What Is the Angle Between Velocity Vector and Radius-Vector of Part A at 4RE?

  • Thread starter Thread starter assaftolko
  • Start date Start date
  • Tags Tags
    Gravity
AI Thread Summary
The discussion centers on the motion of a body in circular orbit around Earth, which splits into two parts, A and B, with A moving at a 34-degree angle. The key question is determining the angle between the velocity vector and the radius-vector of part A when it reaches a distance of 4RE from Earth's center. Participants emphasize the importance of angular momentum conservation, noting that gravity does not produce torque on an orbiting mass when considering Earth as a point mass. They clarify that while A detaches, it remains influenced by Earth's gravitational field, which dominates its motion. The conversation concludes with recommendations for resources on angular momentum in astrodynamics.
assaftolko
Messages
171
Reaction score
0
A body moves in circular motion around the Earth with orbit radius of 3RE
At a certain time the body detaches to 2 identical parts, each one with a mass of m: A and B. A moves in an angle of 34 deg and B moves straight to the center of the earth.

What is the angle between the velocity vector and the radius-vector of part A when it gets to a distance of 4RE from the center of the earth?

I think I'm suppose to use angular momentum conservation with respect to the center of the Earth for part A and calculate it's angular momentum just after the detachment and when it's at 4RE. The problem is that I don't really know how can I justify angular momentum conservation... I think that gravity produces torque and so this is a problem...
 

Attachments

  • Ex 8.jpg
    Ex 8.jpg
    18.1 KB · Views: 386
Physics news on Phys.org
Gravity produced by a point source will not produce a torque on an orbiting point mass. Assuming that the Earth is taken to be a spherically symmetric distribution of mass, it will produce the same field as a point mass located at its center. Angular momentum is always conserved.
 
gneill said:
Gravity produced by a point source will not produce a torque on an orbiting point mass. Assuming that the Earth is taken to be a spherically symmetric distribution of mass, it will produce the same field as a point mass located at its center. Angular momentum is always conserved.

I'm sorry I got confused for a second... r and the gravity force lay on the same line for every moment so of course gravity doesn't produce torque... thanks
 
assaftolko said:
1. After the detachment - can you still say A is "orbiting" something?
2. Can you reffer me to a source that shows why this is true for the angular momentum?

1. Assuming that the mass of the Earth is much greater than that of A, then Earth's field will dominate the motion of A with respect to the Earth. Whether or not the orbit is closed is another matter (check the velocity of A versus escape velocity at R = 3Re).

2. Angular momentum is ALWAYS conserved. For orbiting objects its a constant of the motion. For a proof, any text on astrodynamics should have a derivation of the equation of motion for the two-body system. One of my favorites is "Fundamentals of Astrodynamics" by Bate, Mueller, and White (very inexpensive yet amazingly good).
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Back
Top