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Gravitation question

2022 · 29 Jun · Shift 2 · Q52
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  5. /2022 · 29 Jun · Shift 2 · Q52

Gravitation question

2022 · 29 Jun · Shift 2 · Q52

JEE MainPhysicsGravitationMCQ+4 / −1
The time period of a satellite revolving around earth in a given orbit is 7 hours. If the radius of orbit is increased to three times its previous value, then approximate new time period of the satellite will be
  1. A
    40 hours
  2. B
    36 hours
  3. C
    30 hours
  4. D
    25 hours
View written solutionFree

Correct answer: B

  1. For a satellite revolving around Earth, Kepler’s third law gives T2∝r3T^2 \propto r^3T2∝r3 So, T∝r3/2T \propto r^{3/2}T∝r3/2

  2. Let the initial time period be T1=7 hoursT_1 = 7\text{ hours}T1​=7 hours and the initial orbital radius be r1r_1r1​.

  3. The new radius is increased to three times: r2=3r1r_2 = 3r_1r2​=3r1​

  4. Using T2T1=(r2r1)3/2\frac{T_2}{T_1} = \left(\frac{r_2}{r_1}\right)^{3/2}T1​T2​​=(r1​r2​​)3/2 we get T27=33/2=33\frac{T_2}{7} = 3^{3/2} = 3\sqrt{3}7T2​​=33/2=33​

  5. Therefore, T2=7⋅33T_2 = 7 \cdot 3\sqrt{3}T2​=7⋅33​ Using 3≈1.732\sqrt{3} \approx 1.7323​≈1.732, T2≈7⋅3⋅1.732T_2 \approx 7 \cdot 3 \cdot 1.732T2​≈7⋅3⋅1.732 T2≈36.37 hoursT_2 \approx 36.37\text{ hours}T2​≈36.37 hours

  6. Approximate value: T2≈36 hoursT_2 \approx 36\text{ hours}T2​≈36 hours

  7. Checking options:

  • A: 404040 hours
  • B: 363636 hours
  • C: 303030 hours
  • D: 252525 hours

Hence, the correct option is B.

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