A question about geosynchronous orbit

  • Thread starter Thread starter Eriance
  • Start date Start date
  • Tags Tags
    Orbit
AI Thread Summary
An object in geosynchronous orbit that thrusts straight down will not land directly beneath its original position due to Earth's rotation. As the satellite descends, it retains its horizontal orbital velocity, causing it to land east of the initial point. Significant thrust would be required to alter its trajectory, and without additional maneuvers, it would enter an elliptical orbit before potentially crashing. The Coriolis effect also plays a role in this deviation from the expected landing point. Thus, careful calculations and thrust applications are essential for precise landings from geosynchronous orbit.
Eriance
Messages
6
Reaction score
0
I a question regarding an object, that is originally in geosynchronous orbit, but then thrusted straight down. Will the object land directly on where it was hovering over while in the geosynchronous orbit? Or will it land off target because, as it falls, it's relative position to the Earth is no longer synchronized (earth rotates)? I'm having a hard time visualizing this.
 
Physics news on Phys.org
Welcome to PF!

Earth rotates...and the satellite is also in orbit, revolving around the earth. So no, you can't land by thrusting straight toward Earth.
 
Eriance said:
I a question regarding an object, that is originally in geosynchronous orbit, but then thrusted straight down.
This would require an huge amount of fuel (energy) to accomplish, but since you specifically stated "thrusted straight down", I'm assuming that the thrust is applied in the amount and direction to keep the descending satellite directly above the same point on Earth as the Earth rotates.

If instead a burst of thrust was applied perpendicular to the direction of gravity in order to slow down the satellite, it would enter an elliptical orbit. At least a second burst (again to decrease velocity) would be required to establish a lower circular orbit (at which point the satellite would orbit the Earth in less than 24 hours. Low Earth orbits take about 1 1/2 hours).

http://en.wikipedia.org/wiki/Hohmann_transfer_orbit
 
Ah thanks, for some reason, I thought with an initial thrust the satellite will just drop out of orbit and crash into the planet. Thanks for the feedback.
 
Eriance said:
Ah thanks, for some reason, I thought with an initial thrust the satellite will just drop out of orbit and crash into the planet.
It could if so much velocity was removed that the resulting elliptical path was so narrow that the satellite would crash into the Earth instead of orbiting around it.
 
Let us assume that "geosynchronous" means a geostationary equatorial orbit in this case.

If "thrusted straight down" is taken to mean that the satellite fired its thrusters in a vertical direction then the satellite would surely miss the Earth -- or at least miss the point directly beneath it.

The satellite's orbit is about 22,000 miles above sea level. Add in the radius of the Earth and that's about 26,000 miles total. The circumference is two pi times that for something like 160,000 miles. It completes one orbit in just a little under 24 hours. So that's around 7,000 miles per hour.

A similar calculation shows that a point on the surface of the Earth at the equator is moving in a circular path at a little over 1,000 miles per hour.

If you fire your thusters, pushing the satellite straight down, it will retain its full orbital velocity in the horizontal direction. If you fire hard enough, it will crash into the earth. But that original horizontal velocity will mean that it will crash somewhere to the east of the point directly under where it started.

In the rotating frame of reference in which both Earth and satellite start out stationary, this is seen to be due to the Coriolis force.
 
Thread 'Gauss' law seems to imply instantaneous electric field propagation'
Imagine a charged sphere at the origin connected through an open switch to a vertical grounded wire. We wish to find an expression for the horizontal component of the electric field at a distance ##\mathbf{r}## from the sphere as it discharges. By using the Lorenz gauge condition: $$\nabla \cdot \mathbf{A} + \frac{1}{c^2}\frac{\partial \phi}{\partial t}=0\tag{1}$$ we find the following retarded solutions to the Maxwell equations If we assume that...
Thread 'A scenario of non-uniform circular motion'
(All the needed diagrams are posted below) My friend came up with the following scenario. Imagine a fixed point and a perfectly rigid rod of a certain length extending radially outwards from this fixed point(it is attached to the fixed point). To the free end of the fixed rod, an object is present and it is capable of changing it's speed(by thruster say or any convenient method. And ignore any resistance). It starts with a certain speed but say it's speed continuously increases as it goes...
Maxwell’s equations imply the following wave equation for the electric field $$\nabla^2\mathbf{E}-\frac{1}{c^2}\frac{\partial^2\mathbf{E}}{\partial t^2} = \frac{1}{\varepsilon_0}\nabla\rho+\mu_0\frac{\partial\mathbf J}{\partial t}.\tag{1}$$ I wonder if eqn.##(1)## can be split into the following transverse part $$\nabla^2\mathbf{E}_T-\frac{1}{c^2}\frac{\partial^2\mathbf{E}_T}{\partial t^2} = \mu_0\frac{\partial\mathbf{J}_T}{\partial t}\tag{2}$$ and longitudinal part...
Back
Top