TA 2019 Physics 1 Question 5: Geosynchronous Orbit

  • Thread starter kppc1407
  • Start date
  • Tags
    Orbits
In summary, communications satellites are placed in a circular orbit above the equator, with a radius of 6.37 x 106m and an altitude of 3.58 x 107m. To calculate the speed and acceleration of a satellite in a geosynchronous orbit, we can use the formula GMm = mv2/r2, derived from Newton's 2nd law. The units used are N, kg, and m, and the equation can be applied using the rotational speed of the Earth, \omega_e, and the known values for mass and radius.
  • #1
kppc1407
19
0

Homework Statement



Communications satellites are placed in a circular orbit where they stay directly over a fixed point on the equator as the Earth rotates. The radius of the Earth is 6.37 x 106m, and the altitude of a geosynchronous orbit is 3.58 x 107m. What are (a) the speed and (b) the magnitude of the acceleration of a satellite in a geosynchronous orbit?

Homework Equations



M (mass of the earth) = 5.98 x 1024
G (gravitational constant) = 6.67 x10-11
GMm = mv2
r2... r

The Attempt at a Solution



I understand the problem and how to get the answer. I just do not understand where the units go on the above formula and where it is derived from.
 
Physics news on Phys.org
  • #2
Use MKS (SI) units, so units used are N, kg, and m. You may need to look up the units that go with G; it should be in your textbook.

The formula is a result of Newton's 2nd law, F = ma.

Hope that helps.
 
  • #3
kppc1407 said:
I understand the problem and how to get the answer. I just do not understand where the units go on the above formula and where it is derived from.
Just elaborating on what Redbelly has said, the correct equation is better written as:

[tex]\frac{GMm}{r^2} = m\omega_{e}^2r[/tex]

where [itex]\omega_e[/itex] is the rotational (angular) speed of the earth. Since we know the rotational speed of the Earth and m cancels out, the only unknown is r.

The left side is the force of gravity on a body of mass m at a distance r from the Earth's center. The right side is mass x the (centripetal) acceleration on the body. So the equation is simply an application of F=ma, as Redbelly has stated.

AM
 
Last edited:

FAQ: TA 2019 Physics 1 Question 5: Geosynchronous Orbit

1. What is a geosynchronous orbit?

A geosynchronous orbit is an orbit around the Earth where a satellite's orbital period matches the Earth's rotational period, resulting in the satellite appearing to remain stationary in the sky as seen from a fixed location on Earth.

2. How high is a geosynchronous orbit?

A geosynchronous orbit is approximately 35,786 kilometers (22,236 miles) above the Earth's surface.

3. What are the benefits of a geosynchronous orbit?

Geosynchronous orbits are useful for telecommunications and weather satellites, as they can maintain a constant position over a specific location on Earth, providing continuous coverage and communication.

4. How are geosynchronous orbits maintained?

Geosynchronous orbits require precise calculations and adjustments to account for small changes in the Earth's gravitational pull and solar and lunar influences. This is typically done through small thruster burns to maintain the satellite's position.

5. Are there any drawbacks to geosynchronous orbits?

One potential drawback of geosynchronous orbits is the large distance from Earth's surface, which can result in a longer signal delay for communications. Additionally, there is a limited number of geosynchronous orbital slots available, so satellite operators must carefully plan and coordinate their use.

Similar threads

Replies
11
Views
2K
Replies
2
Views
8K
Replies
1
Views
2K
Replies
2
Views
10K
Replies
12
Views
11K
Replies
2
Views
3K
Back
Top