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Gravitation question

2019 · 11 Jan · Shift 1 · Q66
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Gravitation question

2019 · 11 Jan · Shift 1 · Q66

JEE MainPhysicsGravitationMCQ+4 / −1
A satellite is revolving in a circular orbit at a height h form the earth surface, such that h < < R where R is the earth. Assuming that the effect of earth's atmosphere can be neglected the minimum increase in the speed required so that the satellite could escape from the gravitational field of earth is :
  1. A
    gR(2−1)\sqrt {gR} \left( {\sqrt 2 - 1} \right)gR​(2​−1)
  2. B
    2gR\sqrt {2gR}2gR​
  3. C
    gR\sqrt {gR}gR​
  4. D
    gR2{{\sqrt {gR} } \over 2}2gR​​
View written solutionFree

Correct answer: A

  1. Orbital speed of the satellite

For a satellite in a circular orbit of radius r=R+hr=R+hr=R+h its orbital speed is vo=GMrv_o=\sqrt{\frac{GM}{r}}vo​=rGM​​

Since h≪Rh \ll Rh≪R, we can take r≈Rr \approx Rr≈R So, vo≈GMRv_o \approx \sqrt{\frac{GM}{R}}vo​≈RGM​​

Using g=GMR2g=\dfrac{GM}{R^2}g=R2GM​, we get GM=gR2GM=gR^2GM=gR2 Hence, vo=gR2R=gRv_o=\sqrt{\frac{gR^2}{R}}=\sqrt{gR}vo​=RgR2​​=gR​

  1. Escape speed from that orbit

Escape speed from distance rrr from the center of Earth is ve=2GMrv_e=\sqrt{\frac{2GM}{r}}ve​=r2GM​​ Again using r≈Rr\approx Rr≈R, ve≈2GMR=2gRv_e \approx \sqrt{\frac{2GM}{R}}=\sqrt{2gR}ve​≈R2GM​​=2gR​

  1. Minimum increase in speed required

The satellite already has speed vov_ovo​, so the extra speed needed is Δv=ve−vo\Delta v=v_e-v_oΔv=ve​−vo​

Substitute the values: Δv=2gR−gR\Delta v=\sqrt{2gR}-\sqrt{gR}Δv=2gR​−gR​ Δv=gR(2−1)\Delta v=\sqrt{gR}(\sqrt{2}-1)Δv=gR​(2​−1)

  1. Matching with options

This matches Option A: gR(2−1)\boxed{\sqrt{gR}(\sqrt{2}-1)}gR​(2​−1)​

  1. Comparison with stored correct answer

Stored correct answer: A

Our derived answer is also A, so they agree.

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