Solving Satellite Orbit Problem: Finding Time in Hours

In summary, the conversation discusses the problem of determining the time it takes for a communication satellite placed by NASA in Earth's orbit with a radius of 42000 km to make one complete orbit. The equations used to solve this problem are V(speed)=2πr(radius)÷T(period), Ac=V(speed)^2÷r(radius), and Ac(centripetal acceleration)=4π^2r÷T^2. The issue lies in using the incorrect units for the radius, which should be in meters instead of kilometers. The correct answer is 24 hours.
  • #1
Giu1iano
18
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
  • #2
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.
 
  • #3
I converted the m/s^2 to km/h/s. Haven't tried converting the radius to m.
 
  • #4
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
 

1. How do you calculate the time in hours for a satellite's orbit?

To calculate the time in hours for a satellite's orbit, you need to know the satellite's orbital period, which is the time it takes for the satellite to complete one full orbit around the Earth. The formula for calculating orbital period is T = 2π√(a^3/GM), where T is the orbital period, a is the semi-major axis of the orbit, G is the gravitational constant, and M is the mass of the Earth. Once you have the orbital period, you can convert it to hours by multiplying it by 24.

2. What is the semi-major axis of a satellite's orbit?

The semi-major axis of a satellite's orbit is the longest radius of the elliptical orbit. It is half of the major axis, which is the longest diameter of the ellipse. In simpler terms, it is the distance from the center of the orbit to the farthest point of the orbit.

3. Can you use the same formula to calculate the time in hours for any satellite's orbit?

Yes, the formula T = 2π√(a^3/GM) can be used to calculate the time in hours for any satellite's orbit. The only variables that will change are the semi-major axis and the mass of the Earth, which are specific to each satellite's orbit.

4. What is the difference between a geostationary and a polar orbit?

A geostationary orbit is when a satellite orbits the Earth in the same direction and at the same speed as the Earth's rotation. This results in the satellite appearing to remain stationary in the sky. A polar orbit, on the other hand, is when a satellite orbits the Earth from pole to pole, passing over different parts of the Earth on each orbit. The orbital period and semi-major axis will be different for these two types of orbits.

5. How does the mass of the Earth affect a satellite's orbit?

The mass of the Earth affects a satellite's orbit through the gravitational force it exerts on the satellite. The more massive the Earth, the stronger the gravitational force, and the faster the satellite will need to orbit to maintain its position. This is why the formula for orbital period includes the mass of the Earth. Changes in the Earth's mass, such as due to the presence of other objects like the Moon, can also affect a satellite's orbit.

Similar threads

  • Introductory Physics Homework Help
Replies
7
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
2K
  • Introductory Physics Homework Help
Replies
28
Views
340
  • Introductory Physics Homework Help
Replies
1
Views
830
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
11
Views
1K
  • Introductory Physics Homework Help
Replies
6
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
977
  • Introductory Physics Homework Help
Replies
2
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
1K
Back
Top