Compare the weight of a $5kg$ body $10km$ above and $10km$ below the surface of earth. Given: radius of earth $=6400km$
Answer
603.9k+ views
Hint: Here we need to apply the formulas for change in acceleration due to gravity in a height and depth from the surface of the earth. Then we will compare the weights in both the cases. In both the cases the value of acceleration due to gravity is less than that on the surface of the earth.
Formula used: ${g}'=g(1-\dfrac{2h}{{{R}_{e}}}),{g}''=g(1-\dfrac{d}{{{R}_{e}}})$
Complete step by step answer:
If $g$ and ${g}'$ are the value of acceleration due to gravity on the surface of the earth and at a height$h$ respectively and ${{R}_{e}}$ be the radius of the earth then
${g}'=g{{(1+\dfrac{h}{{{R}_{e}}})}^{-2}}$Now if $h$ is negligible compared to ${{R}_{e}}$ from binomial theorem we can write
${g}'=g(1-\dfrac{2h}{{{R}_{e}}})$, Now if ${W}'$ is the weight of a body a height $h$ ,it is given by
${W}'=mg(1-\dfrac{2h}{{{R}_{e}}})..........(1)$
Now if $g$ and ${g}''$ are the value of acceleration due to gravity on the surface of the earth and at a depth$d$ respectively and ${{R}_{e}}$ be the radius of the earth then
${g}''=g(1-\dfrac{d}{{{R}_{e}}})$, Now if ${W}''$ is the weight of a at a depth $h$,then it is given by
${W}''=mg(1-\dfrac{d}{{{R}_{e}}})..........(2)$
Now dividing $(1)$ by$(2)$ we have
$\dfrac{{{W}'}}{{{W}''}}=\dfrac{1-\dfrac{2h}{{{R}_{e}}}}{1-\dfrac{d}{{{R}_{e}}}}=\dfrac{{{R}_{e}}-2h}{{{R}_{e}}-d}$
Now putting the values of all the quantities we get
$\begin{align}
& \dfrac{{{W}'}}{{{W}''}}=\dfrac{6400-2\times 10}{6400-10}=\dfrac{638}{639} \\
& or{W}':{W}''=638:639 \\
\end{align}$
Note: The value of acceleration due to gravity decreases both for a height and a depth from the surface of the earth. If the height is not negligible compared to the radius of the earth then we need to use the general formula ${g}'=g{{(1+\dfrac{h}{{{R}_{e}}})}^{-2}}$ for a depth of $h$ the body experiences gravitational attraction only due to the inner solid sphere, that is why the acceleration due to gravity also decreases with depth. For a height from the surface of the earth, the acceleration due to gravity decreases due to a larger distance.
Formula used: ${g}'=g(1-\dfrac{2h}{{{R}_{e}}}),{g}''=g(1-\dfrac{d}{{{R}_{e}}})$
Complete step by step answer:
If $g$ and ${g}'$ are the value of acceleration due to gravity on the surface of the earth and at a height$h$ respectively and ${{R}_{e}}$ be the radius of the earth then
${g}'=g{{(1+\dfrac{h}{{{R}_{e}}})}^{-2}}$Now if $h$ is negligible compared to ${{R}_{e}}$ from binomial theorem we can write
${g}'=g(1-\dfrac{2h}{{{R}_{e}}})$, Now if ${W}'$ is the weight of a body a height $h$ ,it is given by
${W}'=mg(1-\dfrac{2h}{{{R}_{e}}})..........(1)$
Now if $g$ and ${g}''$ are the value of acceleration due to gravity on the surface of the earth and at a depth$d$ respectively and ${{R}_{e}}$ be the radius of the earth then
${g}''=g(1-\dfrac{d}{{{R}_{e}}})$, Now if ${W}''$ is the weight of a at a depth $h$,then it is given by
${W}''=mg(1-\dfrac{d}{{{R}_{e}}})..........(2)$
Now dividing $(1)$ by$(2)$ we have
$\dfrac{{{W}'}}{{{W}''}}=\dfrac{1-\dfrac{2h}{{{R}_{e}}}}{1-\dfrac{d}{{{R}_{e}}}}=\dfrac{{{R}_{e}}-2h}{{{R}_{e}}-d}$
Now putting the values of all the quantities we get
$\begin{align}
& \dfrac{{{W}'}}{{{W}''}}=\dfrac{6400-2\times 10}{6400-10}=\dfrac{638}{639} \\
& or{W}':{W}''=638:639 \\
\end{align}$
Note: The value of acceleration due to gravity decreases both for a height and a depth from the surface of the earth. If the height is not negligible compared to the radius of the earth then we need to use the general formula ${g}'=g{{(1+\dfrac{h}{{{R}_{e}}})}^{-2}}$ for a depth of $h$ the body experiences gravitational attraction only due to the inner solid sphere, that is why the acceleration due to gravity also decreases with depth. For a height from the surface of the earth, the acceleration due to gravity decreases due to a larger distance.
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