Special Ratio problem (not a math problem. It's the physic problem)

In summary, the ratio of the periods of orbit TA/TB for the binary star system Krell is 1, as the two stars orbit around the common center of mass at the same rate regardless of their individual masses or distances apart. This is due to the conservation of momentum in an isolated system. The center of mass lies along the line joining the two stars and does not move, as there are no external forces acting on the system.
  • #1
NasuSama
326
3

Homework Statement



The binary star system Krell consists of two stars, Krell A and Krell B, orbiting each other, with no other ojects nearby. Assume Krell A, the smaller star of mass mA, moves in a circle of radius rA at speed vA. Krell B, the larger star of mass mB = 2.51mA), moves in a circle of radius rB at speed vB. The two stars are separated by a distance d.

Find the ratio of the periods of orbit TA/TB

Homework Equations



I would assume these formulas:

→T²/r³
→T²/r³ = 4π²/(G(M + m))

The Attempt at a Solution



I believe the approach is...

T_A²/R_A³ = T_B²/R_B³
T_A²/T_B² = R_A³/R_B³
(T_A/T_B) = √((R_A/R_B)³)

From the previous part of the problem, I've found that R_A/R_B = 2.51. Then, I substitute that for the above expression and got the answer incorrect.

T_A/T_B = √(2.51³) ≈ 3.98

But the answer is incorrect.
 
Physics news on Phys.org
  • #2
In a few days from now Sun will enter the Sagittarius. How long till it will enter the Sagittarius again?

From the Sun point of view, Earth will - again in a few days - enter the Gemini. How long till the Earth will be visible in the same direction (on the background of the distant stars) from the Sun?
 
  • #3
Borek said:
In a few days from now Sun will enter the Sagittarius. How long till it will enter the Sagittarius again?

From the Sun point of view, Earth will - again in a few days - enter the Gemini. How long till the Earth will be visible in the same direction (on the background of the distant stars) from the Sun?

Ah... Seems like there is no way to figure that out. :D

Now, help me with the question, or I will wait for someone to assist me.
 
  • #4
Maybe approach the problem from a different perspective... It's an isolated system of two objects that don't interact with anything external. What characteristics can you ascribe to the center of mass of such a system?
 
  • #5
gneill said:
Maybe approach the problem from a different perspective... It's an isolated system of two objects that don't interact with anything external. What characteristics can you ascribe to the center of mass of such a system?

The masses are distance d apart, orbiting the common center of the mass. This is by the modified form of Kepler's third law.
 

Attachments

  • diagram.JPG
    diagram.JPG
    19.2 KB · Views: 367
  • #6
NasuSama said:
The masses are distance d apart, orbiting the common center of the mass. This is by the modified form of Kepler's third law.

Where does the center of mass lie with respect to the orbiting bodies (in terms of the geometry of the system)? Does it move?
 
  • #7
gneill said:
Where does the center of mass lie with respect to the orbiting bodies (in terms of the geometry of the system)? Does it move?

The center of mass lies along the path where the larger mass will rotate. I believe it moves.
 
  • #8
NasuSama said:
The center of mass lies along the path where the larger mass will rotate. I believe it moves.

Umm, no. If you have two masses and you want to find the center of mass of the pair, do you expect to find it to the left of them both? Above them both? Or perhaps along the line segment joining the two?

If no external forces are acting on the system, what would cause the center of mass to move? This is introductory stuff regarding the motion of the center of mass of an isolated system (relies on conservation of momentum).
 
  • #9
gneill said:
Umm, no. If you have two masses and you want to find the center of mass of the pair, do you expect to find it to the left of them both? Above them both? Or perhaps along the line segment joining the two?

If no external forces are acting on the system, what would cause the center of mass to move? This is introductory stuff regarding the motion of the center of mass of an isolated system (relies on conservation of momentum).

Then, the center of mass lies along the line segment joining the two? I believe that the center lies somewhere closer to the larger mass.
 
  • #10
Nvm. The answer is 1. The velocity and the radius of the orbits are inversely proportional to each other. If the distance between the masses changes, then the periods for both masses are not all the same. If both masses are distance d apart, orbiting the common center of mass at the opposite directions, then the center of mass is "consistent" and so the change of momentum is 0.
 
