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Gravitation question

2016 · Shift 0 · Q60
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Gravitation question

2016 · Shift 0 · Q60

JEE MainPhysicsGravitationMCQ+4 / −1
A satellite is revolving in a circular orbit at a height ′h′'h'′h′ from the earth's surface (radius of earth R;h<<RR;h \lt \lt RR;h<<R). The minimum increase in its orbital velocity required, so that the satellite could escape from the earth's gravitational field, is close to : (Neglect the effect of atmosphere.)
  1. A
    2gR\sqrt{2 g R}2gR​
  2. B
    gR\sqrt{g R}gR​
  3. C
    gR/2\sqrt{g R / 2}gR/2​
  4. D
    gR(2−1)\sqrt{g R}(\sqrt{2}-1)gR​(2​−1)
View written solutionFree

Correct answer: D

  1. Orbital speed at height hhh

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

Since h≪Rh \ll Rh≪R, we can take r≈R,r \approx R,r≈R, and using g=GMR2  ⟹  GM=gR2,g = \frac{GM}{R^2} \implies GM = gR^2,g=R2GM​⟹GM=gR2, we get vo≈gR2R=gR.v_o \approx \sqrt{\frac{gR^2}{R}} = \sqrt{gR}.vo​≈RgR2​​=gR​.

  1. Escape speed from the same point

Escape speed from distance rrr from Earth’s center is ve=2GMr.v_e = \sqrt{\frac{2GM}{r}}.ve​=r2GM​​. Again, for r≈Rr \approx Rr≈R, ve≈2gR2R=2gR.v_e \approx \sqrt{\frac{2gR^2}{R}} = \sqrt{2gR}.ve​≈R2gR2​​=2gR​.

  1. Minimum increase in speed required

The satellite already has orbital speed vov_ovo​, so the extra speed needed is Δv=ve−vo.\Delta v = v_e - v_o.Δv=ve​−vo​. Thus, Δv=2gR−gR=gR(2−1).\Delta v = \sqrt{2gR} - \sqrt{gR} = \sqrt{gR}(\sqrt{2}-1).Δv=2gR​−gR​=gR​(2​−1).

  1. Match with options

This corresponds to: gR(2−1)\boxed{\sqrt{gR}(\sqrt{2}-1)}gR​(2​−1)​ which is Option D.

  1. Verification with stored answer

Stored correct answer: D

Our derived answer: D

So they agree.

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