Question
Question: Calculate the change in the value of g at angle of 45 degree. Take radius of earth \( = 6.37 \times ...
Calculate the change in the value of g at angle of 45 degree. Take radius of earth =6.37×103km
A) −0.064m/s2
B) −0.0168m/s2
C) −0.023m/s2
D) −0.0548m/s2
Solution
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
Δg=g′−g=−Rω2cos2θ
Where g’ is the new acceleration due to gravity at an angle of θ
g is the acceleration due to gravity on the surface of earth whose formula is R2GM
R is the radius of earth
ω is the angular frequency
θ 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=6.37×103km Where R is the radius
θ=45o θ 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 24×60×60 seconds
Now w=T2π where T is the time period which is total seconds in a day and ω is the angular frequency mentioned also.
Hence w=24×36002π=7.2×10−5rads−1
Now let us recall the that we have discussed earlier
Δg=−Rω2cos2θ
Putting the values given,
Δg=−(6.37×106)(7.2×10−5)2cos245
On simplifying further, we get,
Δg=1.65×10−2
Which is the required answer, therefore the correct option is B) −0.0168m/s2
Note: We also have different regarding depth and altitude
For depth g′=g(1−Rd) where d is the depth. The formula
For altitude g′=g(1−R2h) 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, x2 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 Rd and R2h have squares, cubes and further powers are very small hence, can be neglected.