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The critical velocity of a satellite is inversely proportional to the square root of the_______
but it is independent of mass of____________
(A) radius of the orbit, the satellite
(B) mass of the earth, the satellite
(C) mass of the satellite, the earth
(D) radius of the earth, the earth

seo-qna
Last updated date: 13th Jun 2024
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Answer
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Hint To answer this question, we need to use the expression of the critical velocity. Then we have to manipulate that expression in terms of the quantities mentioned in the options.
Formula Used The formulae used to solve this question are
${v_c} = \sqrt {g({R_e} + h)} $
$g = \dfrac{{G{M_e}}}{{{{\left( {{R_e} + h} \right)}^2}}}$
${v_c}$ is the critical velocity of a satellite, ${R_e}$ is the radius of earth, ${M_e}$is the mass of the earth,$h$ is the height of the satellite above the earth.

Complete step-by-step answer
We know that the critical velocity of a satellite revolving around the earth is given by
${v_c} = \sqrt {g({R_e} + h)} $
We know that$g = \dfrac{{G{M_e}}}{{{{\left( {{R_e} + h} \right)}^2}}}$
Substituting this in the above equation, we get
${v_c} = \sqrt {\dfrac{{G{M_e}}}{{{{\left( {{R_e} + h} \right)}^2}}}\left( {{R_e} + h} \right)} $
On simplifying, we get
${v_c} = \sqrt {\dfrac{{G{M_e}}}{{\left( {{R_e} + h} \right)}}} $
As we can see from the above expression for the critical velocity of a satellite, that it is directly proportional to the square root of the mass of the earth. So, this means that the critical velocity is not independent of the mass of the earth.
So, options C and D are incorrect.
Also, from the above expression we can see that the critical velocity of the satellite is inversely proportional to the square root of the radius of the orbit. And there is no term containing the mass of the satellite in the right hand side of the above expression. Therefore, the critical velocity is independent of the mass of the satellite.

Hence, the correct answer is option A.

Note: You might be wondering why the mass of the satellite is not there in the final expression of the critical velocity. The answer to this question comes from the derivation of the critical velocity. We know that the critical velocity is the minimum horizontal velocity given to a satellite to keep it revolving in the earth’s orbit. So, it is obtained by keeping the gravitational force of the earth on the satellite equal to the centripetal force required to keep it moving in the earth’s orbit. Since, both the forces are proportional to the mass of the satellite, so it gets cancelled out of the final expression of the critical velocity.