# Orbital radius and speed of a satellite?

• Nov 22nd 2008, 12:43 PM
realintegerz
Orbital radius and speed of a satellite?
Radio and t.v. signals are sent from continent to continent by "bouncing" them from geosynchronous satellites. These satellites circle the Earth once every 24 hours, and so if the satellite circles eastward above the equator, it always stays over the same spot on the earth because the earth is rotating at the same rate. Weather satellites are also designed to hover in this way. What is the orbital radius for such a satellite? What is its orbital speed?

I know that the period T = 86,400 seconds/1 revolution

and I know a few centripetal acceleration equations...but I still don't know what to do here...

I tried solving

2 pi r/ T = sq. root (( G - mass of earth)/r)

• Nov 22nd 2008, 12:59 PM
skeeter
$\displaystyle m$ = satellite mass in kg
$\displaystyle M$ = earth mass in kg
$\displaystyle G$ = universal gravitational constant
$\displaystyle \omega$ = angular speed of the satellite in rad/sec
$\displaystyle T$ = orbital period in seconds
$\displaystyle r$ = orbital radius from the earth's center

$\displaystyle F_c = mr\omega^2$

$\displaystyle \frac{GMm}{r^2} = mr\omega^2$

$\displaystyle \frac{GM}{\omega^2} = r^3$

since $\displaystyle \omega = \frac{2\pi}{T}$ ...

$\displaystyle \frac{GMT^2}{4\pi^2} = r^3$

$\displaystyle \sqrt[3]{\frac{GMT^2}{4\pi^2}} = r$
• Nov 23rd 2008, 11:19 AM
realintegerz
So I substituted values into the last equation for r with the cube root

I got 4397375 m, but the answer my teacher has is 4.22 x 10^7 m...

And also...there is no specific weight for the satellite...(Thinking)
• Nov 23rd 2008, 11:25 AM
skeeter
I get $\displaystyle 4.23 \times 10^7$ meters

what values are you using?

btw ... mass of the satellite doesn't matter
• Nov 23rd 2008, 12:21 PM
realintegerz
I used

G = 6.67 x 10^-11 N m^2/kg^2
M = 5.98 x 10^24 kg
T = 86400s/1 rev

Am I wrong with T?
• Nov 23rd 2008, 12:33 PM
o_O
Those numbers should work.

$\displaystyle r = \sqrt[3]{\frac{(6.67 \times 10^{-11} \ \text{m}^3 \text{kg}^{-1} \text{s}^{-2}) (5.98 \times 10^{24} \ \text{kg})(86400 s)^2}{4\pi^2}} \approx 4.23 \times 10^{7} \ \text{m}$
• Nov 23rd 2008, 06:47 PM
realintegerz
oh i see my problem now, i didnt square the period T