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

Explanation

Height of the satellite above the
earth 's surface (h) =35780 km
`=35780xx10^(3)`
We know that :
Graviational constant (G)
`=6.67xx10^(-11)Nm^(2)//kg^(2)`
mass of earth (M)= `6xx10^(24)` kg,
redius of earth (R)= 6400 km
`=6400xx10^(3)m`
To find Tangential velocity of satellite `(v_(c))`
Formula : `v_(c)=sqrt((GM)/(R+h))`
Calculation : From formula,
` v_(c)=sqrt(((6.67xx10^(-11))xx(6xx10^(24)))/((6400+35780)xx10^(3)))`
`sqrt((40.02xx10^(13))/(42180xx10^(3)))`
` sqrt((40.02)/(42180)xx10^(10))`
`=sqrt(0.0009487909xx10^(10))`
`=sqrt(9.487909xx10^(6))`
`~~sqrt(9.5)xx10^(3)`
`=3.08xx10^(3)` m/s
3.08 km/s
The tangential velocity of the satellite is 3.08 km/ s.