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Gravitation question

2023 · 13 Apr · Shift 1 · Q47
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Gravitation question

2023 · 13 Apr · Shift 1 · Q47

JEE MainPhysicsGravitationMCQ+4 / −1
A planet having mass 9Me9 \mathrm{Me}9Me and radius 4Re4 \mathrm{R}_{\mathrm{e}}4Re​, where Me\mathrm{Me}Me and Re\mathrm{Re}Re are mass and radius of earth respectively, has escape velocity in km/s\mathrm{km} / \mathrm{s}km/s given by: (Given escape velocity on earth Ve=11.2×103 m/s\mathrm{V}_{\mathrm{e}}=11.2 \times 10^{3} \mathrm{~m} / \mathrm{s}Ve​=11.2×103 m/s )
  1. A
    33.6
  2. B
    11.2
  3. C
    16.8
  4. D
    67.2
View written solutionFree

Correct answer: C

  1. Use the escape velocity formula

    Escape velocity of a planet is ve=2GMRv_e = \sqrt{\frac{2GM}{R}}ve​=R2GM​​

  2. Compare with Earth

    For Earth, ve,Earth=2GMeRe=11.2×103 m/s=11.2 km/sv_{e,\text{Earth}} = \sqrt{\frac{2GM_e}{R_e}} = 11.2\times 10^3\ \text{m/s} = 11.2\ \text{km/s}ve,Earth​=Re​2GMe​​​=11.2×103 m/s=11.2 km/s

    For the given planet: M=9Me,R=4ReM = 9M_e, \qquad R = 4R_eM=9Me​,R=4Re​

  3. Form the ratio

    ve,planetve,Earth=M/RMe/Re\frac{v_{e,\text{planet}}}{v_{e,\text{Earth}}} = \sqrt{\frac{M/R}{M_e/R_e}}ve,Earth​ve,planet​​=Me​/Re​M/R​​

    Substituting, ve,planetve,Earth=9Me/4ReMe/Re=94=32\frac{v_{e,\text{planet}}}{v_{e,\text{Earth}}} = \sqrt{\frac{9M_e/4R_e}{M_e/R_e}} = \sqrt{\frac{9}{4}} = \frac{3}{2}ve,Earth​ve,planet​​=Me​/Re​9Me​/4Re​​​=49​​=23​

  4. Calculate the planet's escape velocity

    ve,planet=32×11.2=16.8 km/sv_{e,\text{planet}} = \frac{3}{2}\times 11.2 = 16.8\ \text{km/s}ve,planet​=23​×11.2=16.8 km/s

  5. Match with the options

    16.8 km/s16.8\ \text{km/s}16.8 km/s corresponds to Option C.

  6. Comparison with stored answer

    Stored correct answer: C

    Derived answer: C

    So, they agree.

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