Assuming that the satellite for surveying is placed in a circular orbit of radius #R# around earth, measured from centre of earth; the centripetal force acting on it is given by the relationship
#F_"cen" = ( M_"sat" v^2 ) / R# ........(1)
The Gravitational force that attracts the satellite towards earth is represented as
#F_"grav" = ( G M_"sat" M_"earth" ) / R^2# ......(2), where #G# is Universal Gravitational constant #=6.673 xx 10^-11 Nm^2kg^(-2)#.
In the event of equilibrium centripetal force and gravitational force balance each other, #=> F_"grav" = F_"cen"#
From (1) and (2)
# ( G M_"sat" M_"earth" ) / R^2 =( M_"sat" v^2 ) / R#
Observe that mass of satellite #M_"sat"# being present on both sides of the equation vanishes. Solving for velocity #v# in the orbit which is independent of the mass of satellite:
# v^2 = ( G M_"earth" ) / R#
#=>|vecv|=sqrt(( G M_"earth" ) / R)# .....(3)
We also know that for circular motion period for one rotation #T#, and angular velocity, #ω# are related as:
#ω = (2 π)/ T#
Also the speed #v# of the object traveling the circle is:
#v = (2 π r) /T = ω r# ....(4)
Inserting from (4) value of #v# in terms of #T# in (3) and solving for #T#
#(2 π R) /T=sqrt(( G M_"earth" ) / R)#
# T=(2 π R)/(sqrt(( G M_"earth" ) / R))#
#=> T=2pisqrt(( R^3)/( G M_"earth" )# .....(5)
Equation (5) is Newton's form of Kepler's third law.
We see that #R=R_e+h#,
where #R_e# is the radius of earth #=6.371xx10^6m# and #h# is the height of satellite from earth's surface. #M_"earth"=5.972 xx 10^24 kg#
Inserting given values in (5)
#T=2pisqrt(( (6.371xx10^6+1.50xx10^5)^3)/( 6.673 xx 10^-11xx 5.972 xx 10^24 )#
#T=5241.2s# rounded to one decimal place.
