Question

The variation of g with height or depth (r) is shown correctly by the graph in figure (where R is the radius of the earth).
A.

B.

C.

D.

Answer 1

Hint: For this question we start with the equation for g with respect to distance r from the center of the earth with mass M and radius R that is $g = \dfrac{{GM}}{{{r^2}}}$ . Then we find the expression of g when r< R that is ${g^1} = g\left( {1 - \dfrac{{2r}}{R}} \right)$ and when r>R that is ${g^1} = g{\left( {1 + \dfrac{r}{R}} \right)^{ - 2}}$ from these we get the relation of g with respect to r as $ \Rightarrow {g^1} \propto \left( {R - 2r} \right)$ and $ \Rightarrow {g^1} \propto \dfrac{1}{{{r^2}}}$ respectively.

Complete Step-by-Step solution:
We know that the acceleration due to gravity is denoted by the symbol g and is defined as the acceleration of an object because of the force acting on that object by the gravitational field of Earth. This force is written as
$F = mg$
Here m is the mass of the object. If $m = 1$, we get
$F = g$
This also says that g is the force on the unit mass. The g is independent of the mass of the body and can be accurately calculated in various ways.
Now we will try to see the variation of g with respect to height h. So let us assume an object of mass m, then the measure of the gravitational field strength $g$, at a distance r from the center of the earth with mass $M$ is given by:
$g = \dfrac{{GM}}{{{r^2}}}$
Now take the object at the surface of the earth that is r=R as shown in figure 1.

Figure 1

Here R is the radius of the earth. We get
$g = \dfrac{{GM}}{{{R^2}}}$-------------------------------- (1)
Now we assume that the object is at a height H from the surface of the earth as shown in figure 2.

Figure 2

${g^1} = \dfrac{{GM}}{{{{\left( {R + H} \right)}^2}}}$--------------------------- (2)
Now dividing equation (2) by (1) we get
${g^1} = \dfrac{{g{R^2}}}{{{{\left( {R + H} \right)}^2}}}$
$ \Rightarrow {g^1} = \dfrac{g}{{{{\left( {1 + \dfrac{H}{R}} \right)}^2}}}$
$ \Rightarrow {g^1} = g{\left( {1 + \dfrac{H}{R}} \right)^{ - 2}}$-------------------------(3)
This expression is used when $\dfrac{H}{R} \geqslant 1$ that is when the object is at a distance greater than the earth radius.
But when $\dfrac{H}{R} < 1$ we can Apply the binomial expansion and we will get

$ \Rightarrow {g^1} = g\left( {1 - \dfrac{{2H}}{R}} \right)$--------------------------- (4)
From these we can conclude that
Region 1: When $r \leqslant R$ we can write
$ \Rightarrow {g^1} = g\left( {1 - \dfrac{{2r}}{R}} \right)$
$ \Rightarrow {g^1} \propto \left( {R - 2r} \right)$
Region 2: When $r \geqslant R$ we can write
$ \Rightarrow {g^1} = g{\left( {1 + \dfrac{r}{R}} \right)^{ - 2}}$
$ \Rightarrow {g^1} \propto \dfrac{1}{{{r^2}}}$
So the graph we get is shown in figure 3

                              Figure 3

Hence option A is correct.

Note: We can also solve this question in a different way that is we get the expression of g with respect to r when $r < R$ as $g = \dfrac{{GM}}{{{R^3}}}r$ and when $r > R$ we get $g = \dfrac{{GM}}{{{r^2}}}$. By using these also we can plot the required graph.