A satellite of mass 200 kg, launched fromthe equator, is placed in Earth orbit a

Question

A satellite of mass 200 kg, launched fromthe equator, is placed in Earth orbit at a height of 100 km above the surface. (a) With a circular orbit, how long does thesatellite take to complete one orbit?
h

(b) What is the satellite's speed?
m/s

(c) What is the minimum energy input necessary to place thissatellite in orbit? Ignore air friction but include the effect ofthe planet's daily rotation.
J (a) With a circular orbit, how long does thesatellite take to complete one orbit?
h

(b) What is the satellite's speed?
m/s

(c) What is the minimum energy input necessary to place thissatellite in orbit? Ignore air friction but include the effect ofthe planet's daily rotation.
J

Explanation / Answer

a) Mass of the satellite is (m) = 200kg Mass of the earth (Me) =5.98*1024kg Radius of the earth (Re) =6.37*106m Where r =Re+h              =6.37*106m + 0.100*106m             = 6.47*106m And G is universal gravitatioanl constant =6.67*10-11N.m2/kg2 Acceleration due to gravity (g) = 9.8m/s2 The time period of the satellite is givenby                                     T =2(Re+h)3/gRe2                = 2r3/gRe2 Now substitute all the above values to get the requiredanswer b) The speed of the satellite  is                    vo =gRe2/(Re+h)                         =gRe2/r Now substitute all the above values to get the requiredanswer Now substitute all the above values to get the requiredanswer b) The speed of the satellite  is                    vo =gRe2/(Re+h)                         =gRe2/r Now substitute all the above values to get the requiredanswer c) The total enery of the orbit when it becomes a satelliteis                            E = -GMem/2r Mass of the earth (Me) =5.98*1024kg Mass of the satellite is (m) = 200kg Radius of the earth (Re) =6.37*106m Where r =Re+h              =6.37*106m + 0.100*106m             = 6.47*106m And G is universal gravitatioanl constant =6.67*10-11N.m2/kg2 Now the on the surface of the earth orbit has the potentialenergy (U1) = -GMem/Re Hence the kinetic energy (K1) to be impoted is given by                           E = K1+U1 Then the minimum energy required is given by                           K1=E -U1                                 =GMem[1/Re - 1/2r] Now substitute all the above values to get the requiredanswer                                 =GMem[1/Re - 1/2r] Now substitute all the above values to get the requiredanswer

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