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Gravitation question

2020 · 8 Jan · Shift 2 · Q57
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Gravitation question

2020 · 8 Jan · Shift 2 · Q57

JEE MainPhysicsGravitationNumerical+4 / −1
An asteroid is moving directly towards the centre of the earth. When at a distance of 10R (R is the radius of the earth) from the earths centre, it has a speed of 12 km/s. Neglecting the effect of earths atmosphere, what will be the speed of the asteroid when it hits the surface of the earth (escape velocity from the earth is 11.2 km/s) ? Give your answer to the nearest integer in kilometer/s ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 16

  1. Use conservation of mechanical energy

Since the asteroid is moving under Earth's gravity and atmospheric effects are neglected, total mechanical energy is conserved:

12mv2−GMmr=constant\frac{1}{2}mv^2 - \frac{GMm}{r} = \text{constant}21​mv2−rGMm​=constant

Let:

  • initial position: r1=10Rr_1 = 10Rr1​=10R
  • initial speed: v1=12 km/sv_1 = 12\,\text{km/s}v1​=12km/s
  • final position: r2=Rr_2 = Rr2​=R
  • final speed: v2=?v_2 = ?v2​=?

So,

12mv12−GMm10R=12mv22−GMmR\frac{1}{2}mv_1^2 - \frac{GMm}{10R} = \frac{1}{2}mv_2^2 - \frac{GMm}{R}21​mv12​−10RGMm​=21​mv22​−RGMm​
  1. Rearrange for v22v_2^2v22​
12m(v22−v12)=GMm(1R−110R)\frac{1}{2}m(v_2^2 - v_1^2) = GMm\left(\frac{1}{R} - \frac{1}{10R}\right)21​m(v22​−v12​)=GMm(R1​−10R1​) 12(v22−v12)=GM⋅910R\frac{1}{2}(v_2^2 - v_1^2) = GM\cdot \frac{9}{10R}21​(v22​−v12​)=GM⋅10R9​ v22=v12+2GM⋅910Rv_2^2 = v_1^2 + 2GM\cdot \frac{9}{10R}v22​=v12​+2GM⋅10R9​
  1. Use escape velocity relation

Escape velocity from Earth's surface is:

ve=2GMR=11.2 km/sv_e = \sqrt{\frac{2GM}{R}} = 11.2\,\text{km/s}ve​=R2GM​​=11.2km/s

Thus,

2GMR=(11.2)2\frac{2GM}{R} = (11.2)^2R2GM​=(11.2)2

So,

2GM⋅910R=910⋅2GMR=910(11.2)22GM\cdot \frac{9}{10R} = \frac{9}{10}\cdot \frac{2GM}{R} = \frac{9}{10}(11.2)^22GM⋅10R9​=109​⋅R2GM​=109​(11.2)2

Hence,

v22=122+910(11.2)2v_2^2 = 12^2 + \frac{9}{10}(11.2)^2v22​=122+109​(11.2)2
  1. Calculate numerically
122=14412^2 = 144122=144 (11.2)2=125.44(11.2)^2 = 125.44(11.2)2=125.44 910(125.44)=112.896\frac{9}{10}(125.44) = 112.896109​(125.44)=112.896

Therefore,

v22=144+112.896=256.896v_2^2 = 144 + 112.896 = 256.896v22​=144+112.896=256.896 v2=256.896≈16.03 km/sv_2 = \sqrt{256.896} \approx 16.03\,\text{km/s}v2​=256.896​≈16.03km/s
  1. Nearest integer
v2≈16 km/sv_2 \approx 16\,\text{km/s}v2​≈16km/s

Final integer answer: 161616

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