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At what height from the surface of earth, value of acceleration due to gravity becomes th of its 9
value at surface of earth? (R is radius of earth)?
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At what height from the surface of earth, value of acceleration due to...
Understanding the problem:
The acceleration due to gravity at a height h above the surface of the Earth is given by the formula:
\[ g' = \frac{GM}{(R+h)^2} \]
where:
- \( g' \) is the acceleration due to gravity at height h,
- G is the universal gravitational constant,
- M is the mass of the Earth,
- R is the radius of the Earth.

Given condition:
We are given that the acceleration due to gravity at a height h is one-third of its value at the surface of the Earth. Mathematically, this can be represented as:
\[ \frac{GM}{(R+h)^2} = \frac{1}{3} \times \frac{GM}{R^2} \]

Solving for h:
Solving the above equation for h, we get:
\[ (R+h)^2 = 3R^2 \]
\[ R^2 + 2Rh + h^2 = 3R^2 \]
\[ h^2 + 2Rh - 2R^2 = 0 \]
Using the quadratic formula, we find:
\[ h = \frac{-2R \pm \sqrt{(2R)^2 + 4(1)(-2R^2)}}{2(1)} \]
\[ h = \frac{-2R \pm \sqrt{4R^2 + 8R^2}}{2} \]
\[ h = \frac{-2R \pm \sqrt{12R^2}}{2} \]
\[ h = \frac{-2R \pm 2\sqrt{3}R}{2} \]
\[ h = -R(1 \pm \sqrt{3}) \]

Conclusion:
The height at which the acceleration due to gravity becomes one-third of its value at the surface of the Earth is given by \( h = -R(1 \pm \sqrt{3}) \). Since height cannot be negative, the only valid solution is:
\[ h = -R(1 + \sqrt{3}) \]
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At what height from the surface of earth, value of acceleration due to gravity becomes th of its 9value at surface of earth? (R is radius of earth)?
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