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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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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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