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Hint
The Relation between $g$ that is gravitational acceleration and the height which is distance from Centre of the earth $(R + h)$ where $R$ is radius of earth. Compare $2$ gravitational values to obtain for height $h$.
Complete Step By Step Solution
Let’s first begin with, what is the relation between the gravitational acceleration and the radius of the earth. the gravitational acceleration is given by ${\text{g}} = \dfrac{{{\text{GM}}}}{{{{\text{R}}^{\text{2}}}}},$ where $G$ is Gravitational constant equals $6.67 \times {10^{ - 11}} \dfrac{N}{{kg^2}{m^2}}$.
Whereas, $M$ is mass and $R$ is the distance of an object from core Centre, here it’s Radius as the height is termed as negligible. If we have a height $h$, the near gravitational acceleration for that body will act as-
${{\text{g}}_1} = \dfrac{{{\text{GM}}}}{{{{\left( {{\text{R}} + {\text{h}}} \right)}^{\text{2}}}}}$
If we compare both the value, we get-
$\dfrac{{\text{g}}}{{{\text{g1}}}} = \dfrac{{{\text{GM}}}}{{{{\text{R}}^{\text{2}}}}} \times \dfrac{{{{\left( {{\text{R}} + {\text{h}}} \right)}^{\text{2}}}}}{{{\text{GM}}}} = {\left[ {{\text{1}} + \dfrac{{\text{h}}}{{\text{R}}}} \right]^2}$
We can say, ${{\text{g}}_{\text{1}}} = {\text{g}}{\left( {1 + \dfrac{{\text{h}}}{{\text{g}}}} \right)^{ - 2}}$
By cross multiplication, as we can see the value of $R$ that is radius of the earth is large to get $\dfrac{{\text{h}}}{{\text{R}}}$ a small value enough. So the binomial expansion of ${\left( {1 + y} \right)^{ - 2}}$ for small y is $\left( {1 - 2y} \right)$ Similarly, here it would be equal to-
${{\text{g}}_{\text{1}}} = {\text{g}}\left( {1 - \dfrac{{{\text{2h}}}}{{\text{R}}}} \right)$
Hence the value of gravitational acceleration will decrease with the increase in height of the object.
Note
We should know the relation between gravitational acceleration, Mass, Radius and it’s Constant which is $6.67 \times {10^{ - 11}} N {kg}^{−2} {m}^{−2}$. The binomial expansion for small value tending to zero is mainly used in many forms to get a reduced form equation.
The Relation between $g$ that is gravitational acceleration and the height which is distance from Centre of the earth $(R + h)$ where $R$ is radius of earth. Compare $2$ gravitational values to obtain for height $h$.
Complete Step By Step Solution
Let’s first begin with, what is the relation between the gravitational acceleration and the radius of the earth. the gravitational acceleration is given by ${\text{g}} = \dfrac{{{\text{GM}}}}{{{{\text{R}}^{\text{2}}}}},$ where $G$ is Gravitational constant equals $6.67 \times {10^{ - 11}} \dfrac{N}{{kg^2}{m^2}}$.
Whereas, $M$ is mass and $R$ is the distance of an object from core Centre, here it’s Radius as the height is termed as negligible. If we have a height $h$, the near gravitational acceleration for that body will act as-
${{\text{g}}_1} = \dfrac{{{\text{GM}}}}{{{{\left( {{\text{R}} + {\text{h}}} \right)}^{\text{2}}}}}$
If we compare both the value, we get-
$\dfrac{{\text{g}}}{{{\text{g1}}}} = \dfrac{{{\text{GM}}}}{{{{\text{R}}^{\text{2}}}}} \times \dfrac{{{{\left( {{\text{R}} + {\text{h}}} \right)}^{\text{2}}}}}{{{\text{GM}}}} = {\left[ {{\text{1}} + \dfrac{{\text{h}}}{{\text{R}}}} \right]^2}$
We can say, ${{\text{g}}_{\text{1}}} = {\text{g}}{\left( {1 + \dfrac{{\text{h}}}{{\text{g}}}} \right)^{ - 2}}$
By cross multiplication, as we can see the value of $R$ that is radius of the earth is large to get $\dfrac{{\text{h}}}{{\text{R}}}$ a small value enough. So the binomial expansion of ${\left( {1 + y} \right)^{ - 2}}$ for small y is $\left( {1 - 2y} \right)$ Similarly, here it would be equal to-
${{\text{g}}_{\text{1}}} = {\text{g}}\left( {1 - \dfrac{{{\text{2h}}}}{{\text{R}}}} \right)$
Hence the value of gravitational acceleration will decrease with the increase in height of the object.
Note
We should know the relation between gravitational acceleration, Mass, Radius and it’s Constant which is $6.67 \times {10^{ - 11}} N {kg}^{−2} {m}^{−2}$. The binomial expansion for small value tending to zero is mainly used in many forms to get a reduced form equation.
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