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Question: The weight of a body at the equator is equal to the body’s weight at a height h above the Earth’s su...

The weight of a body at the equator is equal to the body’s weight at a height h above the Earth’s surface. The value of h will be
A. ω2R2g\dfrac{{{\omega }^{2}}{{R}^{2}}}{g}
B. ω2R22g\dfrac{{{\omega }^{2}}{{R}^{2}}}{2g}
C. 2ω2R2g\dfrac{2{{\omega }^{2}}{{R}^{2}}}{g}
D. 2ω2R23g\dfrac{2{{\omega }^{2}}{{R}^{2}}}{3g}

Explanation

Solution

As a first step, you could recall the expression for acceleration due to gravity in terms of angular speed and then multiply it with mass m to get the weight of the body at equator. Now follow the same steps for finding the weight at height h and then equate it to the above expression and hence get the answer.

Formula used:
Acceleration due to gravity,
g′=g−ω2Rg'=g-{{\omega }^{2}}R
g′=g(1−2hR)g'=g\left( 1-\dfrac{2h}{R} \right)

Complete step-by-step answer:
As a first step, we will have to identify the physical quantity of which we are to find the expression. We see that the options suggest that the quantity h is related to gravitation. So, it would probably be the height above the earth.
Now, let us recall the relation between the acceleration due to gravity and the angular speed. This would be given by,
g′=g−ω2Rg'=g-{{\omega }^{2}}R
Multiplying both sides by mass m we get the weight of the body at the equator as,
mg′=m(g−ω2R)mg'=m\left( g-{{\omega }^{2}}R \right)………………………………………………………. (1)
Now, we have the expression for acceleration due to gravity at a height h above the surface of the earth given by,
g′=g(1−2hR)g'=g\left( 1-\dfrac{2h}{R} \right)
On multiplying both sides by m, we get,
mg′=mg(1−2hR)mg'=mg\left( 1-\dfrac{2h}{R} \right)………………………………………………………… (2)
As per the question, the weight of the body of mass m at equator is equal to that at a height h above the ground. So, we could equate the RHS of both equations (1) and (2),
mg(1−2hR)=m(g−ω2R)mg\left( 1-\dfrac{2h}{R} \right)=m\left( g-{{\omega }^{2}}R \right)
⇒g−ω2R=g−2ghR\Rightarrow g-{{\omega }^{2}}R=g-\dfrac{2gh}{R} ………………………………………………. (3)
∴h=ω2R22g\therefore h=\dfrac{{{\omega }^{2}}{{R}^{2}}}{2g}
Therefore, we found the value of h to be,
h=ω2R22gh=\dfrac{{{\omega }^{2}}{{R}^{2}}}{2g}

So, the correct answer is “Option B”.

Note: Though we are not specifically given that the letter h implies the height at which the mass m is kept. It is quite obvious that h indicates the height from the ground as per the question. You could actually equate the acceleration due to gravity at both these points directly as we know that mass is the same.