Question

Q12. An artificial satellite is getting around the earth at a distance of
1,600 km. Calculate the period of revolution and orbital velocity.
Given radius of the earth = 6,400 Km and g = 9.8ms-2
​

Answer 1

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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=\ ?Vo= ?

Period of revolution, T = ?

(i) Orbital velocity

V_o= $\sqrt{ \frac{GM}{r} }= \sqrt{ \frac{GM}{r} \times \frac{R^2}{R^2} }= \sqrt{ \frac{GM}{R^2} \times \frac{R^2}{r} }= \sqrt{g\times \frac{R^2}{r} }Vo =rGM = rGM ×R 2R2 = R 2GM × rR 2 = g× rR 2$

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

⇒Vo= 9.8×10^7(6.4×10^6)2

⇒Vo = 40.14×10^6Vo = 6.33×10^3 m/s

(ii) Time period

T= $\frac{2 \pi r}{V_o} = \frac{2 \times 3.14 \times 10^7}{6.33 \times 10^3} =\boxed{9.93 \times 10^3\ seconds}T= Vo2πr = 6.33×10^32×3.14×10^7 = 9.93×10^3seconds$

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