Which is the following is/are true about the value of $g$
A) The value of $g$ is minimum at the equator
B) The value of $g$ is maximum at the poles
C) Average value of $g$ is $9.8\,m{s^{ - 2}}$
D) All the above
Answer
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Hint:The standard value of $g$ on the surface of earth is calculated considering earth as a homogeneous sphere. But in reality, earth is not an ideal homogeneous sphere. The earth’s radius is greater at the equator than at the poles. Due to this, acceleration due to gravity is variable.
Complete step by step solution:
We know that the value of $g$ is zero at the centre of the earth and then increases uniformly till the surface of the earth and then decreases exponentially.
Also, our earth is not an ideal homogeneous sphere. It has different radii for different points on its surface. Due to this, the radius of earth at the equator is greater than the radius of earth at the poles. Due to this variation in distance from the centre of the earth, the acceleration due to gravity also changes as it is a function dependent on the distance from the centre of the earth itself.
To solve this question, let's look at all these options individually.
For option A, we know that the equator of the earth has the greatest radius among all the points on the surface of the earth, in other words, the earth is the widest at the equator. At equator, the radius of the earth is even greater than the average value of radius of earth that we use in physical or mathematical calculations. Since, acceleration due to gravity is inversely proportional to the distance from the centre of the earth, equators will have the minimum value of $g$ compared to any other point on the surface of earth. Its value at the equator is $9.78\,m{s^{ - 2}}$. This option is correct.
For option B, it is opposite to the case in option A. Poles have a minimum radius compared to any other point on the surface of the earth. Therefore, acceleration due to gravity is maximum at this point. Its value is $9.83\,m{s^{ - 2}}$. This option is also correct.
For option C, we need to calculate the average value of $g$ on the surface of earth. It can be calculated by adding the above two extreme values of $g$ and dividing them by 2.
This gives us an average value of $g$ as $\dfrac{{9.78 + 9.83}}{2} = 9.805 \simeq 9.80\,m{s^{ - 2}}$. This option is also correct.
Hence, option D is the correct answer as the above options are correct.
Note: Earth’s radius is not constant throughout its surface. An average of its value is used in physical and mathematical calculations to make calculations easier. For an in-depth study of a particular place on earth, individual values of radius of earth and acceleration due to gravity are still used. Since we are not doing such a detailed study of earth, we will use the average values as provided to us.
Complete step by step solution:
We know that the value of $g$ is zero at the centre of the earth and then increases uniformly till the surface of the earth and then decreases exponentially.
Also, our earth is not an ideal homogeneous sphere. It has different radii for different points on its surface. Due to this, the radius of earth at the equator is greater than the radius of earth at the poles. Due to this variation in distance from the centre of the earth, the acceleration due to gravity also changes as it is a function dependent on the distance from the centre of the earth itself.
To solve this question, let's look at all these options individually.
For option A, we know that the equator of the earth has the greatest radius among all the points on the surface of the earth, in other words, the earth is the widest at the equator. At equator, the radius of the earth is even greater than the average value of radius of earth that we use in physical or mathematical calculations. Since, acceleration due to gravity is inversely proportional to the distance from the centre of the earth, equators will have the minimum value of $g$ compared to any other point on the surface of earth. Its value at the equator is $9.78\,m{s^{ - 2}}$. This option is correct.
For option B, it is opposite to the case in option A. Poles have a minimum radius compared to any other point on the surface of the earth. Therefore, acceleration due to gravity is maximum at this point. Its value is $9.83\,m{s^{ - 2}}$. This option is also correct.
For option C, we need to calculate the average value of $g$ on the surface of earth. It can be calculated by adding the above two extreme values of $g$ and dividing them by 2.
This gives us an average value of $g$ as $\dfrac{{9.78 + 9.83}}{2} = 9.805 \simeq 9.80\,m{s^{ - 2}}$. This option is also correct.
Hence, option D is the correct answer as the above options are correct.
Note: Earth’s radius is not constant throughout its surface. An average of its value is used in physical and mathematical calculations to make calculations easier. For an in-depth study of a particular place on earth, individual values of radius of earth and acceleration due to gravity are still used. Since we are not doing such a detailed study of earth, we will use the average values as provided to us.
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