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Gravitation question

2012 · Shift 0 · Q57
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Gravitation question

2012 · Shift 0 · Q57

JEE MainPhysicsGravitationMCQ+4 / −1
The mass of a spaceship is 1000kg.1000kg.1000kg. It is to be launched from the earth's surface out into free space. The value of ggg and RRR(radius of earth ) are 10 m/s210\,m/{s^2}10m/s2 and 6400km6400km6400km respectively. The required energy for this work will be:
  1. A
    6.4×1011 6.4 \times {10^{11}}\,6.4×1011 Joules
  2. B
    6.4×108 6.4 \times {10^8}\,6.4×108 Joules
  3. C
    6.4×109 6.4 \times {10^9}\,6.4×109 Joules
  4. D
    6.4×1010 6.4 \times {10^{10}}\,6.4×1010 Joules
View written solutionFree

Correct answer: D

  1. Required concept

To take the spaceship from the Earth's surface to free space (i.e. to infinity) with zero final speed, the minimum required energy is the gravitational potential energy needed to escape Earth’s field:

E=GMmRE = \frac{GMm}{R}E=RGMm​

Since on Earth’s surface,

g=GMR2  ⟹  GM=gR2g = \frac{GM}{R^2} \implies GM = gR^2g=R2GM​⟹GM=gR2

So,

E=gR2mR=mgRE = \frac{gR^2 m}{R} = mgRE=RgR2m​=mgR

  1. Substitute the given values
  • m=1000 kgm = 1000\,\text{kg}m=1000kg
  • g=10 m/s2g = 10\,\text{m/s}^2g=10m/s2
  • R=6400 km=6.4×106 mR = 6400\,\text{km} = 6.4 \times 10^6\,\text{m}R=6400km=6.4×106m

Thus,

E=mgR=1000×10×6.4×106E = mgR = 1000 \times 10 \times 6.4 \times 10^6E=mgR=1000×10×6.4×106

  1. Calculate

E=104×6.4×106=6.4×1010 JE = 10^4 \times 6.4 \times 10^6 = 6.4 \times 10^{10}\,\text{J}E=104×6.4×106=6.4×1010J

  1. Match with options

6.4×1010 J6.4 \times 10^{10}\,\text{J}6.4×1010J

This corresponds to Option D.

  1. Comparison with stored answer

Stored correct answer: D

Our derived answer: D

So they agree.

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