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A gravity meter can detect change in acceleration due to gravity (g) of the order of 10-9%. Calculate the smallest change in altitude near the surface of the earth that results in a detectable change in g. Radius of the earth R = 6.4 x 106m.
- a)22μm
- b)32μm
- c)12μm
- d)30μm
Correct answer is option 'B'. Can you explain this answer?
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A gravity meter can detect change in acceleration due to gravity (g)of...
To calculate the smallest change in altitude that results in a detectable change in acceleration due to gravity, we can use the formula for acceleration due to gravity:
g = GM/R^2
where G is the gravitational constant, M is the mass of the Earth, and R is the radius of the Earth.
Let's assume that the initial altitude is h, and the change in altitude is Δh. The new altitude would be h + Δh.
The new radius of the Earth would be R + Δh.
The new acceleration due to gravity would be:
g' = GM / (R + Δh)^2
The change in acceleration due to gravity is then:
Δg = g' - g
Now, we can use the given information that a gravity meter can detect a change of the order of 10^-9% in g. This means that:
Δg / g = 10^-9
Substituting the expressions for g, g', and Δg:
(GM / (R + Δh)^2 - GM / R^2) / (GM / R^2) = 10^-9
Simplifying the equation:
(R^2 / (R + Δh)^2 - 1) = 10^-9
(R^2 - (R + Δh)^2) / (R + Δh)^2 = 10^-9
Expanding and simplifying:
R^2 - (R^2 + 2RΔh + Δh^2) = 10^-9(R^2 + 2RΔh + Δh^2)
R^2 - R^2 - 2RΔh - Δh^2 = 10^-9R^2 + 10^-9(2RΔh) + 10^-9Δh^2
-2RΔh - Δh^2 = 10^-9R^2 + 2(10^-9)RΔh + 10^-9Δh^2
Multiplying through by -1:
2RΔh + Δh^2 = -10^-9R^2 - 2(10^-9)RΔh - 10^-9Δh^2
Rearranging the terms:
(1 + 2(10^-9)R)Δh + (10^-9R^2 + 10^-9Δh^2) = 0
Since the term with Δh is multiplied by a small factor (2(10^-9)R), we can neglect it compared to the term with Δh^2. Thus, we have:
10^-9R^2 + 10^-9Δh^2 = 0
Simplifying:
R^2 + Δh^2 = 0
Δh^2 = -R^2
Taking the square root:
Δh = ±iR
Since altitude cannot be an imaginary number, we discard the imaginary part and conclude that the smallest detectable change in altitude near the surface of the Earth is Δh = R.
Therefore, the smallest change in altitude that results in a detectable change in acceleration due to gravity is equal to the radius of the Earth, which is approximately 6.4 x 10^6 meters.
g = GM/R^2
where G is the gravitational constant, M is the mass of the Earth, and R is the radius of the Earth.
Let's assume that the initial altitude is h, and the change in altitude is Δh. The new altitude would be h + Δh.
The new radius of the Earth would be R + Δh.
The new acceleration due to gravity would be:
g' = GM / (R + Δh)^2
The change in acceleration due to gravity is then:
Δg = g' - g
Now, we can use the given information that a gravity meter can detect a change of the order of 10^-9% in g. This means that:
Δg / g = 10^-9
Substituting the expressions for g, g', and Δg:
(GM / (R + Δh)^2 - GM / R^2) / (GM / R^2) = 10^-9
Simplifying the equation:
(R^2 / (R + Δh)^2 - 1) = 10^-9
(R^2 - (R + Δh)^2) / (R + Δh)^2 = 10^-9
Expanding and simplifying:
R^2 - (R^2 + 2RΔh + Δh^2) = 10^-9(R^2 + 2RΔh + Δh^2)
R^2 - R^2 - 2RΔh - Δh^2 = 10^-9R^2 + 10^-9(2RΔh) + 10^-9Δh^2
-2RΔh - Δh^2 = 10^-9R^2 + 2(10^-9)RΔh + 10^-9Δh^2
Multiplying through by -1:
2RΔh + Δh^2 = -10^-9R^2 - 2(10^-9)RΔh - 10^-9Δh^2
Rearranging the terms:
(1 + 2(10^-9)R)Δh + (10^-9R^2 + 10^-9Δh^2) = 0
Since the term with Δh is multiplied by a small factor (2(10^-9)R), we can neglect it compared to the term with Δh^2. Thus, we have:
10^-9R^2 + 10^-9Δh^2 = 0
Simplifying:
R^2 + Δh^2 = 0
Δh^2 = -R^2
Taking the square root:
Δh = ±iR
Since altitude cannot be an imaginary number, we discard the imaginary part and conclude that the smallest detectable change in altitude near the surface of the Earth is Δh = R.
Therefore, the smallest change in altitude that results in a detectable change in acceleration due to gravity is equal to the radius of the Earth, which is approximately 6.4 x 10^6 meters.
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A gravity meter can detect change in acceleration due to gravity (g)of the order of 10-9%.Calculate the smallest change in altitude near the surface of the earth that results in a detectable change in g.Radius of the earth R = 6.4 x 106m.a)22μmb)32μmc)12μmd)30μmCorrect answer is option 'B'. Can you explain this answer?
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A gravity meter can detect change in acceleration due to gravity (g)of the order of 10-9%.Calculate the smallest change in altitude near the surface of the earth that results in a detectable change in g.Radius of the earth R = 6.4 x 106m.a)22μmb)32μmc)12μmd)30μmCorrect answer is option 'B'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A gravity meter can detect change in acceleration due to gravity (g)of the order of 10-9%.Calculate the smallest change in altitude near the surface of the earth that results in a detectable change in g.Radius of the earth R = 6.4 x 106m.a)22μmb)32μmc)12μmd)30μmCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A gravity meter can detect change in acceleration due to gravity (g)of the order of 10-9%.Calculate the smallest change in altitude near the surface of the earth that results in a detectable change in g.Radius of the earth R = 6.4 x 106m.a)22μmb)32μmc)12μmd)30μmCorrect answer is option 'B'. Can you explain this answer?.
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