A satellite is revolving in a circular orbit at a height $h$ from the earth surface, such that $h< < R$ where $R$ is the radius of the earth. Assuming that the effect of earth's atmosphere can be neglected the minimum increase in the speed required so that the satellite could escape from the gravitational field of earth is

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Solution

Formula used:$v_0=\sqrt{Rg},v_e=\sqrt{2Rg}$

Orbital velocity of a satellite revolving in a circular orbit at a height$(h< < R),v_0=\sqrt{Rg}$
and the velocity required to escape is $v_e=\sqrt{2Rg}$
So increase in velocity,

$\Delta v=v_e−v_0=(\sqrt2-1)\sqrt{Rg}$

Final Answer: (a)

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Formula used:\(v_0=\sqrt{Rg},v_e=\sqrt{2Rg}\) Orbital velocity of a satellite revolving in a circular orbit at a height\((h< < R),v_0=\sqrt{Rg}\) and the velocity required to escape is \(v_e=\sqrt{2Rg}\) So increase in velocity, \(\Delta v=v_e−v_0=(\sqrt2-1)\sqrt{Rg}\) Final Answer: (a)

Formula used:\(v_0=\sqrt{Rg},v_e=\sqrt{2Rg}\)

Orbital velocity of a satellite revolving in a circular orbit at a height\((h< < R),v_0=\sqrt{Rg}\)
and the velocity required to escape is \(v_e=\sqrt{2Rg}\)
So increase in velocity,

\(\Delta v=v_e−v_0=(\sqrt2-1)\sqrt{Rg}\)

Final Answer: (a)