Answer
347.4k+ views
Hint: We know that the weight of a body (or a person in this case) is the product of its mass and the acceleration due to gravity. The acceleration due to gravity changes when it moves from the surface of the earth, leading us to take a new value for the acceleration due to gravity.
Formula used:
\[W = mg\]
Complete step by step answer:
Let the new weight of the person be \[W'\]. It is given that the new value of weight will be one fourth the value of the initial weight and let the initial weight be \[W\]. This gives us the relation between them as: \[W' = \dfrac{W}{4}\].We know that the formula to find weight is: \[W = mg\]. This gives us the value of the new weight to be: \[W' = mg'\]. Substituting the values in the equation above, we have, \[mg' = \dfrac{{mg}}{4}\].Cancelling the mass of the person from both the sides, we arrive at the new value for the acceleration of gravity when the weight is one fourth that is, \[g' = \dfrac{g}{4}\].
Now, we need another equation to find the value of \[g'\] . Let \[d\] be the depth from the surface of the earth at which the weight is one fourth of its actual value. And the radius of earth is \[R = 6400\,km\].
\[g' = \dfrac{g}{4} = g\left( {1 - \dfrac{d}{R}} \right)\]
Cancelling out g from both of the sides we get,
\[\dfrac{1}{4} = 1 - \dfrac{d}{R}\]
The above equation leads us to believe that the value off \[\dfrac{d}{R} = \dfrac{3}{4}\]
We have the value of R to be \[R = 6400km\] . Substituting this for R we get the value of d as:
\[d = \dfrac{3}{4}R \\
\Rightarrow d= \dfrac{3}{4} \times 6400 \\
\therefore d= 4800\,km\]
Hence, the depth below the surface of the earth at which the weight of a person will be one-fourth of his weight at the surface of the earth will be \[4800\,km\].
Note: The value of acceleration due to gravity is changing as it moves from the surface to the interior. This leads us to take a new value for the acceleration due to gravity. Some of us often forget to assign the new value and that in conclusion gives us the wrong answer.
Formula used:
\[W = mg\]
Complete step by step answer:
Let the new weight of the person be \[W'\]. It is given that the new value of weight will be one fourth the value of the initial weight and let the initial weight be \[W\]. This gives us the relation between them as: \[W' = \dfrac{W}{4}\].We know that the formula to find weight is: \[W = mg\]. This gives us the value of the new weight to be: \[W' = mg'\]. Substituting the values in the equation above, we have, \[mg' = \dfrac{{mg}}{4}\].Cancelling the mass of the person from both the sides, we arrive at the new value for the acceleration of gravity when the weight is one fourth that is, \[g' = \dfrac{g}{4}\].
Now, we need another equation to find the value of \[g'\] . Let \[d\] be the depth from the surface of the earth at which the weight is one fourth of its actual value. And the radius of earth is \[R = 6400\,km\].
\[g' = \dfrac{g}{4} = g\left( {1 - \dfrac{d}{R}} \right)\]
Cancelling out g from both of the sides we get,
\[\dfrac{1}{4} = 1 - \dfrac{d}{R}\]
The above equation leads us to believe that the value off \[\dfrac{d}{R} = \dfrac{3}{4}\]
We have the value of R to be \[R = 6400km\] . Substituting this for R we get the value of d as:
\[d = \dfrac{3}{4}R \\
\Rightarrow d= \dfrac{3}{4} \times 6400 \\
\therefore d= 4800\,km\]
Hence, the depth below the surface of the earth at which the weight of a person will be one-fourth of his weight at the surface of the earth will be \[4800\,km\].
Note: The value of acceleration due to gravity is changing as it moves from the surface to the interior. This leads us to take a new value for the acceleration due to gravity. Some of us often forget to assign the new value and that in conclusion gives us the wrong answer.
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