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

Answer:

7661.06 m/s

Explanation:

R = Radius of Earth = $6.38\times 10^6\ m$

h = Distance from the Earth = 420000 m

G = Gravitational constant = $6.674\times 10^{-11} N m^2/kg^2$

M = Mass of Earth = $5.98\times 10^{24}\ kg$

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

The orbital speed of the satellite is 7661.06 m/s