For a circular orbit, gravity supplies inward force: r2GMm=rmv2. Therefore v=√rGM and T2=4π2GMr3. Satellite mass cancels. A larger orbit has a lower speed but a longer period. Radius r is measured from the planet centre, so altitude h=r− Rplanet.
A geostationary satellite must orbit in the equatorial plane, in the same direction as Earth’s rotation, with one sidereal-day period. It then appears fixed above one longitude. A near-polar orbit passes high latitudes while Earth rotates beneath it. Under the circular central-gravity model, the orbit centre must coincide with the planet centre.
Example 1
For GM=4.00×1014m3s−2 and r=1.00×107m, find circular orbital speed.
v=√rGM=√4.00×107=6.32×103ms−1.
Example 2
A circular orbit has radius 7.00×106m around a planet of radius 6.40×106m. Find altitude.
h=r−R=0.600×106m=600km.
Watch out:It must lie in an equatorial orbit; an inclined orbit changes its apparent latitude during each day.
Circular satellites and useful orbits practice | Year 12 Physics | Ascent Arcade