Consider Earth to be a homogeneous sphere. Scientist A goes deep down in a mine and Scientist B goes high up in a balloon. The gravitational field measured by:
A) A goes on decreasing and that of B goes on increasing
B) B goes on increasing and that of A goes on increasing
C) Each decreases at the same rate
D) Each decreases at a different rate
Answer
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Hint: Let us get a brief idea about the gravitational field first. In physics, a gravitational field is a model used to explain the influence that a massive body extends into the space around itself, producing a force on another massive body. Thus, a gravitational field is used to explain gravitational phenomena and is measured in newtons per kilogram.
Formula Used:
\[{{g}_{h}}=g\left( 1-\dfrac{2h}{R} \right)\], \[{{g}_{d}}=g\left( 1-\dfrac{d}{R} \right)\]
Complete step by step solution:
Variation in the gravitational field of the earth is equivalent to the variation in the acceleration due to gravity and hence we will find out the variation in the value of acceleration due to gravity as one goes high up from the earth’s surface or as one goes deep down.
The variation of acceleration due to gravity with height above the earth’s surface is given as
\[{{g}_{h}}=\dfrac{g}{{{\left( 1+\dfrac{h}{R} \right)}^{2}}}\]
where \[{{g}_{h}}\] is the acceleration due to gravity at height \[h\], \[g\] is the acceleration due to gravity at the earth’s surface and \[R\] is the earth’s radius.
When the height is very less as compared to the earth’s radius, we can apply binomial approximation to the above equation, and we get
\[{{g}_{h}}=g\left( 1-\dfrac{2h}{R} \right)\] for h < < < R
Now, for variation in acceleration due to gravity with depth, we have
\[{{g}_{d}}=g\left( 1-\dfrac{d}{R} \right)\]
where \[{{g}_{d}}\] is the acceleration due to gravity at a depth \[d\] below the earth’s surface and the meaning of the other symbols have been discussed above.
Hence we can see that the variation with height as well as depth is decreasing but their rate of variation is different.
Hence (D) is the correct option.
Note: Besides variation due to height and depth from the earth’s surface, the acceleration due to gravity also varies with the rotation of the earth about its axis and the shape of the earth. Due to these variations, the gravitational field at the poles is maximum and the gravitational field at the equator is minimum.
Formula Used:
\[{{g}_{h}}=g\left( 1-\dfrac{2h}{R} \right)\], \[{{g}_{d}}=g\left( 1-\dfrac{d}{R} \right)\]
Complete step by step solution:
Variation in the gravitational field of the earth is equivalent to the variation in the acceleration due to gravity and hence we will find out the variation in the value of acceleration due to gravity as one goes high up from the earth’s surface or as one goes deep down.
The variation of acceleration due to gravity with height above the earth’s surface is given as
\[{{g}_{h}}=\dfrac{g}{{{\left( 1+\dfrac{h}{R} \right)}^{2}}}\]
where \[{{g}_{h}}\] is the acceleration due to gravity at height \[h\], \[g\] is the acceleration due to gravity at the earth’s surface and \[R\] is the earth’s radius.
When the height is very less as compared to the earth’s radius, we can apply binomial approximation to the above equation, and we get
\[{{g}_{h}}=g\left( 1-\dfrac{2h}{R} \right)\] for h < < < R
Now, for variation in acceleration due to gravity with depth, we have
\[{{g}_{d}}=g\left( 1-\dfrac{d}{R} \right)\]
where \[{{g}_{d}}\] is the acceleration due to gravity at a depth \[d\] below the earth’s surface and the meaning of the other symbols have been discussed above.
Hence we can see that the variation with height as well as depth is decreasing but their rate of variation is different.
Hence (D) is the correct option.
Note: Besides variation due to height and depth from the earth’s surface, the acceleration due to gravity also varies with the rotation of the earth about its axis and the shape of the earth. Due to these variations, the gravitational field at the poles is maximum and the gravitational field at the equator is minimum.
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