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Question

(D) The minimum velocity required for an object to escape the gravitational pull of a planet is known as the escape velocity $(v_e)$.The formula for e

Vedclass · Accepted Answer

$11.2 	imes 10^3 \ m/s$

Vedclass · Answer

(D) The minimum velocity required for an object to escape the gravitational pull of a planet is known as the escape velocity $(v_e)$.The formula for escape velocity is given by $v_e = \sqrt{2gR}$,where $g$ is the acceleration due to gravity and $R$ is the radius of the planet.Given: $g = 9.8 \ m/s^2$ and $R = 6.4 	imes 10^6 \ m$.Substituting these values into the formula:$v_e = \sqrt{2 	imes 9.8 	imes 6.4 	imes 10^6}$$v_e = \sqrt{19.6 	imes 6.4 	imes 10^6}$$v_e = \sqrt{125.44 	imes 10^6}$$v_e = 11.2 	imes 10^3 \ m/s$.Thus,the correct option is $D$.

The least velocity required to throw a body away from the surface of a planet so that it may not return is (radius of the planet is $6.4 \times 10^6 \ m$,$g = 9.8 \ m/s^2$).

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