
At what altitude will the acceleration due to gravity be $25\% $ of that at the earth's surface (given radius of earth is $R$ )?
A) $\dfrac{R}{4}$
B) $R$
C) $\dfrac{{3R}}{8}$
D) $\dfrac{R}{2}$
Answer
572.1k+ views
Hint: You can easily solve the question by trying to recall how we define the acceleration due to gravity of earth and then replacing the distance from $R$ to $\left( {R + h} \right)$ . If you are having trouble recalling the mathematical definition of gravity, let us tell you that:
$g = G\dfrac{{{M_E}}}{{{R^2}}}$
Complete step by step answer:
We will be trying to solve the question exactly like we told in the hint section of the solution to this question. But for that, we first need to define acceleration due to gravity of the Earth and its mathematical expression.
Acceleration due to gravity of the Earth is the acceleration with which the Earth is pulling any or every body on its surface.
Mathematically, we can write acceleration due to gravity as:
$g = G\dfrac{{{M_E}}}{{{R^2}}}$
Where, $G$ is the gravitational constant,
${M_E}$ is the mass of Earth and,
$R$ is the radius of Earth.
The question has asked us about the altitude at which the acceleration due to gravity is only $25\% $ of that at earth’s surface. This can be mathematically represented as:
$g' = \dfrac{{25}}{{100}}g = \dfrac{g}{4}$
But we can also write $g'$ as:
$g' = G\dfrac{{{M_E}}}{{{{\left( {R + h} \right)}^2}}}$
Equating both the equations, we get:
$\dfrac{g}{4} = G\dfrac{{{M_E}}}{{{{\left( {R + h} \right)}^2}}}$
We have already seen that $g = G\dfrac{{{M_E}}}{{{R^2}}}$
Substituting that here, we get:
$G\dfrac{{{M_E}}}{{4{R^2}}} = G\dfrac{{{M_E}}}{{{{\left( {R + h} \right)}^2}}}$
After cancelling out common terms on both sides, we get:
${\left( {R + h} \right)^2} = 4{R^2}$
Solving this, we get:
$
R + h = 2R \\
h = R \\
$
We can see that the value of altitude for each of the said conditions meet is the same as that of option (B). Hence, option (B) is the correct answer.
Note: Many students tick the option with the value of $\left( {R + h} \right)$ instead of $h$ which causes them to lose marks even if they solved the question correctly, so do not make the same mistake. Also, always remember that acceleration due to gravity is the maximum at the surface of Earth and decreases with increase of height as well as increase in depth.
$g = G\dfrac{{{M_E}}}{{{R^2}}}$
Complete step by step answer:
We will be trying to solve the question exactly like we told in the hint section of the solution to this question. But for that, we first need to define acceleration due to gravity of the Earth and its mathematical expression.
Acceleration due to gravity of the Earth is the acceleration with which the Earth is pulling any or every body on its surface.
Mathematically, we can write acceleration due to gravity as:
$g = G\dfrac{{{M_E}}}{{{R^2}}}$
Where, $G$ is the gravitational constant,
${M_E}$ is the mass of Earth and,
$R$ is the radius of Earth.
The question has asked us about the altitude at which the acceleration due to gravity is only $25\% $ of that at earth’s surface. This can be mathematically represented as:
$g' = \dfrac{{25}}{{100}}g = \dfrac{g}{4}$
But we can also write $g'$ as:
$g' = G\dfrac{{{M_E}}}{{{{\left( {R + h} \right)}^2}}}$
Equating both the equations, we get:
$\dfrac{g}{4} = G\dfrac{{{M_E}}}{{{{\left( {R + h} \right)}^2}}}$
We have already seen that $g = G\dfrac{{{M_E}}}{{{R^2}}}$
Substituting that here, we get:
$G\dfrac{{{M_E}}}{{4{R^2}}} = G\dfrac{{{M_E}}}{{{{\left( {R + h} \right)}^2}}}$
After cancelling out common terms on both sides, we get:
${\left( {R + h} \right)^2} = 4{R^2}$
Solving this, we get:
$
R + h = 2R \\
h = R \\
$
We can see that the value of altitude for each of the said conditions meet is the same as that of option (B). Hence, option (B) is the correct answer.
Note: Many students tick the option with the value of $\left( {R + h} \right)$ instead of $h$ which causes them to lose marks even if they solved the question correctly, so do not make the same mistake. Also, always remember that acceleration due to gravity is the maximum at the surface of Earth and decreases with increase of height as well as increase in depth.
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