
The weight of the body at the center of the earth is:
A. Zero
B. Infinite
C. Same as on the surface on earth
D. None of these
Answer
575.1k+ views
Hint:We know that weight of the body is a force by which it is attracted towards the center of the earth. Here, in this problem we need to use two important concepts. The first is the relation between force (weight) and gravitational acceleration which is as per Newton’s second law of motion. Second, we have to consider how the gravitational acceleration changes inside the earth surface.
Formulas used:
$W = mg$
Where $W$ is weight of the body, $m$ is mass of the body and is the gravitational acceleration
${g_r} = g\left( {1 - \dfrac{r}{R}} \right)$
Where ${g_r}$ is gravitational acceleration at distance $r$ from the surface of the earth, $r$ is the distance of the body from the surface of the earth and $R$ is the radius of the earth.
Complete step by step solution:
We know that as per the definition of weight $W = mg$ which clearly indicates that weight of the body is dependent on mass of the body and gravitational acceleration.
Now mass of the body remains constant at any place. However, gravitational acceleration inside the earth’s surface changes with respect to the distance from the earth surface, which is given by ${g_r} = g\left( {1 - \dfrac{r}{R}} \right)$.
Here, we are given that the body is at the center of the earth. Therefore, its distance from the surface of the earth will be the same as the radius of the earth.
\[
r = R \\
\Rightarrow {g_r} = g\left( {1 - \dfrac{R}{R}} \right) \\
\Rightarrow {g_r} = g\left( {1 - 1} \right) \\
\therefore {g_r} = 0m/{s^2}\]
Thus, there is no gravitational acceleration at the center of the earth.
Now, $W = mg$
Putting $g = {g_r}$ for the given case, we get
$W = m{g_r}$
But we have seen that ${g_r} = 0m/{s^2}$. Therefore, $W = 0N$. Thus, at the center of the earth, the body has no weight.
Hence, option A is the right answer.
Note:We have used the formula ${g_r} = g\left( {1 - \dfrac{r}{R}} \right)$ which indicates that as we go below the surface of the earth, gravitational acceleration decreases which in turn decreases the weight of the body. From this, we can say that the weight of the body is maximum on the earth’s surface and zero at the center of the earth.
Formulas used:
$W = mg$
Where $W$ is weight of the body, $m$ is mass of the body and is the gravitational acceleration
${g_r} = g\left( {1 - \dfrac{r}{R}} \right)$
Where ${g_r}$ is gravitational acceleration at distance $r$ from the surface of the earth, $r$ is the distance of the body from the surface of the earth and $R$ is the radius of the earth.
Complete step by step solution:
We know that as per the definition of weight $W = mg$ which clearly indicates that weight of the body is dependent on mass of the body and gravitational acceleration.
Now mass of the body remains constant at any place. However, gravitational acceleration inside the earth’s surface changes with respect to the distance from the earth surface, which is given by ${g_r} = g\left( {1 - \dfrac{r}{R}} \right)$.
Here, we are given that the body is at the center of the earth. Therefore, its distance from the surface of the earth will be the same as the radius of the earth.
\[
r = R \\
\Rightarrow {g_r} = g\left( {1 - \dfrac{R}{R}} \right) \\
\Rightarrow {g_r} = g\left( {1 - 1} \right) \\
\therefore {g_r} = 0m/{s^2}\]
Thus, there is no gravitational acceleration at the center of the earth.
Now, $W = mg$
Putting $g = {g_r}$ for the given case, we get
$W = m{g_r}$
But we have seen that ${g_r} = 0m/{s^2}$. Therefore, $W = 0N$. Thus, at the center of the earth, the body has no weight.
Hence, option A is the right answer.
Note:We have used the formula ${g_r} = g\left( {1 - \dfrac{r}{R}} \right)$ which indicates that as we go below the surface of the earth, gravitational acceleration decreases which in turn decreases the weight of the body. From this, we can say that the weight of the body is maximum on the earth’s surface and zero at the center of the earth.
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