
If mass of earth decreases by $25\%$ and its radius is increased by $50\%$, then the acceleration due to gravity at its surface decreases by nearly?
A. 89%
B. 67 %
C. 33 %
D. 11 %
Answer
546.3k+ views
Hint:Acceleration due to the gravity although a constant but varies with respect to height and depth. The universal law of gravitation holds for two bodies anywhere in the universe. It is due to the gravitational force of the earth that we are bound to its surface. If a body is allowed to fall freely from a certain height then it falls under the influence of gravity and the acceleration of the body is g whose value is 9.8 \[m/{{s}^{2}}\].
Complete step by step answer:
We know on earth, \[g=\dfrac{GM}{{{R}^{2}}}\], so,
Now mass is decreased by 25%, new mass will be
$M'=M-\dfrac{25M}{100} \\
\Rightarrow M'=M-0.25M \\
\therefore M'=0.75M \\$
Also, the radius is increased by 50%, so new radius becomes,
$\Rightarrow R'=R+\dfrac{50R}{100} \\
\Rightarrow R'=R+0.5R \\
\therefore R'=1.5R \\$
Now putting this in the main formula, we get,
$g'=\dfrac{GM'}{{{R}^{'2}}} \\
\Rightarrow g'=\dfrac{0.75GM}{{{(1.5R)}^{2}}} \\
\Rightarrow g'=0.33\dfrac{GM}{{{R}^{2}}} \\
\Rightarrow g'=0.33g $
\[\therefore g'=33%\] of g
So, the correct option is C.
Note:Acceleration due to gravity is directly proportional to the mass of the earth and inversely proportional to the square of distance from the center of the earth. At the center the value of g is 0. The angular velocity of rotation of earth affects the value of g. it is clear that the value of g is independent of the mass of the body. It is dependent on the mass of the earth and the distance between the two bodies.
Complete step by step answer:
We know on earth, \[g=\dfrac{GM}{{{R}^{2}}}\], so,
Now mass is decreased by 25%, new mass will be
$M'=M-\dfrac{25M}{100} \\
\Rightarrow M'=M-0.25M \\
\therefore M'=0.75M \\$
Also, the radius is increased by 50%, so new radius becomes,
$\Rightarrow R'=R+\dfrac{50R}{100} \\
\Rightarrow R'=R+0.5R \\
\therefore R'=1.5R \\$
Now putting this in the main formula, we get,
$g'=\dfrac{GM'}{{{R}^{'2}}} \\
\Rightarrow g'=\dfrac{0.75GM}{{{(1.5R)}^{2}}} \\
\Rightarrow g'=0.33\dfrac{GM}{{{R}^{2}}} \\
\Rightarrow g'=0.33g $
\[\therefore g'=33%\] of g
So, the correct option is C.
Note:Acceleration due to gravity is directly proportional to the mass of the earth and inversely proportional to the square of distance from the center of the earth. At the center the value of g is 0. The angular velocity of rotation of earth affects the value of g. it is clear that the value of g is independent of the mass of the body. It is dependent on the mass of the earth and the distance between the two bodies.
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