Tell me more ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.
up vote 2 down vote favorite

Could you please tell, Why all the geo-stationary satellites are to be dropped at same height from earth? Why can't it be closer or away from its regular orbit(ie, 35,000 km)? If all satellites are dropped in the same orbit, then will not those collide one another?

share|improve this question
3  
Also, if they're all geo-stationary... then how will they collide? – Flimzy Oct 21 '11 at 5:42
ya, there is a space crunch der in geosynch orbit. but they won't collide because its like all moving at same speed!!! Two cars will never collide if they are traveling along a straight single lane road in the same direction at exactly same velocity!!! – Vineet Menon Oct 21 '11 at 6:50
4  
Wikipedia already has a perfect explanation for this... – oezi Oct 21 '11 at 8:46
The length of the equatorial geostationary orbit is 35mm (mega-meters) times two pi =~ 220mm long, so there's lots of room. That plus (as Vineet says) they're all going the same speed in the same direction. – Mike Dunlavey Oct 21 '11 at 14:08
@MikeDunlavey Think of editing your comment as m is milli and M is Mega. Hence, 35mm is 35 millimeters, not mega meters ;-) – ChrisR Oct 22 '11 at 23:53

migrated from skeptics.stackexchange.com Oct 21 '11 at 6:12

2 Answers

up vote 4 down vote

About the collision question; By definition, a geostationary satellite has a frequency of rotation equal to earth's frequency of rotation. That means it's a specific angular velocity. If two satellites orbit with the same angular velocity they will always maintain the same distance.

About the radius for geostationary satellites; The velocity of the satellite is a function of the radius. So in order to have a specific period you need a specific radius.

From Newton's second law, we have that the centripetal acceleration of the satellite is equal to the gravitational force $m\vec{a_c}=\frac{GMm}{r^2}\hat{r}$.

$|\vec{a_c}|=\frac{GM}{r^2}|\hat{r}|$

$\omega^2 r=\frac{GM}{r^2}$

$r=(\frac{GM}{\omega ^2})^{1/3}$

If you put the frequency equivalent to a geostationary orbit into this equation, you'll have the appropriate radius.

share|improve this answer
2  
Geostationary also presumes an orbital plane through the equator. Otherwise, your conclusion about equal distance wouldn't hold. E.g. GPS uses 4 orbital planes, and therefore its satellites don't keep the same distance, but all are in a 12 hour orbit. – MSalters Oct 21 '11 at 12:53
3  
GPS satellites are not geostationary. – Colin K Oct 21 '11 at 13:37
@MSalters you're right about the assumption. I'll add that. – Diego Oct 22 '11 at 2:02
@ColinK: That's rather obvious from my comment, isn't it? The geostationary orbit has a single plane, not 4, and it's a 24 hour orbit, not 12. I'm poiting out a difference between geostationary satellites (which keep fixed distances amongst themselves) and other satellites (which may have fluctuating distances despite fixed orbital speeds) – MSalters Oct 24 '11 at 11:10
up vote 4 down vote

Geostationary satellites are less likely to collide since they are all moving in the same direction. But they aren't all perfectly positioned and so do drift - even 'perfectly' positioned ones do drift in a figure of 8 around their intended point.

But if there ever is a collision, or an explosion, the debris is going to stay there essentially for ever - unlike stuff in LEO which will re-enter the atmosphere. And you don't have an alternative geostationary to choose from. If there is ever a large cloud of debris in the GSO slot for N. America then you are all going to have to switch to cable.

On the other hand GSO is become less important as more and more comms switches to fibre.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.