Question

an orbital satillite revolves around the earth at a distance of 3400km calculate its orbital velocity and period of revolution. radius of earth is 6400km g is 9.8 ms-1

Answer 1

Distance of artificial satellite (d)= 3400 km
Radius of earth(R) = 6400 km = 6.4×10⁶ m
Radius of orbit(r) = R+d = 10,000 km = 10⁷ m
g = 9.8 m/s²
Orbital velocity, $V_o=\ ?$
Period of revolution, T = ?

(i) Orbital velocity

$V_o= \sqrt{ \frac{GM}{r} }= \sqrt{ \frac{GM}{r} \times \frac{R^2}{R^2} }= \sqrt{ \frac{GM}{R^2} \times \frac{R^2}{r} }= \sqrt{g\times \frac{R^2}{r} }$

$\Rightarrow V_o= \sqrt{9.8 \times \frac{(6.4 \times 10^6)^2}{10^7} } \\ \\\Rightarrow V_o= \sqrt{40.14 \times 10^6 } \\ \\ V_o=6.33 \times 10^3\ m/s$

(ii) Time period
$T= \frac{2 \pi r}{V_o} = \frac{2 \times 3.14 \times 10^7}{6.33 \times 10^3} =9.93 \times 10^3\ seconds$