V=√(GM/(R+h) at any direction?

  • Thread starter luckis11
  • Start date
  • Tags
    Direction
In summary, an artificial satellite set at geostationary orbit must have a speed of v=√(GM/(R+h)). It is possible for it to have the opposite direction of motion compared to traditional geostationary satellites, but it will not be geostationary. The air resistance at different heights must also be taken into account when determining the initial speed for an orbit. Rocket scientists carefully consider these factors and may use booster rockets to adjust the satellite's orbit.
  • #1
luckis11
272
2
Helpful notes:
The speed v=√(GM/(R+h)) which is required for an artificial satellite to be set at orbit, is one that its Δx is drawn on the reference frame (call it FR2) that does not move together with the self-rotating movement of the earth. Because whereas it has the speed v=√(GM/(R+h)), the Δx of its speed which is drawn on the reference frame (call it FR1) that moves together with the self-rotating movement of the earth, is zero.
The geostationary orbit can only happen at the height of ~35,000km above the ground (see http://imagine.gsfc.nasa.gov/docs/ask_astro/answers/970408d.html).

My question is:
Is it possible to set an artificial satellite at an orbit at the geostαtionary height (35,000), and the direction of its motion (its motion Δx that is drawn on the FR2) to be the opposite of the geostationary satellites? (It IS possible as it seems at the moving drawing at http://en.wikipedia.org/wiki/Satellite). If-since it is possible, then the speed that it must have is again v=√(GM/(R+h))?
 
Last edited:
Astronomy news on Phys.org
  • #2
luckis11 said:
Is it possible to set an artificial satellite at an orbit at the geostαtionary height (35,000), and the direction of its motion (its motion Δx that is drawn on the FR2) to be the opposite of the geostationary satellites?

Of course. But it won't be geostationary, since it's going in the wrong direction.
 
  • #3
The answer I ended up with so far, is:
The speed required should or might be v=√(GM/(R+h) for both cases of a satellite which moves at the same direction as the self-rotation of the earth, and of a sattellite moving at the opposite direction. But providing initial speed v=√(GM/(R+h) alone, cannot result in an orbit for both cases, because the air resistence those two meet, is not the same. And that air resistence difference is not small. The motion of the air drawn on FR2 at sea level is a wind of 465metres/sec=1,674km/hour. Same speed drawn on FR2=>different speed drawn on FR1=> different speed in relation to the air. Now, at the height of the geostationaries 35,000km, the air or eather resistence might be almost zero, but the v=√(GM/(R+h) refers to all heights.
 
Last edited:
  • #4
Huh? Rocket scientists do not overlook air resistance in satellite launches. They also generally include booster rockets on the payload to nudge satellites into the desired orbit.
 
  • #5


I would like to clarify a few points about orbital mechanics and the specific scenario mentioned in the content. Firstly, the equation v=√(GM/(R+h)) represents the speed required for a satellite to maintain a circular orbit at a given altitude (h) above the surface of a central body (such as the Earth), where G is the gravitational constant and M is the mass of the central body. This equation holds true for any direction of motion, as long as the satellite remains in a circular orbit.

Regarding the reference frames mentioned (FR1 and FR2), it is important to note that the speed of a satellite in orbit is always measured relative to its central body. In this case, the speed v=√(GM/(R+h)) is measured from the perspective of the Earth, which is the central body. Therefore, the speed of the satellite will be the same in both reference frames, as it is measured relative to the Earth.

Furthermore, the geostationary orbit is a specific type of circular orbit located at an altitude of approximately 35,000km above the Earth's surface. This orbit is unique because it has a period of 24 hours, which allows a satellite to remain fixed above a specific location on Earth. It is not possible for a satellite to have an opposite direction of motion in a geostationary orbit, as this would require it to have a period of less than 24 hours.

Finally, I would like to address the question of whether it is possible to have a satellite in a geostationary orbit with a direction of motion opposite to other geostationary satellites. The answer is no, it is not possible. All satellites in a geostationary orbit must have the same direction of motion in order to maintain their fixed position above the Earth. If a satellite were to have an opposite direction of motion, it would eventually drift away from its designated location.

In conclusion, the equation v=√(GM/(R+h)) applies to any satellite in orbit, regardless of its direction of motion. The geostationary orbit is a special type of orbit with a specific altitude and period, and it is not possible for a satellite to have an opposite direction of motion in this orbit.
 

1. What does the equation V=√(GM/(R+h) represent?

The equation V=√(GM/(R+h) represents the velocity required for an object to maintain a circular orbit at a certain altitude (h) above the surface of a planet with a given mass (M) and radius (R).

2. How is this equation derived?

This equation is derived from the law of universal gravitation and Newton's second law of motion. It takes into account the gravitational force between two objects and the centripetal force required for an object to maintain a circular orbit.

3. What is the significance of the variables in this equation?

The variable V represents the orbital velocity, G is the universal gravitational constant, M is the mass of the planet, and R+h is the distance between the center of the planet and the object's orbit.

4. Can this equation be applied to any object in space?

Yes, this equation can be applied to any object in space as long as the object is orbiting a central body with a known mass and radius. It is commonly used to calculate the orbital velocity of spacecraft and satellites.

5. What is the significance of the direction component in this equation?

The direction component in this equation represents the direction of the object's velocity vector in relation to the central body. This allows the equation to be used for objects in both prograde (moving in the same direction as the planet's rotation) and retrograde (moving in the opposite direction of the planet's rotation) orbits.

Similar threads

  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
19
Views
723
Replies
10
Views
8K
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
848
  • Introductory Physics Homework Help
Replies
12
Views
11K
  • Introductory Physics Homework Help
Replies
7
Views
2K
  • Introductory Physics Homework Help
Replies
7
Views
3K
  • Astronomy and Astrophysics
Replies
4
Views
5K
  • Special and General Relativity
2
Replies
42
Views
4K
Back
Top