
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}$
Answer
569.4k+ views
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
Complete step by step answer:
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.
$\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
Complete step by step answer:
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.
Recently Updated Pages
Why are manures considered better than fertilizers class 11 biology CBSE

Find the coordinates of the midpoint of the line segment class 11 maths CBSE

Distinguish between static friction limiting friction class 11 physics CBSE

The Chairman of the constituent Assembly was A Jawaharlal class 11 social science CBSE

The first National Commission on Labour NCL submitted class 11 social science CBSE

Number of all subshell of n + l 7 is A 4 B 5 C 6 D class 11 chemistry CBSE

Trending doubts
What is meant by exothermic and endothermic reactions class 11 chemistry CBSE

10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

What are Quantum numbers Explain the quantum number class 11 chemistry CBSE

