Question

The orbital velocity of an artificial satellite in a circular orbit
just above the centre’s surface is u. For a satellite orbiting
at an altitude of half of the earth’s radius, the orbital velocity is
(a) (√(2/3))v₀
(b) 2/3 v₀
(c) 3/2v₀
(d) √(3/2)v₀

Answer 1

The expression for orbital velocity is,

$\longrightarrow\sf{v_o=\sqrt{\dfrac{GM}{R+h}}}$

where $\sf{h}$ is the altitude of the satellite from the surface of the earth.

Since $\sf{G}$ and $\sf{M}$ are constants,

$\longrightarrow\sf{v_o\propto\dfrac{1}{\sqrt{R+h}}}$

Therefore,

$\longrightarrow\sf{\dfrac{(v_o)_2}{(v_o)_1}=\sqrt{\dfrac{R+h_1}{R+h_2}}}$

$\longrightarrow\sf{(v_o)_2=(v_o)_1\sqrt{\dfrac{R+h_1}{R+h_2}}}$

According to the question,

• $\sf{(v_o)_1=u}$

• $\sf{h_1=0}$

• $\sf{h_2=\dfrac{R}{2}}$

Hence,

$\longrightarrow\sf{(v_o)_2=u\sqrt{\dfrac{R+0}{R+\dfrac{R}{2}}}}$

$\longrightarrow\sf{\underline{\underline{(v_o)_2=u\sqrt{\dfrac{2}{3}}}}}$

Hence (a) is the answer.