Speed of satellite in geosynchronous orbit

Click For Summary
To derive the speed of a satellite in geosynchronous orbit without using altitude or radius, one must focus on forces and uniform circular motion. The discussion highlights concerns about potentially using rotational kinematics, which is prohibited for this assignment. A participant expresses uncertainty about their solution, specifically regarding the use of formulas related to period and centripetal motion. Another contributor reassures that the approach seems valid but advises caution in confirming the absence of rotational kinematics. The emphasis remains on adhering strictly to the assignment guidelines to avoid losing credit.
glennib
Messages
1
Reaction score
0

Homework Statement


Derive an equation and solve for the speed of a satellite in geosynchronous orbit. You may NOT use the value of the satellite's altitude or radius in your calculations. If you do, you will receive no credit.

  • You should use your knowledge of forces and uniform circular motion for this assignment.
  • DO NOT use the rotational kinematics from chapter 10. If you do you will receive NO credit.
  • DO NOT attempt to apply Kepler's Laws. If you do you will receive NO credit.


Homework Equations


I have arrived at a solution, but I suspect that I have used rotational kinematics. Can someone confirm or dismiss this?


The Attempt at a Solution


Here is a screenshot from my Maple work. It is relationship1 (1) and (2) that I am worried about.
jk7lT4o.jpg


Thanks in advance.
 
Physics news on Phys.org
glennib said:
I have arrived at a solution, but I suspect that I have used rotational kinematics. Can someone confirm or dismiss this?

Here is a screenshot from my Maple work. It is relationship1 (1) and (2) that I am worried about.

If I recall, rotational kinematics is full of ω's, and such associated terms. Step 1 is the formula for a period, and step 2 is just the application of centripetal motion.

I think your work in regards to your rules/restrictions looks fine, but I strongly advise you not to take my word for it, and just double check that my suggestions are valid.
 
Actually, I just remembered that rotational velocity = v/r, and that rotational acceleration = a/r. So yeah, if that helps, use it.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
9K
Replies
2
Views
2K
Replies
5
Views
3K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 8 ·
Replies
8
Views
1K
  • · Replies 2 ·
Replies
2
Views
1K