At which height from the surface of the Earth, the acceleration due to gravity is the same as at a depth $200\,km$ below the surface of Earth. Radius of Earth is $6400\,km$.
Answer
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Hint:To solve this question students should know what acceleration due to gravity and how it is related with the earth. Then the equation of acceleration due to gravity at depth and height. The gravity of Earth, denoted by g, is the net acceleration that is imparted to objects due to the combined effect of gravitation (from mass distribution within Earth) and the centrifugal force (from the Earth's rotation).
Complete step by step answer:
Acceleration due gravity at depth is,
${g_d} = g\left( {1 - \dfrac{d}{R}} \right)$
where $d$ is the distance and $R$ is the radius of the earth in km
Acceleration due gravity at height ${g_h} = g{(\dfrac{R}{{R + h}})^2}$.
As the change in depth is too less, that is about 3.125 percent.
We can use the formula
${g_h} = g(1 - \dfrac{{2h}}{R})$
Now,
${g_d} = {g_h}$ from the question
Therefore,
$g\left( {1 - \dfrac{d}{R}} \right) = g(1 - \dfrac{{2h}}{R})$
$\Rightarrow 1 - \dfrac{d}{R} = 1 - \dfrac{{2h}}{R}$
Removing the common terms,
$d = 2h$
Therefore,
$h = \dfrac{d}{2}$
$\Rightarrow h = \dfrac{{200}}{2}$
Therefore, we get
$\therefore h = 100{\text{ km}}$
At the height of 100 km from the surface of the Earth, the acceleration due to gravity is the same as at a depth 200 km below the surface of Earth with the Radius of Earth as 6400 km.
Note:Acceleration due to gravity changes with the variation in height (altitude), depth. It changes with the shape and size of the earth. It also changes with the rotation of earth. As we move into the depth of the earth the acceleration due to gravity decreases and at the Centre of the earth the acceleration due to gravity is zero. Similarly, while we move away from the surface of earth the gravitation due to gravity decreases and we say that it becomes zero at infinite distance. When the depth is zero that is on the surface of earth the acceleration due to gravity is equal to g. acceleration due to gravity is more at poles and less at the equator.
Complete step by step answer:
Acceleration due gravity at depth is,
${g_d} = g\left( {1 - \dfrac{d}{R}} \right)$
where $d$ is the distance and $R$ is the radius of the earth in km
Acceleration due gravity at height ${g_h} = g{(\dfrac{R}{{R + h}})^2}$.
As the change in depth is too less, that is about 3.125 percent.
We can use the formula
${g_h} = g(1 - \dfrac{{2h}}{R})$
Now,
${g_d} = {g_h}$ from the question
Therefore,
$g\left( {1 - \dfrac{d}{R}} \right) = g(1 - \dfrac{{2h}}{R})$
$\Rightarrow 1 - \dfrac{d}{R} = 1 - \dfrac{{2h}}{R}$
Removing the common terms,
$d = 2h$
Therefore,
$h = \dfrac{d}{2}$
$\Rightarrow h = \dfrac{{200}}{2}$
Therefore, we get
$\therefore h = 100{\text{ km}}$
At the height of 100 km from the surface of the Earth, the acceleration due to gravity is the same as at a depth 200 km below the surface of Earth with the Radius of Earth as 6400 km.
Note:Acceleration due to gravity changes with the variation in height (altitude), depth. It changes with the shape and size of the earth. It also changes with the rotation of earth. As we move into the depth of the earth the acceleration due to gravity decreases and at the Centre of the earth the acceleration due to gravity is zero. Similarly, while we move away from the surface of earth the gravitation due to gravity decreases and we say that it becomes zero at infinite distance. When the depth is zero that is on the surface of earth the acceleration due to gravity is equal to g. acceleration due to gravity is more at poles and less at the equator.
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