Assuming that the satellite for surveying is placed in a circular orbit of radius RR around earth, measured from centre of earth; the centripetal force acting on it is given by the relationship
F_"cen" = ( M_"sat" v^2 ) / RFcen=Msatv2R ........(1)
The Gravitational force that attracts the satellite towards earth is represented as
F_"grav" = ( G M_"sat" M_"earth" ) / R^2Fgrav=GMsatMearthR2 ......(2), where GG is Universal Gravitational constant =6.673 xx 10^-11 Nm^2kg^(-2)=6.673×10−11Nm2kg−2.
In the event of equilibrium centripetal force and gravitational force balance each other, => F_"grav" = F_"cen"⇒Fgrav=Fcen
From (1) and (2)
( G M_"sat" M_"earth" ) / R^2 =( M_"sat" v^2 ) / RGMsatMearthR2=Msatv2R
Observe that mass of satellite M_"sat"Msat being present on both sides of the equation vanishes. Solving for velocity vv in the orbit which is independent of the mass of satellite:
v^2 = ( G M_"earth" ) / Rv2=GMearthR
=>|vecv|=sqrt(( G M_"earth" ) / R)⇒∣∣→v∣∣=√GMearthR .....(3)
We also know that for circular motion period for one rotation TT, 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.
