(C) The acceleration due to gravity $g$ on the surface of a planet is given by $g = \frac{GM}{R^2}$.Since $M = \text{density} (\rho) \times \text{volu

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(C) The acceleration due to gravity $g$ on the surface of a planet is given by $g = \frac{GM}{R^2}$.Since $M = \text{density} (\rho) \times \text{volume} (V) = \rho \times \frac{4}{3} \pi R^3$, we can write $g = \frac{G \rho \frac{4}{3} \pi R^3}{R^2} = \frac{4}{3} \pi G \rho R$.For the earth, $g = \frac{4}{3} \pi G \rho R$.For the new planet, $g' = \frac{4}{3} \pi G \rho (2R) = 2 \times (\frac{4}{3} \pi G \rho R) = 2g$.Therefore, $g' = 2g$.

Class 11 Physics Gravitation Acceleration Due to Gravity and its Variation

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Imagine a new planet having the same density as that of the earth, but it is two times bigger than the earth in size (radius). If the acceleration due to gravity on the surface of the new planet is $g'$ and that on the surface of the earth is $g$, then:

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A
$g' = g/4$
B
$g' = 8g$
C
$g' = 2g$
D
$g' = 4g$

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Gravitation

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Acceleration Due to Gravity and its Variation

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Derive the equation for the variation of acceleration due to gravity $g$ with depth $d$ below the surface of the Earth.

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The angular speed of rotation of the Earth about its axis,at which the weight of a man standing on the equator becomes half of his weight at the poles,is given by:

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Match Column-$I$ with Column-$II$.

Column-$I$ Column-$II$

$(1)$ Maximum value of acceleration due to gravity $g$ $(a)$ At the center of the Earth

$(2)$ Minimum value of acceleration due to gravity $g$ $(b)$ At the poles

$(3)$ Zero value of acceleration due to gravity $g$ $(c)$ At the equator

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$A$ 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 $N$)?

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The speed with which the earth would have to rotate about its axis so that a person on the equator would weigh $\frac{3}{5}$ th as much as at present weight is ($g=$ gravitational acceleration,$R=$ equatorial radius of the earth).

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Column-$I$	Column-$II$
$(1)$ Maximum value of acceleration due to gravity $g$	$(a)$ At the center of the Earth
$(2)$ Minimum value of acceleration due to gravity $g$	$(b)$ At the poles
$(3)$ Zero value of acceleration due to gravity $g$	$(c)$ At the equator

Imagine a new planet having the same density as that of the earth, but it is two times bigger than the earth in size (radius). If the acceleration due to gravity on the surface of the new planet is $g'$ and that on the surface of the earth is $g$, then:

Similar Questions

Derive the equation for the variation of acceleration due to gravity $g$ with depth $d$ below the surface of the Earth.

The angular speed of rotation of the Earth about its axis,at which the weight of a man standing on the equator becomes half of his weight at the poles,is given by:

Match Column-$I$ with Column-$II$. Column-$I$ Column-$II$ $(1)$ Maximum value of acceleration due to gravity $g$ $(a)$ At the center of the Earth $(2)$ Minimum value of acceleration due to gravity $g$ $(b)$ At the poles $(3)$ Zero value of acceleration due to gravity $g$ $(c)$ At the equator

$A$ 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 $N$)?

The speed with which the earth would have to rotate about its axis so that a person on the equator would weigh $\frac{3}{5}$ th as much as at present weight is ($g=$ gravitational acceleration,$R=$ equatorial radius of the earth).

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Match Column-$I$ with Column-$II$.

Column-$I$ Column-$II$

$(1)$ Maximum value of acceleration due to gravity $g$ $(a)$ At the center of the Earth

$(2)$ Minimum value of acceleration due to gravity $g$ $(b)$ At the poles

$(3)$ Zero value of acceleration due to gravity $g$ $(c)$ At the equator