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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)?
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:
\[ h = -R(1 + \sqrt{3}) \]
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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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)? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about 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)? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 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)?.
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)? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about 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)? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 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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