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Gravitation question

2023 · 10 Apr · Shift 2 · Q49
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  5. /2023 · 10 Apr · Shift 2 · Q49

Gravitation question

2023 · 10 Apr · Shift 2 · Q49

JEE MainPhysicsGravitationMCQ+4 / −1
The time period of a satellite, revolving above earth's surface at a height equal to R\mathrm{R}R will be (Given g=π2 m/s2,R=g=\pi^{2} \mathrm{~m} / \mathrm{s}^{2}, \mathrm{R}=g=π2 m/s2,R= radius of earth)
  1. A
    32R\sqrt{32 R}32R​
  2. B
    4R\sqrt{4 \mathrm{R}}4R​
  3. C
    8R\sqrt{8 R}8R​
  4. D
    2R\sqrt{2 R}2R​
View written solutionFree

Correct answer: A

  1. Orbital radius

    The satellite is at a height equal to Earth's radius RRR above the surface.

    So its distance from the center of Earth is r=R+R=2R.r = R + R = 2R.r=R+R=2R.

  2. Use the formula for time period of a satellite

    The orbital time period is T=2πr3GM.T = 2\pi \sqrt{\frac{r^3}{GM}}.T=2πGMr3​​.

    Also, at Earth's surface, g=GMR2⇒GM=gR2.g = \frac{GM}{R^2} \Rightarrow GM = gR^2.g=R2GM​⇒GM=gR2.

    Substitute into the time period formula: T=2π(2R)3gR2.T = 2\pi \sqrt{\frac{(2R)^3}{gR^2}}.T=2πgR2(2R)3​​.

  3. Simplify

    T=2π8R3gR2=2π8Rg.T = 2\pi \sqrt{\frac{8R^3}{gR^2}} = 2\pi \sqrt{\frac{8R}{g}}.T=2πgR28R3​​=2πg8R​​.

    Given g=π2 m/s2,g = \pi^2 \text{ m/s}^2,g=π2 m/s2, so T=2π8Rπ2.T = 2\pi \sqrt{\frac{8R}{\pi^2}}.T=2ππ28R​​.

    T=2π⋅8Rπ=28R=32R.T = 2\pi \cdot \frac{\sqrt{8R}}{\pi} = 2\sqrt{8R} = \sqrt{32R}.T=2π⋅π8R​​=28R​=32R​.

  4. Match with options

    T=32RT = \sqrt{32R}T=32R​

    Hence, the correct option is A.

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