# Geostationary Orbit Question.

1. Nov 9, 2007

### Oliviam12

1. The problem statement, all variables and given/known data
What is the orbital speed of a satellite of mass 490 kg in geostationary orbit?

2. Relevant equations
Fg= G m1m2/r

3. The attempt at a solution

I don't really know where to begin... I know the period is 24 hours because it is geostationary, and the height is 35880, because that is the altitude that they orbit.... What am I suppost to do now?

2. Nov 9, 2007

### Natique

Well to start you want to find Fg. m1 is given, and m2 is the mass of the earth, so you can substitute all that into the equation and you'll get Fg. Btw isn't the rule Fg = Gm1m2/r^2? I might be wrong on the rule, it's been a long time since I took this stuff but you might wanna double check it

Last edited: Nov 9, 2007
3. Nov 9, 2007

### mgb_phys

In orbit the gravity force inwards balances the centrifugal force outwards

F_gravity = GMm/r^2 , where M = mass earth m=mass satelite
F_out = m w^2 r , where w = angular velocity in rad/s , m is the mass of the satelite

so r^3 = G M / w^2 and you have the height.
Remember that r is from the centre of the earth so you have to subtract the radius of the earth to get height above ground.

The angular velocity has to be one full circle in 24 hours, there are 2pi radians in a circle.
w = 2 pi / (24 * 60 * 60 )

Notice that the mass of the satelite cancels - a general feature of orbits is that as long as the body is much less massive than the thing it is orbiting, it's own mass doesn't matter.

Last edited: Nov 9, 2007
4. Nov 7, 2008

### JumpinJohny

I don't understand how you arrived at the answer mgb_phys.

5. Nov 7, 2008

### mgb_phys

Which bit?

Orbit is like swinging a weight on a string.
The outward force depends on the lenght of the string (ie radius of orbit) and the speed (rev/s). The inward force is gravity which depends on the mass of the central body and the radius. In orbit these two forces are equal.

The equations are F_gravity = GMm/r^2 and F_out = m w^2 r
So just set these two equal to each other and do a bit of rearranging.
w is really$$\varpi$$ which is rotation rate in radians/sec.

6. Feb 1, 2009

### AROD

angular velocity of the satellite and the earth will always be the same, w = 2pi/(time in a day)

orbital velocity must mean the angular velocity x the radius, [w x (height above earth + radius of earth)]

the satellite will follow a certain point on the earth at this point.
my question is:

how precisely must the height be stabilized (+/- how much) to keep the relative change between the satellite's position with respect to a certain point on the earth surface less than 100 m per day?

7. Jun 11, 2010

### gmlobdell

If the altitude is 35880 km, then the orbit is a circle with radius 35880 + 6387 km (radius of the earth). The geostationary satellite travels that circle in 23h 56m (google "sidereal day" for an explanation of why it's not 24 hours).

The circumference of the circle (orbit) is 2*pi*r = 2 * 3.14159 * 42267 = 265571.4 km

The length of the day, in seconds, = 23 * 3600 + 56 * 60 = 86160 seconds

Speed of the satellite = distance / time = 265571.4 km / 86160 sec = 3.082 km/sec = 11096.3 km/hr = 6894.9 m/hr

To determine the altitude may take the fancy math, but given the altitude, the speed is simple.