Question

A satellite is in circular orbit around the earth. The orbit the satellite is at a height of 420 km above the earth's surface. Find the orbital speed for the satellite. mE= 5.98 x 10^24 kg rE= 6.38 x 10^6 m G=6.674 x 10^-11 N m^2/kg^2

Answer 1

7661.06 m/s

Step-by-step explanation:

R = Radius of Earth =
`6.38* 10^6\ m`

h = Distance from the Earth = 420000 m

G = Gravitational constant =
`6.674* 10^(-11) N m^2/kg^2`

M = Mass of Earth =
`5.98* 10^(24)\ kg`

`V=\sqrt{g{(R^2)/(R + h)}}\\\Rightarrow V=\sqrt{(GM)/(R^2){(R^2)/(R + h)}}\\\Rightarrow V=\sqrt{(6.674* 10^(-11)* 5.98* 10^(24))/((6.38* 10^6)^2){((6.38* 10^6)^2)/(6.38* 10^6 + 420000)}}\\\Rightarrow V=7661.06\ m/s`

The orbital speed of the satellite is 7661.06 m/s