Fictitious force in a binary stars

Click For Summary
Fictitious forces arise in non-inertial frames of reference, complicating the understanding of motion in binary star systems. The discussion emphasizes the need to clarify which frame of reference is being considered to address the question effectively. It suggests focusing on the frame of reference of mass 'm' to determine the fictitious force that, when combined with the normal force, yields the instantaneous acceleration of 'm'. Understanding these forces is crucial for accurately analyzing dynamics in such systems. Clarity in the frame of reference is essential for resolving the complexities of fictitious forces in binary stars.
Rikudo
Messages
120
Reaction score
26
Homework Statement
Two planets with mass M1 and M2 revolve around each other in circular orbits. The distance between them is r. A mass m that is located on the surface of M1 with radius R feels fictitious force that modifies the mass's acceleration. Assume that mass m is very small and ignore the effect from M1's rotation. (see figure)

Write down the vector of the fictitious force in m, M2, r, gravity constant G, and unit vector!
Relevant Equations
-
1661309980439.png


I have a difficulty in understanding the question.
Fictitious force is a force whose motion is described using a non-inertial frame of reference. Which frame is the question referring to?
 
Physics news on Phys.org
Rikudo said:
Which frame is the question referring to?
Excellent question!
I suggest m's frame of reference, i.e. find that force which, added to the normal force, results in m's instantaneous acceleration.
 
Thread 'Correct statement about size of wire to produce larger extension'
The answer is (B) but I don't really understand why. Based on formula of Young Modulus: $$x=\frac{FL}{AE}$$ The second wire made of the same material so it means they have same Young Modulus. Larger extension means larger value of ##x## so to get larger value of ##x## we can increase ##F## and ##L## and decrease ##A## I am not sure whether there is change in ##F## for first and second wire so I will just assume ##F## does not change. It leaves (B) and (C) as possible options so why is (C)...

Similar threads

  • · Replies 114 ·
4
Replies
114
Views
5K
Replies
3
Views
1K
  • · Replies 44 ·
2
Replies
44
Views
5K
  • · Replies 5 ·
Replies
5
Views
913
Replies
5
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
Replies
24
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K