Calculate the percentage decrease in the weight of the body when it is taken 32 km below the surface of the earth. (Radius of the earth = $6400$km).
Answer
645.3k+ views
Hint: Percentage Decrease in the weight is given by
= $\dfrac{{mg - mg'}}{{mg'}} \times 100$
We will initially determine the value of g’ and then substitute this value in the above equation to get the value of percentage decrease in the value of weight.
Complete step-by-step answer:
Given height (h) = $32\,km$and radius of earth (R) = $6400\,km$ and we also know that value for the acceleration of gravity (g) is $9.8\,m/{s^2}$.
We will first determine the value of acceleration due to gravity at the given height.
To calculate the gravity at certain height (g’), we can use below formula
g’ = $g \times \left( {1 - \dfrac{h}{R}} \right)$
Substituting all these values we get
$\Rightarrow$g’ = $9.8 \times \left( {1 - \dfrac{{32}}{{6400}}} \right)$
$\Rightarrow$g’ = $9.8 \times \left( {\dfrac{{6400 - 32}}{{6400}}} \right)$
On Subtracting
$\Rightarrow$g’ = $9.8 \times \left( {\dfrac{{6368}}{{6400}}} \right)$
$\Rightarrow$g’ = $9.8 \times 0.995$ $m/{s^2}$
On multiplying above equation, finally we get
$\Rightarrow$g’ = \[9.751\,m/{s^2}\]
We already know that percentage decrease in the weight is given by
$\dfrac{{mg - mg'}}{{mg}} \times 100$
Taking “m” common from above equation
$\dfrac{{g - g'}}{g} \times 100$
On substituting value of g = \[9.8\,m/{s^2}\] and g’ value derived from above
$\Rightarrow$$\dfrac{{9.8 - 9.751}}{{9.8}} \times 100\, = \,0.5\% $
Hence, the percentage decrease in the value of weight of a body is 0.5%.
Additional Information: The value of acceleration due to gravity is given by
$g \times \left( {1 - \dfrac{h}{R}} \right)$
from the formula above we can see that the value of g depends upon the mass enclosed by the Earth and inversely depends upon the square of the radius of this sphere. As we go inside the earth the radius of the sphere decreases so g should increase but as we go deep in the earth the mass enclosed by the sphere also decreases hence the overall effect on the value of acceleration due to gravity is decreased.
Note: The value of acceleration due to gravity changes due to height, due to depth and also due to the shape of the earth. Value of acceleration due to gravity is maximum on the surface of the earth As we go above or below the surface of the earth the value of acceleration due to gravity goes on decreasing.
= $\dfrac{{mg - mg'}}{{mg'}} \times 100$
We will initially determine the value of g’ and then substitute this value in the above equation to get the value of percentage decrease in the value of weight.
Complete step-by-step answer:
Given height (h) = $32\,km$and radius of earth (R) = $6400\,km$ and we also know that value for the acceleration of gravity (g) is $9.8\,m/{s^2}$.
We will first determine the value of acceleration due to gravity at the given height.
To calculate the gravity at certain height (g’), we can use below formula
g’ = $g \times \left( {1 - \dfrac{h}{R}} \right)$
Substituting all these values we get
$\Rightarrow$g’ = $9.8 \times \left( {1 - \dfrac{{32}}{{6400}}} \right)$
$\Rightarrow$g’ = $9.8 \times \left( {\dfrac{{6400 - 32}}{{6400}}} \right)$
On Subtracting
$\Rightarrow$g’ = $9.8 \times \left( {\dfrac{{6368}}{{6400}}} \right)$
$\Rightarrow$g’ = $9.8 \times 0.995$ $m/{s^2}$
On multiplying above equation, finally we get
$\Rightarrow$g’ = \[9.751\,m/{s^2}\]
We already know that percentage decrease in the weight is given by
$\dfrac{{mg - mg'}}{{mg}} \times 100$
Taking “m” common from above equation
$\dfrac{{g - g'}}{g} \times 100$
On substituting value of g = \[9.8\,m/{s^2}\] and g’ value derived from above
$\Rightarrow$$\dfrac{{9.8 - 9.751}}{{9.8}} \times 100\, = \,0.5\% $
Hence, the percentage decrease in the value of weight of a body is 0.5%.
Additional Information: The value of acceleration due to gravity is given by
$g \times \left( {1 - \dfrac{h}{R}} \right)$
from the formula above we can see that the value of g depends upon the mass enclosed by the Earth and inversely depends upon the square of the radius of this sphere. As we go inside the earth the radius of the sphere decreases so g should increase but as we go deep in the earth the mass enclosed by the sphere also decreases hence the overall effect on the value of acceleration due to gravity is decreased.
Note: The value of acceleration due to gravity changes due to height, due to depth and also due to the shape of the earth. Value of acceleration due to gravity is maximum on the surface of the earth As we go above or below the surface of the earth the value of acceleration due to gravity goes on decreasing.
Recently Updated Pages
Write any three differences between metals and nonmetals class 10 social science CBSE

Amit standing on a horizontal plane finds a bird flying class 10 maths CBSE

Two circles of radii 5 cm and 3 cm intersect at two class 10 maths CBSE

Solve the following i John and Jivanti together have class 10 maths CBSE

What is the relation between orthocenter circumcentre class 10 maths CBSE

Two plane mirrors are inclined at 70circ A ray incident class 10 physics CBSE

Trending doubts
Explain the Treaty of Vienna of 1815 class 10 social science CBSE

Which country is known as "The land of Fire and Ice"?

1 GB equals how many MB?

10 examples of evaporation in daily life with explanations

What is the full form of POSCO class 10 social science CBSE

Make a sketch of the human nerve cell What function class 10 biology CBSE

