Answer
Verified
387.9k+ views
Hint: Gravity is the force which pulls the objects down towards the ground and acceleration produced by this force is called the acceleration due to gravity.Gravitational acceleration is a quantity of vector, that is it has both magnitude and direction.
Complete step by step answer:
Acceleration due to gravity at the surface of earth is approximately \[9.8\,m/{{s}^{2}}\]. It is not the same everywhere. It keeps on decreasing as we go above the surface of earth.
As we go below the surface of earth, the variation in g(acceleration due to gravity) is given by:
\[g'=g\left( 1-\dfrac{h}{{{R}_{e}}} \right)\]
where,
\[g'=\] Actual acceleration due to gravity at a depth ‘h’ from the surface of earth
\[g=\] Acceleration due to gravity at the surface of earth \[=9.8m/{{s}^{2}}\]
\[h=\] Distance from the surface of the earth
\[\operatorname{R_e}=\] Radius of the earth
Now, at the centre of earth, \[h=\operatorname{Re}\]
\[g'=g\left( 1-\dfrac{{{R}_{e}}}{{{R}_{e}}} \right)\]
\[\therefore g'=0\]
Therefore, acceleration due to gravity at the centre of the earth is \[0\] and it keeps on increasing till we reach at the surface of the earth and after that it keeps on decreasing.
Note: Logically, we can understand this by, when we move inside the earth, the mass that exerts gravitational force on us decreases and hence at the centre of the earth the acceleration due to gravity becomes zero. Variation of g as we go above the surface of the earth is given by \[g'=\dfrac{g}{{{\left( 1+\dfrac{h}{{{R}_{e}}} \right)}^{2}}}\].
Complete step by step answer:
Acceleration due to gravity at the surface of earth is approximately \[9.8\,m/{{s}^{2}}\]. It is not the same everywhere. It keeps on decreasing as we go above the surface of earth.
As we go below the surface of earth, the variation in g(acceleration due to gravity) is given by:
\[g'=g\left( 1-\dfrac{h}{{{R}_{e}}} \right)\]
where,
\[g'=\] Actual acceleration due to gravity at a depth ‘h’ from the surface of earth
\[g=\] Acceleration due to gravity at the surface of earth \[=9.8m/{{s}^{2}}\]
\[h=\] Distance from the surface of the earth
\[\operatorname{R_e}=\] Radius of the earth
Now, at the centre of earth, \[h=\operatorname{Re}\]
\[g'=g\left( 1-\dfrac{{{R}_{e}}}{{{R}_{e}}} \right)\]
\[\therefore g'=0\]
Therefore, acceleration due to gravity at the centre of the earth is \[0\] and it keeps on increasing till we reach at the surface of the earth and after that it keeps on decreasing.
Note: Logically, we can understand this by, when we move inside the earth, the mass that exerts gravitational force on us decreases and hence at the centre of the earth the acceleration due to gravity becomes zero. Variation of g as we go above the surface of the earth is given by \[g'=\dfrac{g}{{{\left( 1+\dfrac{h}{{{R}_{e}}} \right)}^{2}}}\].
Recently Updated Pages
Who among the following was the religious guru of class 7 social science CBSE
what is the correct chronological order of the following class 10 social science CBSE
Which of the following was not the actual cause for class 10 social science CBSE
Which of the following statements is not correct A class 10 social science CBSE
Which of the following leaders was not present in the class 10 social science CBSE
Garampani Sanctuary is located at A Diphu Assam B Gangtok class 10 social science CBSE
Trending doubts
Derive an expression for drift velocity of free electrons class 12 physics CBSE
Which are the Top 10 Largest Countries of the World?
Write down 5 differences between Ntype and Ptype s class 11 physics CBSE
The energy of a charged conductor is given by the expression class 12 physics CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Derive an expression for electric field intensity due class 12 physics CBSE
How do you graph the function fx 4x class 9 maths CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
Derive an expression for electric potential at point class 12 physics CBSE