1.

Suppose the orbit of a satellite is exactly 35780 km above the earth's surface. Determine the tangential velocity of the satellite.

Answer»

Solution :GIVEN : `G=6.67xx10^(-11) Nm^2//kg^2, M=6xx10^24` kg (for earth)
R=6400 KM (for earth) = `6.4xx10^6` m, h=height of the SATELLITE above the earth's surface 35780 km.
To find: TANGENTIAL velocity (v)
Solution: R+h =6400 + 35780 = 42180x `10^3` m
`v=SQRT((G4M)/(R+h))`
`=sqrt(((6.67xx10^(-11))xx(4xx6xx10^24))/(42180xx10^3 m))`
`=sqrt((40.02xx10^13)/(42e180xx10^3 m))=sqrt((40.02xx10^10)/(42180))`
`=sqrt(0.0009487909xx10^10)=sqrt(9487909)`
=3080.245 m/s = 3.08 km/s
The tangential velocity of the satellite is 3.08 km/s.


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