Radious of the geosynchronous ORBIT

  • Thread starter Thread starter 8parks11
  • Start date Start date
  • Tags Tags
    Orbit
AI Thread Summary
The radius of a geosynchronous orbit is determined by the balance between gravitational and centripetal forces acting on the satellite. A satellite in this orbit must maintain a specific angular velocity to remain above a fixed point on Earth, which is influenced by the Earth's rotation period of approximately 24 hours. While the radius can vary, the crucial factor is ensuring the satellite's orbital speed matches the Earth's rotational speed. The gravitational force acting on the satellite is inversely proportional to the square of its distance from the Earth's center. Thus, while the radius can be adjusted, the satellite's ability to remain geosynchronous is fundamentally linked to its orbital dynamics.
8parks11
Messages
63
Reaction score
0
a)

What is the radius of the orbit of a communications relay satellite that always reamins above one point on the Earth's surface? Such an orbit is called a geosynchronous orbit.


- I could find this on google but what is an another way to solve this?


b) Can such a satellite be placed in geosynchronous orbit over ANY point on Earth's surface? WHY?
 
Physics news on Phys.org
You can find it out by calculating it really. Are you sure that is the only important point? Radius is simply the height from the centre of the Earth and you can have almost any radius you want. What is important is angular velocity.
(Hint: How long does it take fo the Earth to make 1 full rotation?)

b) Yes and no really. The point is not as important as the actual orbit itself.
 
We assume that the satellite is in a circular orbit. This means that it is experiencing a centripetal force F_C supplied by the gravitational attractive force F_G of the earth

F_C = F_G

the centripetal force is given by

F_C = m\frac{v^2}{r}

with the mass and orbital speed of the satellite. r is its distance from the centre of the earth. The gravitational attractive force is given by

F_G = G \frac{Mm}{r^2}
 
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 'Minimum mass of a block'
Here we know that if block B is going to move up or just be at the verge of moving up ##Mg \sin \theta ## will act downwards and maximum static friction will act downwards ## \mu Mg \cos \theta ## Now what im confused by is how will we know " how quickly" block B reaches its maximum static friction value without any numbers, the suggested solution says that when block A is at its maximum extension, then block B will start to move up but with a certain set of values couldn't block A reach...
Thread 'Calculation of Tensile Forces in Piston-Type Water-Lifting Devices at Elevated Locations'
Figure 1 Overall Structure Diagram Figure 2: Top view of the piston when it is cylindrical A circular opening is created at a height of 5 meters above the water surface. Inside this opening is a sleeve-type piston with a cross-sectional area of 1 square meter. The piston is pulled to the right at a constant speed. The pulling force is(Figure 2): F = ρshg = 1000 × 1 × 5 × 10 = 50,000 N. Figure 3: Modifying the structure to incorporate a fixed internal piston When I modify the piston...
Back
Top