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The height at which the acceleration due to gravity becomes g/9 (where g = the acceleration due to gravity on the surface of the earth) in terms of R, the radius of the earth, is :
- a)R/√2
- b)R/2
- c)√2R
- d)2 R
Correct answer is option 'D'. Can you explain this answer?
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The acceleration due to gravity at a height h above the surface of the earth can be given by the formula:
g' = (g * R^2) / (R + h)^2
where g' is the acceleration due to gravity at height h and R is the radius of the earth.
We are given that g' = (8/9)g. Substituting this into the formula, we get:
(8/9)g = (g * R^2) / (R + h)^2
Cross multiplying and simplifying, we get:
8(R + h)^2 = 9R^2
Expanding and rearranging the terms, we get:
8R^2 + 16Rh + 8h^2 = 9R^2
Simplifying further, we get:
8h^2 + 16Rh + (R^2 - 9R^2) = 0
8h^2 + 16Rh - 8R^2 = 0
Dividing through by 8, we get:
h^2 + 2Rh - R^2 = 0
This is a quadratic equation in h. Using the quadratic formula, we can solve for h:
h = (-2R ± √(4R^2 + 4R^2)) / 2
h = (-2R ± √(8R^2)) / 2
h = (-2R ± 2√2R) / 2
h = -R ± √2R
Since height cannot be negative, we take the positive solution:
h = √2R
Therefore, the height at which the acceleration due to gravity becomes 8/9 in terms of R is √2R.
g' = (g * R^2) / (R + h)^2
where g' is the acceleration due to gravity at height h and R is the radius of the earth.
We are given that g' = (8/9)g. Substituting this into the formula, we get:
(8/9)g = (g * R^2) / (R + h)^2
Cross multiplying and simplifying, we get:
8(R + h)^2 = 9R^2
Expanding and rearranging the terms, we get:
8R^2 + 16Rh + 8h^2 = 9R^2
Simplifying further, we get:
8h^2 + 16Rh + (R^2 - 9R^2) = 0
8h^2 + 16Rh - 8R^2 = 0
Dividing through by 8, we get:
h^2 + 2Rh - R^2 = 0
This is a quadratic equation in h. Using the quadratic formula, we can solve for h:
h = (-2R ± √(4R^2 + 4R^2)) / 2
h = (-2R ± √(8R^2)) / 2
h = (-2R ± 2√2R) / 2
h = -R ± √2R
Since height cannot be negative, we take the positive solution:
h = √2R
Therefore, the height at which the acceleration due to gravity becomes 8/9 in terms of R is √2R.
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The height at which the acceleration due to gravity becomes g/9(where g = the acceleration due to gravity on the surface of the earth) in terms of R, the radius of the earth, is :a)R/√2b)R/2c)√2Rd)2 RCorrect answer is option 'D'. Can you explain this answer?
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