Question

In: Physics

Suppose two bodies are orbiting about their common center of mass, and the distance between the...

Suppose two bodies are orbiting about their common center of mass, and the distance between the bodies is a fixed distance d. Surely one possible configuration is the following: each body moves in a circle, the circles are concentric (centered at the center of mass of the system), and they have the same angular speed. Does it hold that this is the only possible configuration?

Note: I believe the answer is yes but I am clueless about how to prove it.

Solutions

Expert Solution

Yes, you are correct. Here is an easy way to see this:

Suppose you have to bodies (A and B), which do not necessarily have the same mass. Now suppose these bodies exist far from any other disturbances and that we are sitting in a reference frame stationary with respect to their center of mass. Then in our reference frame, the center of mass cannot move, and our reference frame is inertial because there are no external forces (we are far from disturbances). Thus the center of mass is fixed.

Now draw an imaginary line segment between the two masses. This line should always have length d, since we specified that the bodies are always at a fixed distance d from each other. Furthermore, the center of mass of the system is located on this line, and in particular, its location is given by the ratio of the two masses (i.e. if they are equal then it will be located at the midpoint of this line). Since the center of mass does not move in our reference frame and the bodies are always at a fixed d from each other, we can thus imagine the bodies as actually being connected to a rod that is fixed at the center of mass and can pivot around it. In this case it is obvious that the bodies must move in concentric circles and that this is the only possible motion. Any other motion will result in one of our requirements breaking down (either the center of mass will move, in which case there must be some external force, or distance between the bodies will change)


Related Solutions

Three asteroids of identical mass (M = 1.03×1012 kg) are orbiting their common center of mass...
Three asteroids of identical mass (M = 1.03×1012 kg) are orbiting their common center of mass in a perfect circle of radius R=75.8 km. a. What is the period of orbit of one of these asteroids? You are standing on one of the asteroids. You are standing on the side of the asteroid which faces out from the circle. Your goal is to jump up off the asteroid and escape the entire three asteroid system. The radius of the asteroid...
Q: In an elastic collision between two bodies of equal mass, with body 2 initially at...
Q: In an elastic collision between two bodies of equal mass, with body 2 initially at rest, body 1 moves off at angle θ relative to the direction of its initial velocity and body 2 at angle φ. The sine of the sum of θ and φ, sin(θ + φ), is equal to..................... Question options: a). 0. b). 0.500. c). 0.707. d). 0.866. e). 1.00. => I tried but it's not A or D so maybe someone can help. =>...
Two celestial bodies are  apart. One body has a mass of  , and the second has a mass...
Two celestial bodies are  apart. One body has a mass of  , and the second has a mass of  Calculate the strength of the gravitational force exerted by one body on the other.
Find the mass, the center of mass, and the moment of inertia about the z-axis for...
Find the mass, the center of mass, and the moment of inertia about the z-axis for the hemisphere x^2+y^2+z^2=1, z >(greater than or equal to) 0 if density is sqrt(x^2+y^2+z^2)
Two stars of mass M1 and M2<M1 are orbiting eachother in a circular orbit. The heavy...
Two stars of mass M1 and M2<M1 are orbiting eachother in a circular orbit. The heavy star experiences a supernova explosion, losing most of its mass in a spherically symmetric outflow (i.e. without losing angular momentum) and leaving behind a small neutron star of mass MNS. Show that if the mass lost is larger then half of the total mass of the system, the binary is disrupted.
Suppose that the mean value of interpupillary distance (the distance between the pupils of the left...
Suppose that the mean value of interpupillary distance (the distance between the pupils of the left and right eyes) for adult males is 65 mm and that the population standard deviation is 5 mm. (a) If the distribution of interpupillary distance is normal and a random sample of n = 25 adult males is to be selected, what is the probability that the sample mean distance x for these 25 will be between 63 and 67 mm? (Round all your...
Suppose that the mean value of interpupillary distance (the distance between the pupils of the left...
Suppose that the mean value of interpupillary distance (the distance between the pupils of the left and right eyes) for adult males is 65 mm and that the population standard deviation is 5 mm. The distribution of interpupillary distance is unknown. a.Suppose that a sample of size 100 is taken from this population. What is the expected value of the sample mean? The standard deviation (called standard error) of the mean? What distribution does the sample mean follow? Justify your...
Suppose that the mean value of interpupillary distance (the distance between the pupils of the left...
Suppose that the mean value of interpupillary distance (the distance between the pupils of the left and right eyes) for adult males is 65 mm and that the population standard deviation is 5 mm. (a) If the distribution of interpupillary distance is normal and a random sample of n = 25 adult males is to be selected, what is the probability that the sample mean distance x for these 25 will be between 63 and 67 mm? (Round all your...
Find the center of mass of two identical rod
Two identical rods each of mass (m) and length (L) are connected . Locate the centre of mass of the system.  
The distance from earth to the center of the galaxy is about 22500 ly (1 ly...
The distance from earth to the center of the galaxy is about 22500 ly (1 ly = 1 light-year = 9.47 ? 1015 m), as measured by an earth-based observer. A spaceship is to make this journey at a speed of 0.9950c. According to a clock on board the spaceship, how long will it take to make the trip? Express your answer in years (1 yr = 3.16 ? 107 s). Please be as detailed as possible in your response,...
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT