Answer
450.6k+ views
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.
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.
${g^1} = \dfrac{{GM}}{{{{\left( {R + H} \right)}^2}}}$--------------------------- (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.
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.
![seo images](https://www.vedantu.com/question-sets/bd1aa226-a07a-4ad7-a67c-e6839c60007f6476190980754643710.png)
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.
![seo images](https://www.vedantu.com/question-sets/60e4bde9-17e2-433d-a85c-baecf8d78ce12962137295021860158.png)
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
![seo images](https://www.vedantu.com/question-sets/042d72d5-d8be-48ea-93b5-07bc2ca24ed61832712440779795695.png)
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.
Recently Updated Pages
Mark and label the given geoinformation on the outline class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
When people say No pun intended what does that mea class 8 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Name the states which share their boundary with Indias class 9 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give an account of the Northern Plains of India class 9 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Advantages and disadvantages of science
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
10 examples of evaporation in daily life with explanations
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
How do you graph the function fx 4x class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
What are the monomers and polymers of carbohydrate class 12 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)