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Gravitation question

2002 · Shift 0 · Q184
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  5. /2002 · Shift 0 · Q184

Gravitation question

2002 · Shift 0 · Q184

JEE MainPhysicsGravitationMCQ+4 / −1
Energy required to move a body of mass mmm from an orbit of radius 2R2R2R to 3R3R3R is
  1. A
    GMm12R2{{GMm} \over {12{R^2}}}12R2GMm​
  2. B
    GMm3R2{{GMm} \over {3{R^2}}}3R2GMm​
  3. C
    GMm8R{{GMm} \over {8R}}8RGMm​
  4. D
    GMm6R{{GMm} \over {6R}}6RGMm​
View written solutionFree

Correct answer: NONE OF THE GIVEN OPTIONS; CORRECT VALUE IS $\DFRAC{GMM}{12R}$

  1. Total energy of a satellite in circular orbit

For a body of mass mmm revolving around a planet of mass MMM in a circular orbit of radius rrr, the total mechanical energy is

E=−GMm2rE = -\frac{GMm}{2r}E=−2rGMm​

  1. Initial energy at orbit radius 2R2R2R

E1=−GMm2(2R)=−GMm4RE_1 = -\frac{GMm}{2(2R)} = -\frac{GMm}{4R}E1​=−2(2R)GMm​=−4RGMm​

  1. Final energy at orbit radius 3R3R3R

E2=−GMm2(3R)=−GMm6RE_2 = -\frac{GMm}{2(3R)} = -\frac{GMm}{6R}E2​=−2(3R)GMm​=−6RGMm​

  1. Energy required to move from 2R2R2R to 3R3R3R

Required energy is the increase in total energy:

ΔE=E2−E1\Delta E = E_2 - E_1ΔE=E2​−E1​

Substitute:

ΔE=−GMm6R−(−GMm4R)\Delta E = -\frac{GMm}{6R} - \left(-\frac{GMm}{4R}\right)ΔE=−6RGMm​−(−4RGMm​)

ΔE=GMm4R−GMm6R\Delta E = \frac{GMm}{4R} - \frac{GMm}{6R}ΔE=4RGMm​−6RGMm​

Taking LCM 12R12R12R:

ΔE=3GMm−2GMm12R=GMm12R\Delta E = \frac{3GMm - 2GMm}{12R} = \frac{GMm}{12R}ΔE=12R3GMm−2GMm​=12RGMm​

  1. Compare with options
  • A: GMm12R2\dfrac{GMm}{12R^2}12R2GMm​ ❌ wrong dimensions
  • B: GMm3R2\dfrac{GMm}{3R^2}3R2GMm​ ❌ wrong dimensions
  • C: GMm8R\dfrac{GMm}{8R}8RGMm​ ❌ not equal
  • D: GMm6R\dfrac{GMm}{6R}6RGMm​ ❌ not equal

So the correct value is

GMm12R\boxed{\frac{GMm}{12R}}12RGMm​​

This is not present among the listed options, and therefore the stored answer appears incorrect.

Previous

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