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 $g={\displaystyle \frac{GM}{r^2}}$ . Then we find the expression of g when r< R that is $g^1=g\left(1-{\displaystyle \frac{2r}{R}}\right)$ and when r>R that is $g^1=g{\left(1+{\displaystyle \frac{r}{R}}\right)}^{-2}$ from these we get the relation of g with respect to r as $\Rightarrow g^1\propto \left(R-2r\right)$ and $\Rightarrow g^1\propto {\displaystyle \frac{1}{r^2}}$ 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:
$g={\displaystyle \frac{GM}{r^2}}$
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
$g={\displaystyle \frac{GM}{R^2}}$-------------------------------- (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

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

$\Rightarrow g^1=g\left(1-{\displaystyle \frac{2H}{R}}\right)$--------------------------- (4)
From these we can conclude that
Region 1: When $r⩽R$ we can write
$\Rightarrow g^1=g\left(1-{\displaystyle \frac{2r}{R}}\right)$
$\Rightarrow g^1\propto \left(R-2r\right)$
Region 2: When $r⩾R$ we can write
$\Rightarrow g^1=g{\left(1+{\displaystyle \frac{r}{R}}\right)}^{-2}$
$\Rightarrow g^1\propto {\displaystyle \frac{1}{r^2}}$
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 $g={\displaystyle \frac{GM}{R^3}}r$ and when $r>R$ we get $g={\displaystyle \frac{GM}{r^2}}$. By using these also we can plot the required graph.