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प्रश्न
Explain the variation of g with altitude.
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उत्तर
Consider an object of mass m at a height h from the surface of the Earth. Acceleration experienced by the object due to Earth is
g’ = `("GM")/(("R"_"e" + "h")^2)`
Mass at a height h from the center of the Earth
g’ = `("GM")/("R"_"e"^2 (1 + "h"/"R"_"e")^2)`
g’ = `("GM")/("R"_"e"^2) (1 + "h"/"R"_"e")^-2`
If h << Re
We can use Binomial expansion. Taking the terms up to first order
g’ = `("GM")/("R"_"e"^2) (1 - 2"h"/"R"_"e")`
g’ = `"g" (1 - 2"h"/"R"_"e")`
We find that g’ < g. This means that as altitude h increases the acceleration due to gravity g decreases.
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संबंधित प्रश्न
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Find the acceleration due to gravity in a mine of depth 640 m if the value at the surface is 9.800 m s−2. The radius of the earth is 6400 km.
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Which of the following options are correct?
- Acceleration due to gravity decreases with increasing altitude.
- Acceleration due to gravity increases with increasing depth (assume the earth to be a sphere of uniform density).
- Acceleration due to gravity increases with increasing latitude.
- Acceleration due to gravity is independent of the mass of the earth.
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If the radius of the earth shrinks by 2% while its mass remains the same. The acceleration due to gravity on the earth's surface will approximately ______.
The mass of a planet is `1/10`th that of the earth and its diameter is half that of the earth. The acceleration due to gravity on that planet is ______.
