(A) The formula for gravitational acceleration at a height $h$ above the earth's surface is given by $g' = g \left( \frac{R}{R+h} \right)^2$.Given tha

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(A) The formula for gravitational acceleration at a height $h$ above the earth's surface is given by $g' = g \left( \frac{R}{R+h} \right)^2$.Given that $g' = \frac{g}{4}$,we substitute this into the equation:$\frac{g}{4} = g \left( \frac{R}{R+h} \right)^2$.Dividing both sides by $g$,we get $\frac{1}{4} = \left( \frac{R}{R+h} \right)^2$.Taking the square root of both sides,we obtain $\frac{1}{2} = \frac{R}{R+h}$.Cross-multiplying gives $R + h = 2R$.Therefore,$h = 2R - R = R$.

Class 11 Physics Gravitation Acceleration Due to Gravity and its Variation

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The value of gravitational acceleration $g'$ at a height $h$ above the earth's surface is $\frac{g}{4}$. Then,what is the value of $h$ in terms of the earth's radius $R$?

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A
$h=R$
B
$h=\frac{R}{2}$
C
$h=\frac{R}{3}$
D
$h=\frac{R}{4}$

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Gravitation

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Acceleration Due to Gravity and its Variation

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The value of gravitational acceleration $g'$ at a height $h$ above the earth's surface is $\frac{g}{4}$. Then,what is the value of $h$ in terms of the earth's radius $R$?

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If the value of $g$ at the surface of the earth is $9.8 \, m/s^2$,then the value of $g$ at a place $480 \, km$ above the surface of the earth will be .......... $m/s^2$. (Radius of the earth is $6400 \, km$)

If the value of acceleration due to gravity on the surface of the Earth is $g$,what will be the value of $g$ at a height equal to the radius of the Earth from the surface?

$A$ body weighs $72 \ N$ on the surface of the earth. What is the gravitational force on it at a height equal to half the radius of the earth (in $N$)?

$A$ body weighs $250\,N$ on the surface of the earth. How much will it weigh halfway down to the center of the earth (in $,N$)?

Find the angular speed of rotation of the Earth,so that the apparent $g$ at the equator becomes $(1/6)$th of its original value. $(R = 6.4 \times 10^6 \ m)$

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