(C) The acceleration due to gravity at the surface of the Earth is given by $g = \frac{GM}{R^2} = 9.8\, m\,s^{-2}$.At an altitude $h$,the acceleration

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(C) The acceleration due to gravity at the surface of the Earth is given by $g = \frac{GM}{R^2} = 9.8\, m\,s^{-2}$.At an altitude $h$,the acceleration due to gravity $g'$ is given by $g' = \frac{GM}{(R+h)^2}$.We are given $g' = 4.9\, m\,s^{-2}$,which is $\frac{g}{2}$.So,$\frac{GM}{(R+h)^2} = \frac{1}{2} \frac{GM}{R^2}$.Taking the reciprocal and square root on both sides: $\frac{R+h}{R} = \sqrt{2}$.$1 + \frac{h}{R} = 1.414$.$\frac{h}{R} = 0.414$.$h = 0.414 \times R = 0.414 \times 6.4 \times 10^6\, m$.$h \approx 2.65 \times 10^6\, m$.

Class 11 Physics Gravitation Acceleration Due to Gravity and its Variation

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The value of acceleration due to gravity at Earth's surface is $9.8\, m\,s^{-2}$. The altitude above its surface at which the acceleration due to gravity decreases to $4.9\, m\,s^{-2}$,is close to: (Radius of Earth $= 6.4 \times 10^6\, m$)

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A
$6.4 \times 10^6\, m$
B
$9.0 \times 10^6\, m$
C
$2.6 \times 10^6\, m$
D
$1.6 \times 10^6\, m$

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The height of the point vertically above the earth's surface,at which the acceleration due to gravity becomes $1\%$ of its value at the surface is (Radius of the earth $= R$). (in $,R$)

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Assertion $(A)$: $A$ particle of mass $m$ dropped into a hole made along the diameter of the Earth from one end to the other possesses simple harmonic motion.
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The value of acceleration due to gravity at Earth's surface is $9.8\, m\,s^{-2}$. The altitude above its surface at which the acceleration due to gravity decreases to $4.9\, m\,s^{-2}$,is close to: (Radius of Earth $= 6.4 \times 10^6\, m$)

Similar Questions

The depth at which acceleration due to gravity becomes $\frac{g}{2n}$ is ($R=$ radius of earth,$g=$ acceleration due to gravity on earth's surface,$n$ is an integer).

$A$ tunnel is dug through the centre of the earth. Show that a body of mass $m$ when dropped from rest from one end of the tunnel will execute simple harmonic motion.

The radii of two planets $A$ and $B$ are $R$ and $4R$ and their densities are $\rho$ and $\rho/3$ respectively. The ratio of acceleration due to gravity at their surfaces $(g_A : g_B)$ will be

The height of the point vertically above the earth's surface,at which the acceleration due to gravity becomes $1\%$ of its value at the surface is (Radius of the earth $= R$). (in $,R$)

Assertion $(A)$: $A$ particle of mass $m$ dropped into a hole made along the diameter of the Earth from one end to the other possesses simple harmonic motion.Reason $(R)$: Gravitational force between any two particles is inversely proportional to the square of the distance between them.

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Assertion $(A)$: $A$ particle of mass $m$ dropped into a hole made along the diameter of the Earth from one end to the other possesses simple harmonic motion.
Reason $(R)$: Gravitational force between any two particles is inversely proportional to the square of the distance between them.