Motion in fields - Space Shuttle orbiting the Earth

AI Thread Summary
The discussion focuses on solving a problem related to uniform circular motion, specifically for a Space Shuttle orbiting Earth. The initial calculation provided is an energy value of 1.9 * 10^11 J. Participants are encouraged to explore the equations that govern uniform circular motion to further tackle the problem. The request emphasizes the need for clarity in understanding the physics concepts involved. Overall, the thread seeks assistance in applying relevant equations to analyze the motion of the Space Shuttle.
KrisG
Messages
1
Reaction score
0
I don;t really know how should I "tackle" this problem. The only thing I got is the answer for a) i) 1.9 * 10^11 J.

It's a problem from IB Oxford Study Guide.
http://img59.imageshack.us/img59/4986/fiz6.jpg

I would be very grateful if You could help :)
Thanks!
 
Last edited by a moderator:
Physics news on Phys.org
Start with a.ii. What equations describe uniform circular motion?
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top