
State expression for acceleration due to gravity at depth ‘d’ and altitude ‘h’. Hence show that ratio is equal to $\dfrac{{R - d}}{{R - 2h}}$ , assuming that $h
Answer
485.1k+ views
Hint To solve this problem, the first step is to find the acceleration due to gravity at a depth $d$ and a height $h$ from the earth's surface. In both cases, using the universal law of gravitation obtains the relation between the force experienced by the body at a height and depth at which the body is relative to the surface of the earth in terms of the acceleration due to gravity.
Complete Step-by-step solution
Since Force acting on the test mass due to gravity at depth d is:
$F = \dfrac{{GM\left( {R - d} \right)m}}{{{R^3}}}$
$ \Rightarrow m.\dfrac{{GM}}{{{R^2}}}\left( {1 - \dfrac{d}{R}} \right)$
$ \Rightarrow m.g\left( {1 - \dfrac{d}{R}} \right)$ $\left[ {\because g = \dfrac{{GM}}{{{R^2}}}} \right]$
$ \Rightarrow m.{g_{depth}}$
Where the acceleration due to gravity at a depth of d:
${g_{depth}} = g\left( {1 - \dfrac{d}{R}} \right)$
Similarly, Force acting on the test mass due to gravity at height \[h\] is:
$F = \dfrac{{GMm}}{{{{\left( {R + h} \right)}^2}}}$
$ = m.\left\{ {\dfrac{{GM}}{{{{\left( {R + h} \right)}^2}}}} \right\}$
$ = \dfrac{{GMm}}{{{R^2}}}.\dfrac{1}{{{{\left( {1 + \dfrac{h}{R}} \right)}^2}}}$
On solving the above equation we get,
$ \Rightarrow \dfrac{{GMm}}{{{R^2}}}\left( {1 - \dfrac{{2h}}{R}} \right)$
$ \Rightarrow m.g\left( {1 - \dfrac{{2h}}{R}} \right)$ $\left[ {\because g = \dfrac{{GM}}{{{R^2}}}} \right]$
$ \Rightarrow m.{g_{height}}$
Where the acceleration due to gravity at a height h :
${g_{height}} = g\left( {1 - \dfrac{{2h}}{R}} \right)$
Now taking the ratio of the equation $\left( 1 \right)$ and $\left( 2 \right)$we get
$\dfrac{{{g_{depth}}}}{{{g_{height}}}} = \dfrac{{g\left( {1 - \dfrac{d}{R}} \right)}}{{g\left( {1 - \dfrac{{2h}}{R}} \right)}}$
$ = \dfrac{{\left( {1 - \dfrac{d}{R}} \right)}}{{\left( {1 - \dfrac{{2h}}{R}} \right)}}$
$ \Rightarrow \dfrac{{R - d}}{{R - 2h}}$
Hence the ratio is $\dfrac{{R - d}}{{R - 2h}}$
Additional Information
Any two bodies in the universe exert a force of attraction on each other. This force of attraction is called the gravitational force. The two forces have the same magnitude but opposite in direction. The direction of the force exerted by the body $1$ on the body $2$ is always directed towards the body $1$ and vice versa. Therefore, the gravitational force is given $F = G\dfrac{{{M_1}{M_2}}}{{{r^2}}}$ .
Note As we know that earth exerts a force on us known as the gravitational force due to which anything goes above or below the surface of the earth experiences this force of attraction towards the center of the earth and due to which there is an acceleration called acceleration due to gravity whose value is $9.8m{s^{ - 2}}$. But this value varies with the variation of our position below the earth's surface or above it.
Complete Step-by-step solution
Since Force acting on the test mass due to gravity at depth d is:
$F = \dfrac{{GM\left( {R - d} \right)m}}{{{R^3}}}$
$ \Rightarrow m.\dfrac{{GM}}{{{R^2}}}\left( {1 - \dfrac{d}{R}} \right)$
$ \Rightarrow m.g\left( {1 - \dfrac{d}{R}} \right)$ $\left[ {\because g = \dfrac{{GM}}{{{R^2}}}} \right]$
$ \Rightarrow m.{g_{depth}}$
Where the acceleration due to gravity at a depth of d:
${g_{depth}} = g\left( {1 - \dfrac{d}{R}} \right)$
Similarly, Force acting on the test mass due to gravity at height \[h\] is:
$F = \dfrac{{GMm}}{{{{\left( {R + h} \right)}^2}}}$
$ = m.\left\{ {\dfrac{{GM}}{{{{\left( {R + h} \right)}^2}}}} \right\}$
$ = \dfrac{{GMm}}{{{R^2}}}.\dfrac{1}{{{{\left( {1 + \dfrac{h}{R}} \right)}^2}}}$
On solving the above equation we get,
$ \Rightarrow \dfrac{{GMm}}{{{R^2}}}\left( {1 - \dfrac{{2h}}{R}} \right)$
$ \Rightarrow m.g\left( {1 - \dfrac{{2h}}{R}} \right)$ $\left[ {\because g = \dfrac{{GM}}{{{R^2}}}} \right]$
$ \Rightarrow m.{g_{height}}$
Where the acceleration due to gravity at a height h :
${g_{height}} = g\left( {1 - \dfrac{{2h}}{R}} \right)$
Now taking the ratio of the equation $\left( 1 \right)$ and $\left( 2 \right)$we get
$\dfrac{{{g_{depth}}}}{{{g_{height}}}} = \dfrac{{g\left( {1 - \dfrac{d}{R}} \right)}}{{g\left( {1 - \dfrac{{2h}}{R}} \right)}}$
$ = \dfrac{{\left( {1 - \dfrac{d}{R}} \right)}}{{\left( {1 - \dfrac{{2h}}{R}} \right)}}$
$ \Rightarrow \dfrac{{R - d}}{{R - 2h}}$
Hence the ratio is $\dfrac{{R - d}}{{R - 2h}}$
Additional Information
Any two bodies in the universe exert a force of attraction on each other. This force of attraction is called the gravitational force. The two forces have the same magnitude but opposite in direction. The direction of the force exerted by the body $1$ on the body $2$ is always directed towards the body $1$ and vice versa. Therefore, the gravitational force is given $F = G\dfrac{{{M_1}{M_2}}}{{{r^2}}}$ .
Note As we know that earth exerts a force on us known as the gravitational force due to which anything goes above or below the surface of the earth experiences this force of attraction towards the center of the earth and due to which there is an acceleration called acceleration due to gravity whose value is $9.8m{s^{ - 2}}$. But this value varies with the variation of our position below the earth's surface or above it.
Recently Updated Pages
Master Class 9 General Knowledge: Engaging Questions & Answers for Success

Earth rotates from West to east ATrue BFalse class 6 social science CBSE

The easternmost longitude of India is A 97circ 25E class 6 social science CBSE

Write the given sentence in the passive voice Ann cant class 6 CBSE

Convert 1 foot into meters A030 meter B03048 meter-class-6-maths-CBSE

What is the LCM of 30 and 40 class 6 maths CBSE

Trending doubts
What is the difference between superposition and e class 11 physics CBSE

State and prove Bernoullis theorem class 11 physics CBSE

Which one is a true fish A Jellyfish B Starfish C Dogfish class 11 biology CBSE

1 ton equals to A 100 kg B 1000 kg C 10 kg D 10000 class 11 physics CBSE

State the laws of reflection of light

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE
