A geostationary satellite is located directly above the Earth\'s equator and orb

Question

A geostationary satellite is located directly above the Earth's equator and orbits at the same rate as the Earth rotates. So the satellite stays over the same spot on the Earth. This is important for communications. Calculate the altitude of a geostationary satellite above the surface of the Earth. (Do not use Kepler's third law directly. You should start with the fact that the period is 24 hours. Calculate the orbital period from that. Then determine the distance from the center of the Earth.) An 800 kg satellite is placed in a circular orbit at an altitude of 300 km above the Earth's surface. What is What is the orbital speed of the satellite? What is the kinetic energy of the satellite? What is the gravitational potential energy of the satellite? A rocket engine raises the satellite from this orbit to a geostationary orbit. How much energy does the engine have to provide?

Explanation / Answer

a) The period of a satellite is the time it takes it to make one full orbit around an object. The period of the Earth as it travels around the sun is one year. If you know the satellite’s speed and the radius at which it orbits, you can figure out its period.

You can calculate the speed of a satellite around an object using the equation

v=sqrt(GM/r) ................M = mass of earth

The satellite travels around the entire circumference of the circle — which is

2r

if r is the radius of the orbit — in the period, T. This means the orbital speed must be

v=2r/T

giving you

sqrt(GM/r) = 2r/T

If you solve this for the period of the satellite, you get

T = 2*sqrt[r^3/(GM)-----------(1)

T is given as 24hr. Put this in above equation and calculate orbital radius r , ‘

Height h = r – R   ...............R= radius of earth

b) h = 300km = 300*10^3 km

w= 1rev/24hr = (2/(24*60*60)) = 7.3*10^-5 rad/s

w=v/r = v/(R+h)     => v= w*(R+h) = 7.3*10^-5*(6.378*10^6*300*10^3) = 1.4*10^8 m/s

c) KE = 1/2mv^2 = ½*800*(1.4*10^8)^2 = 7.8*10^18 J

d) PE = -GMm/r = - (6.67*10^-11*5.9*10^24*800)/( 6.378*10^6*300*10^3)^2 = -8.6*10^-8 J

e) Use law of conservation of energy

PEi+KEi = PEf + KEf

KEf – KEi = PEf – PEi = -GMm/rf - (-GMm/ri) = GMm(1/ri – 1/rf) = GMm(1/h– 1/R) = (6.67*10^-11*5.9*10^24*800)*(1/(300*10^3) – 1/(6.378*10^6*300*10^3)) = 1.05*10^12 J

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