What will be the acceleration due to gravity at a height $h$ above the Earth's surface,where $R$ is the radius of the Earth and $g$ is the acceleration due to gravity on the surface of the Earth?
- A$\frac{g}{(1 + \frac{h}{R})^2}$
- B$g(1 - \frac{2h}{R})$
- C$\frac{g}{(1 - \frac{h}{R})^2}$
- D$g(1 - \frac{h}{R})$
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Similar Questions
The time period of a simple pendulum on the surface of the earth is $T$. The height above the surface of the earth at which the time period of the pendulum becomes $2T$ is (Radius of the earth $= 6400 \text{ km}$) (in $\text{ km}$)
At what height above the earth's surface does the acceleration due to gravity fall to $1\%$ of its value at the earth's surface (in $,R$)?
Difficult
View SolutionAt a certain depth $d$ below the surface of the earth,the value of acceleration due to gravity becomes four times its value at a height $3R$ above the earth's surface. Where $R$ is the radius of the earth (Take $R = 6400 \ km$). The depth $d$ is equal to $............ \ km$.
The depth $d$ at which the value of acceleration due to gravity becomes $\frac{1}{n}$ times the value at the surface is [$R =$ radius of the earth].
Difficult
View SolutionThe value of acceleration due to gravity at a depth of $ 1600 \,km $ is equal to: (Radius of Earth $ = 6400 \,km $) (in $\,ms^{-2}$)
MediumKCET 2017
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