EXAM APLE 6.6.2 | The altitude of a near circular, low earth orbit (LEO) satellite is about 200 miles. (a) Calculate the period of this satellite. Solution: For circular orbits, we have (GM_(E)m)/(R^2)=m(v^2)/(R)=m(4pi ^2R^2/z^2)/(R) Solving for t tau ^2=(4pi ^2)/(GM_(E))R^3 6.6 Kepler's Third Law: The Harmonic Law 241 which-no surprise-is Kepler's third law for objects in orbit about the Earth. Let R=R_(E)+h where his the altitude of th e satellite above the Earth's surface. Then x^2=(4pi ^2)/(GM_(E))R_(E)^3(1+(h)/(R_(E)))^3 But GM_(E)/R_(E)^2=g so we have z=2pi sqrt ((R_(E))/(g))(1+(h)/(R_(E)))^3/2approx 2pi sqrt ((R_(E))/(g))(1+(3h)/(2R_(E))) Putting in numbers R_(E)=6371km,h=322km we get tau approx 90.8minapprox 1.51 hr. There is another way to do this if you realize that Kepler'; third law, being a deriv- ative of Newton's laws of motion and his law of gravitation , applies to any set of bodies in orbit about another. The Moon orbits the Earth once every 27.3 days at a radius 60.3R_(E) . Thus, scaling Kepler's third law to these values (1 month=27.3 days a nd 1 lunar unit (LU)=60.3R_(E)) we have tau ^2(months)=R^3(LU) Thus, for our LEO satellite R=(6693)/(6371)R_(E)=1.051R_(E)=(1.051R_(E))/(60.3R_(E)IL)=0.01743LU t(mondhs)=R^2a[LU]=(0.01743)^122months=0.002801months=1.51ln (b) A geosynchronous satellite orbits the Earth in its equatorial plane with a period of 24 hr.Thus,it seems to hover above a fixed point on the ground (which is why you can point your TV satellite receiver dish towards a fixed direction in the sky). What is the radius of its orbit? Solution: Using Kepler'; third law again,we get R_(gvo)=x^203=((1)/(273))^2/3=0.110LU=6.65R_(E)approx 42,400km