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The height vertically above the earth’s surface at which the acceleration due to gravity becomes 1% of its value at the surface is (R is the radius of the Earth)​
  • a)
    8 R
  • b)
    9 R
  • c)
    10 R
  • d)
    20 R
Correct answer is option 'B'. Can you explain this answer?
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Calculation for Height at which gravity becomes 1% of surface value

Given:
- Acceleration due to gravity at the Earth's surface, g = 9.8 m/s^2
- Acceleration due to gravity at a height h above the Earth's surface, g' = 0.01 * g

Formula:
- Acceleration due to gravity at a height h above the Earth's surface, g' = g * (R / (R + h))^2

Setting up the equation:
- g' = 0.01 * g
- 0.01 * g = g * (R / (R + h))^2
- 0.01 = (R / (R + h))^2

Solving for h:
- √0.01 = R / (R + h)
- 0.1 = R / (R + h)
- 0.1(R + h) = R
- 0.1R + 0.1h = R
- 0.1h = 0.9R
- h = 9R
Therefore, the height vertically above the Earth's surface at which the acceleration due to gravity becomes 1% of its value at the surface is 9 times the radius of the Earth, which is option (B) 9R.
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The height vertically above the earth’s surface at which the acceleration due to gravity becomes 1% of its value at the surface is (R is the radius of the Earth)a)8 Rb)9 Rc)10 Rd)20 RCorrect answer is option 'B'. Can you explain this answer?
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