Solving Satellite Orbit Problem: Finding Time in Hours

Click For Summary
The discussion revolves around calculating the orbital period of a satellite at a radius of 42,000 km with a centripetal acceleration of 0.22 m/s². The key equations used include the relationships between speed, radius, and period, specifically Ac = 4π²r/T². A common error identified is the incorrect unit conversion of the radius from kilometers to meters, which should be 4.2 x 10^6 m. After correcting the calculations, the period is determined to be approximately 24 hours. Accurate unit conversion is crucial for solving such physics problems effectively.
Giu1iano
Messages
18
Reaction score
0
Member advised to use the homework template for posts in the homework sections of PF.
The problem is as follows. "NASA places communication satellites into Earth orbit with a radius of 42000 km. If the centripetal acceleration of one of the satellite is 0.22m/s^2, how long, in hours, will it take this satellite to make one complete orbit?

For this chapter, these are the following equations I can use. V(speed)=2πr(radius)÷T(period), Ac=V(speed)^2÷r(radius), and Ac(centripetal acceleration)=4π^2r÷T^2I used Ac=4π^2r÷T^2. Now I'm stumped because the attributes are straight forwarded, so when insert 0.22m/s=4π^2*4.2e4÷T^2 I know I need to rearrange the problem.

I get T^2=4π^2r÷Ac. The answer is 24 hours and I don't come close to it. I just want to know if I'm setting up the problem correctly? If not, what am I doing wrong?

Thanks! :)
 
Last edited:
Physics news on Phys.org
Your problem is that you used r in km when it should be in m because the acceleration is given in m/s^2.
r= 4200 km which is 4200000 m. That is 4.2 x 10^6, NOT 4.2 x 10^4 as you have.
 
I converted the m/s^2 to km/h/s. Haven't tried converting the radius to m.
 
HallsofIvy said:
Your problem is that you used r in km when it should be in m because the acceleration is given in m/s^2.
r= 4200 km which is 4200000 m. That is 4.2 x 10^6, NOT 4.2 x 10^4 as you have.
Either way works. My problem is that I didn't convert my time properly lol I got 24 hrs.

T^2=4π^2*42000km÷0.792km/h/s, T^2=2093636, T=√2093636, T=1446.94 min÷60 T=24.11 or 24 hours
 
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 7 ·
Replies
7
Views
2K
Replies
3
Views
3K
Replies
2
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
Replies
1
Views
1K
  • · Replies 12 ·
Replies
12
Views
5K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 30 ·
2
Replies
30
Views
2K