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Gravitation question

2021 · Q166
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Gravitation question

2021 · Q166

NEETPhysicsGravitationMCQ+4 / −1
A particle of mass 'm' is projected with a velocity υ\upsilonυ = kVe = (k < 1) from the surface of the earth. (Ve = escape velocity)

The maximum height above the surface reached by the particle is :
  1. A
    Rk21−k2{{R{k^2}} \over {1 - {k^2}}}1−k2Rk2​
  2. B
    R(k1−k2)2R{\left( {{k \over {1 - {k^2}}}} \right)^2}R(1−k2k​)2
  3. C
    R(k1+k2)2R{\left( {{k \over {1 + {k^2}}}} \right)^2}R(1+k2k​)2
  4. D
    R2k1+k{{{R^2}k} \over {1 + k}}1+kR2k​
View written solutionFree

Correct answer: A

−GMmR+12mk2ve2=−GMmr - {{GMm} \over R} + {1 \over 2}m{k^2}{v_e}^2 = - {{GMm} \over r}−RGMm​+21​mk2ve​2=−rGMm​

−GMmR+12mk22GMR=−GMmr - {{GMm} \over R} + {1 \over 2}m{k^2}{{2GM} \over R} = - {{GMm} \over r}−RGMm​+21​mk2R2GM​=−rGMm​

−1R+k2R=−1r - {1 \over R} + {{{k^2}} \over R} = - {1 \over r}−R1​+Rk2​=−r1​

1r=1R−k2R{1 \over r} = {1 \over R} - {{{k^2}} \over R}r1​=R1​−Rk2​

1r=1−k2R{1 \over r} = {{1 - {k^2}} \over R}r1​=R1−k2​

r=R1−k2r = {R \over {1 - {k^2}}}r=1−k2R​

R+h=R1−k2R + h = {R \over {1 - {k^2}}}R+h=1−k2R​

h=R1−k2−R=k21−k2Rh = {R \over {1 - {k^2}}} - R = {{{k^2}} \over {1 - {k^2}}}Rh=1−k2R​−R=1−k2k2​R

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