JEE MAIN - Physics (2019 - 11th January Morning Slot - No. 25)
A satellite is revolving in a circular orbit at a height h form the earth surface, such that h < < R where R is 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 :
$$\sqrt {gR} \left( {\sqrt 2 - 1} \right)$$
$$\sqrt {2gR} $$
$$\sqrt {gR} $$
$${{\sqrt {gR} } \over 2}$$
Explanation
v0 = $$\sqrt {g(R + h)} \approx \sqrt {gR} $$
ve = $$\sqrt {2g(R + h)} \approx \sqrt {2gR} $$
$$\Delta $$v=ve $$-$$ v0 = $$\left( {\sqrt 2 - 1} \right)\sqrt {gR} $$
ve = $$\sqrt {2g(R + h)} \approx \sqrt {2gR} $$
$$\Delta $$v=ve $$-$$ v0 = $$\left( {\sqrt 2 - 1} \right)\sqrt {gR} $$
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