  • #11
NasuSama said:
Nvm. The answer is 1. The velocity and the radius of the orbits are inversely proportional to each other. If the distance between the masses changes, then the periods for both masses are not all the same. If both masses are distance d apart, orbiting the common center of mass at the opposite directions, then the center of mass is "consistent" and so the change of momentum is 0.

Correct conclusion, but part of your reasoning is flawed. Even if the distance between the bodies changes, so that the orbits are elliptical rather than circular, the periods must still be the same.

The center of mass cannot move (assuming the coordinate system chosen has its origin at the center of mass) because there are no external forces operating. Further, the center of mass must always lie along the line joining the two mutually orbiting bodies. That means that when one body completes a full orbit and returns to its starting position, the other body must also return to its diametrically opposed position at the same instant.
 
  • #12
NasuSama said:
Nvm. The answer is 1.

Now, if you will try to understand the geometry I described in my post, you will see it was leading to this conclusion.

It is just a matter of understanding what the orbital period is - if you are on one celestial body and observing other, orbital period is a time required for the other body to appear in exactly the same place on the sky (in the reference frame of distant stars, which makes local motion of the system irrelevant). That means after a period passes from the body which you are observing you are also seen in exactly the same place (which happens to be in exactly the opposite direction, hence zodiac sings that are exactly 6 months apart).
 
  • #13
Borek said:
Now, if you will try to understand the geometry I described in my post, you will see it was leading to this conclusion.

It is just a matter of understanding what the orbital period is - if you are on one celestial body and observing other, orbital period is a time required for the other body to appear in exactly the same place on the sky (in the reference frame of distant stars, which makes local motion of the system irrelevant). That means after a period passes from the body which you are observing you are also seen in exactly the same place (which happens to be in exactly the opposite direction, hence zodiac sings that are exactly 6 months apart).

Good advice!

How much do you know about astrology and stuff like this? How do you know such thing?
 
  • #14
NasuSama said:
How much do you know about astrology and stuff like this?

Nothing. All I know is that there are 12 zodiac signs and they are placed more or less on the ecliptic.
 

1. What is the Special Ratio problem?

The Special Ratio problem is a physics problem that involves finding the ratio between two quantities in a given system, usually involving forces or distances. It is used to determine the relationship between these quantities and how they affect each other.

2. Why is the Special Ratio problem important?

The Special Ratio problem is important because it helps scientists and engineers understand the behavior of physical systems. By determining the ratio between different forces or distances, they can predict how a system will behave and make informed decisions about design and functionality.

3. How is the Special Ratio problem solved?

The Special Ratio problem is solved using mathematical equations and principles from physics. Scientists use known laws and theories to determine the relationship between the quantities in the system and then calculate the ratio using these equations.

4. What are some real-world applications of the Special Ratio problem?

The Special Ratio problem has many real-world applications, such as in engineering, astronomy, and mechanics. It is used to design structures and machines, understand the motion of planets and satellites, and predict the behavior of systems under different conditions.

5. How does the Special Ratio problem relate to other physics concepts?

The Special Ratio problem is closely related to other physics concepts such as force, distance, and motion. It is also connected to principles such as Newton's laws of motion, the law of gravity, and the conservation of energy. Understanding the Special Ratio problem can help scientists better understand these concepts and how they interact in different systems.

Similar threads

  • Introductory Physics Homework Help
Replies
1
Views
2K
  • Advanced Physics Homework Help
Replies
1
Views
2K
  • Introductory Physics Homework Help
Replies
10
Views
10K
  • Introductory Physics Homework Help
Replies
4
Views
5K
  • Astronomy and Astrophysics
Replies
1
Views
5K
  • Introductory Physics Homework Help
Replies
3
Views
11K
  • Astronomy and Astrophysics
Replies
2
Views
8K
  • Mechanics
Replies
4
Views
11K
  • Mechanics
Replies
3
Views
6K
  • Advanced Physics Homework Help
Replies
1
Views
2K
Back
Top