Question

Where will $'g'$ be greatest when one goes from the centre of earth to an altitude equal to the radius of the earth?
A) At the surface of earth
B) At the centre of earth
C) At the height point
D) None of the above

Answer 1

Hint: Always remember that the value of acceleration due to gravity is different at different places. To solve this question, we have to use the formulae of the acceleration due to gravity at different places of the earth surface. We just have to compare the values so obtained and answer the question accordingly.

Formulae used:
${g_h} = \dfrac{g}{{{{\left( {1 + \dfrac{h}{{{R_e}}}} \right)}^2}}}$
Here ${g_h}$ is the acceleration due to gravity of the earth at a given height above the earth’s, $g$ is the acceleration due to gravity of the earth, $h$ is the height above the earth’s surface at which we have to calculate acceleration due to gravity and ${R_e}$ is the radius of the earth surface.
${g_h} = \dfrac{{GMr}}{{{R_e}^3}}$
Here ${g_h}$ is the acceleration due to gravity of the earth at given height below the earth’s surface, $G$ is the gravitational constant, $M$ is the mass of the earth and ${R_e}$ is the radius of the earth surface.

Complete step by step solution:
Let’s calculate the value of acceleration due to gravity at each of the given surfaces.
At the surface of earth
We know that,
$ \Rightarrow {g_h} = \dfrac{g}{{{{\left( {1 + \dfrac{h}{{{R_e}}}} \right)}^2}}}$
Here ${g_h}$ is the acceleration due to gravity of the earth at given height above the earth’s, $g$ is the acceleration due to gravity of the earth, $h$is the height above the earth’s surface at which we have to calculate acceleration due to gravity and ${R_e}$ is the radius of the earth surface.
As we are calculating the value of $g$ at the surface,
$ \Rightarrow h = 0$
Putting the values of $h$ in the above equation, we get
$ \Rightarrow {g_h} = \dfrac{g}{{{{\left( {1 + \dfrac{0}{{{R_e}}}} \right)}^2}}} = g$
Hence $g$ will remain the same at surface.
At the centre of earth
We know that,
$ \Rightarrow {g_h} = \dfrac{{GMr}}{{{R_e}^3}}$

Here ${g_h}$ is the acceleration due to gravity of the earth at given height below the earth’s surface, $G$ is the gravitational constant, $r$ is the distance from the earth's centre, $M$ is the mass of the sphere with radius $r$ and ${R_e}$ is the radius of the earth surface.
At the centre of the earth surface,
$ \Rightarrow r = 0$
Putting the values of $r$ in the above equation, we get
$ \Rightarrow {g_h} = \dfrac{{GM0}}{{{R_e}^3}} = 0$
Hence $g$ will be 0 at the centre.
At the height point
We know that,
$ \Rightarrow {g_h} = \dfrac{g}{{{{\left( {1 + \dfrac{h}{{{R_e}}}} \right)}^2}}}$
Here ${g_h}$ is the acceleration due to gravity of the earth at given height above the earth’s, $g$ is the acceleration due to gravity of the earth, $h$ is the height above the earth’s surface at which we have to calculate acceleration due to gravity and ${R_e}$ is the radius of the earth surface.
At the highest point,
$ \Rightarrow h = 8.9Km$
Putting the values of $h$ in the above equation, we get
$ \Rightarrow {g_h} = \dfrac{g}{{{{\left( {1 + \dfrac{{8.9}}{{6400}}} \right)}^2}}} = 0.99g$
Hence there will be a slight change in $g$
Hence we can say that $g$ at surface will be max i.e. $g$.

So, option (A) is the correct answer.

Note: While calculating the acceleration due to gravity at any given point, we should always use the correct formulae. Also always remember acceleration due to gravity will be maximum at the surface of the earth i.e. at $6400Km$ from the centre and decreases as the height increases or as we move towards the centre of the earth.