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Gravitation question

2008 · Shift 0 · Q70
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Gravitation question

2008 · Shift 0 · Q70

JEE MainPhysicsGravitationMCQ+4 / −1
A planet in a distant solar system is 101010 times more massive than the earth and its radius is 101010 times smaller. Given that the escape velocity from the earth is 11  km s−1,11\,\,km\,{s^{ - 1}},11kms−1, the escape velocity from the surface of the planet would be
  1. A
    1.1  km s−11.1\,\,km\,{s^{ - 1}}1.1kms−1
  2. B
    100  km s−1100\,\,km\,{s^{ - 1}}100kms−1
  3. C
    110  km s−1110\,\,km\,{s^{ - 1}}110kms−1
  4. D
    0.11  km s−10.11\,\,km\,{s^{ - 1}}0.11kms−1
View written solutionFree

Correct answer: C

  1. Use the formula for escape velocity

The escape velocity from the surface of a planet is

ve=2GMRv_e = \sqrt{\frac{2GM}{R}}ve​=R2GM​​

where:

  • GGG = gravitational constant
  • MMM = mass of the planet
  • RRR = radius of the planet
  1. Compare the given planet with Earth

Let Earth’s escape velocity be

ve,E=11 km s−1v_{e,E} = 11\,\text{km s}^{-1}ve,E​=11km s−1

For the distant planet:

  • Mass Mp=10MEM_p = 10 M_EMp​=10ME​
  • Radius Rp=RE10R_p = \frac{R_E}{10}Rp​=10RE​​
  1. Find the ratio of escape velocities

ve,pve,E=Mp/RpME/RE\frac{v_{e,p}}{v_{e,E}} = \sqrt{\frac{M_p/R_p}{M_E/R_E}}ve,E​ve,p​​=ME​/RE​Mp​/Rp​​​

Substitute the given relations:

ve,pve,E=10MERE/10⋅REME\frac{v_{e,p}}{v_{e,E}} = \sqrt{\frac{10M_E}{R_E/10} \cdot \frac{R_E}{M_E}}ve,E​ve,p​​=RE​/1010ME​​⋅ME​RE​​​

=10×10=100=10= \sqrt{10 \times 10} = \sqrt{100} = 10=10×10​=100​=10

So,

ve,p=10×ve,E=10×11=110 km s−1v_{e,p} = 10 \times v_{e,E} = 10 \times 11 = 110\,\text{km s}^{-1}ve,p​=10×ve,E​=10×11=110km s−1

  1. Check the options
  • A: 1.1 km s−11.1\,\text{km s}^{-1}1.1km s−1 ❌
  • B: 100 km s−1100\,\text{km s}^{-1}100km s−1 ❌
  • C: 110 km s−1110\,\text{km s}^{-1}110km s−1 ✅
  • D: 0.11 km s−10.11\,\text{km s}^{-1}0.11km s−1 ❌

Hence, the correct option is C.

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