, then ∠ AEB =?