Answer
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Hint: Acceleration due to gravity depends on the mass of the planet wherein the value of acceleration due to gravity varies from planet to planet. It is also referred to as gravitational field strength or gravitational acceleration.By using Classical Mechanics formulae, one can estimate the value of acceleration due to gravity.
Complete step by step answer:
Using Newton’s law of Universal Gravitation,
The force of gravity between two objects in space is formulated as,
\[g = G\dfrac{M}{{{R^2}}}\]
Where, $G$ is the gravitational constant
$R$ is the distance between the center of masses
$M$ is the mass of the object
Given criteria,
$M = \dfrac{{{M_e}}}{9},R = \dfrac{R}{2}$
For planet Earth,
${g_e} = \dfrac{{G{M_e}}}{{R_e^2}}$ ..........(1)
For planet Venus,
\[{g_v} = \dfrac{{G{M_v}}}{{R_v^2}}\] ........(2)
Dividing equation(2) upon equation(1) , we get
$
\Rightarrow \dfrac{{{g_v}}}{{{g_e}}} = \dfrac{{\dfrac{{G{M_v}}}{{R_v^2}}}}{{\dfrac{{G{M_e}}}{{R_e^2}}}} \\
\Rightarrow \dfrac{{{g_v}}}{{{g_e}}} = \dfrac{{\dfrac{{{M_v}}}{{R_v^2}}}}{{\dfrac{{{M_e}}}{{R_e^2}}}} \\
$
Using the above constraints given in the question,
$
\dfrac{{{g_v}}}{{{g_e}}} = \dfrac{{\dfrac{{{M_v}}}{{R_v^2}}}}{{\dfrac{{{M_e}}}{{R_e^2}}}}
\\
\Rightarrow\dfrac{{{g_v}}}{{{g_e}}} =
\dfrac{{\dfrac{{\dfrac{{{M_e}}}{9}}}{{{{(2{R_e})}^2}}}}}{{\dfrac{{{M_e}}}{{R_e^2}}}} \\
\Rightarrow\dfrac{{{g_v}}}{{{g_e}}} = \dfrac{{{M_e}}}{{9(4R_e^2)}} \times \dfrac{{R_e^2}}{{{M_e}}} \\
\Rightarrow\dfrac{{{g_v}}}{{{g_e}}} = \dfrac{1}{{36}} \\
\therefore {g_v} = \dfrac{{{g_e}}}{{36}}$
Hence, the value of {g_v} on the surface of venus is \[\dfrac{{{g_e}}}{{36}}\].
Note:The law of universal gravitation of Newton is generally stated as each particle attracts any other particle in the universe with a force directly proportional to the product of its masses and inversely proportional to the square of the distance between its center. Venus spins clockwise on its axis ,unlike the other planets in the solar system.
Complete step by step answer:
Using Newton’s law of Universal Gravitation,
The force of gravity between two objects in space is formulated as,
\[g = G\dfrac{M}{{{R^2}}}\]
Where, $G$ is the gravitational constant
$R$ is the distance between the center of masses
$M$ is the mass of the object
Given criteria,
$M = \dfrac{{{M_e}}}{9},R = \dfrac{R}{2}$
For planet Earth,
${g_e} = \dfrac{{G{M_e}}}{{R_e^2}}$ ..........(1)
For planet Venus,
\[{g_v} = \dfrac{{G{M_v}}}{{R_v^2}}\] ........(2)
Dividing equation(2) upon equation(1) , we get
$
\Rightarrow \dfrac{{{g_v}}}{{{g_e}}} = \dfrac{{\dfrac{{G{M_v}}}{{R_v^2}}}}{{\dfrac{{G{M_e}}}{{R_e^2}}}} \\
\Rightarrow \dfrac{{{g_v}}}{{{g_e}}} = \dfrac{{\dfrac{{{M_v}}}{{R_v^2}}}}{{\dfrac{{{M_e}}}{{R_e^2}}}} \\
$
Using the above constraints given in the question,
$
\dfrac{{{g_v}}}{{{g_e}}} = \dfrac{{\dfrac{{{M_v}}}{{R_v^2}}}}{{\dfrac{{{M_e}}}{{R_e^2}}}}
\\
\Rightarrow\dfrac{{{g_v}}}{{{g_e}}} =
\dfrac{{\dfrac{{\dfrac{{{M_e}}}{9}}}{{{{(2{R_e})}^2}}}}}{{\dfrac{{{M_e}}}{{R_e^2}}}} \\
\Rightarrow\dfrac{{{g_v}}}{{{g_e}}} = \dfrac{{{M_e}}}{{9(4R_e^2)}} \times \dfrac{{R_e^2}}{{{M_e}}} \\
\Rightarrow\dfrac{{{g_v}}}{{{g_e}}} = \dfrac{1}{{36}} \\
\therefore {g_v} = \dfrac{{{g_e}}}{{36}}$
Hence, the value of {g_v} on the surface of venus is \[\dfrac{{{g_e}}}{{36}}\].
Note:The law of universal gravitation of Newton is generally stated as each particle attracts any other particle in the universe with a force directly proportional to the product of its masses and inversely proportional to the square of the distance between its center. Venus spins clockwise on its axis ,unlike the other planets in the solar system.
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