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Last updated date: 26th Nov 2023
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# Calculate the change in the value of g at angle of 45 degree. Take radius of earth $= 6.37 \times {10^3}km$A) $- 0.064m/{s^2}$B) $- 0.0168m/{s^2}$C) $- 0.023m/{s^2}$D) $- 0.0548m/{s^2}$

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Hint: The basic approach to solve this question is by applying the formula of variation of acceleration due to gravity with angle made horizontally with the earth. The formula is given below
$\Delta g = g' - g = - R{\omega ^2}{\cos ^2}\theta$
Where g’ is the new acceleration due to gravity at an angle of $\theta$
g is the acceleration due to gravity on the surface of earth whose formula is $\dfrac{{GM}}{{{R^2}}}$
R is the radius of earth
$\omega$ is the angular frequency
$\theta$ is the angle with the horizontal axis of earth or with equator

According to the question the given quantities are
$R = 6.37 \times {10^3}km$ Where $R$ is the radius
$\theta = {45^o}$ $\theta$ is the angle with the horizontal axis of earth or with equator
Now to calculate angular frequency,
In one day we have 24 hours that is $24 \times 60 \times 60$ seconds
Now $w = \dfrac{{2\pi }}{T}$ where T is the time period which is total seconds in a day and $\omega$ is the angular frequency mentioned also.
Hence $w = \dfrac{{2\pi }}{{24 \times 3600}} = 7.2 \times {10^{ - 5}}rad\,{s^{ - 1}}$
Now let us recall the that we have discussed earlier
$\Delta g = - R{\omega ^2}{\cos ^2}\theta$
Putting the values given,
$\Delta g = - (6.37 \times {10^6}){(7.2 \times {10^{ - 5}})^2}{\cos ^2}45$
On simplifying further, we get,
$\Delta g = 1.65 \times {10^{ - 2}}$
Which is the required answer, therefore the correct option is B) $- 0.0168m/{s^2}$

Note: We also have different regarding depth and altitude
For depth $g' = g\left( {1 - \dfrac{d}{R}} \right)$ where d is the depth. The formula
For altitude $g' = g\left( {1 - \dfrac{{2h}}{R}} \right)$ where h is the altitude.
For both the equations above R and G are constant. Both the above formula comes from a assumption or a proof in Binomial theorem in Mathematics which say
For ${(1 + x)^n}$ if, ${x^2}$ and the further terms are very small then, this expression can be assumed as $1 + nx$. Now in both the equations above whether of altitude or of depth $\dfrac{d}{R}$ and $\dfrac{{2h}}{R}$ have squares, cubes and further powers are very small hence, can be neglected.