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Calculate the distance from the surface of the earth at which the acceleration due to gravity is the same below as well as above the surface of earth.
gh = gs ( R/(R+H))2 where s is surface and h is height
gd = gs ( (R-d)R) where d is depth
gh = gs
gs ( R/(R+H))2 = gs ( (R-d)R)
since H = d
therefore
(R2)R = (R -d)((R+H)2)
R3 = R3 - RH2 + HR2 - H3
H3 = H( R2 - RH )
H2 = R2 - RH
H2 + RH - R2 = 0
Substitute the value of R as 6400 km
then the answer will come as 3200(√6 -1) km.
Hi Arnab,
Let Re be the radius of the earth.
And let x be the distance at which the accn due to gvt is the same (both above and below)
For above,
g = gs*[ Re2/(Re+x)2 ] -----------(1)
For below
g = gs* [ (Re-x)/Re ] --------------(2)
where gs is the accn due to gvt at the surface of the earth.
Both are given to be equal. So (1) = (2) will imply,
[ Re2/(Re+x)2 ] = [ (Re-x)/Re ]
Hence x2 + Rex - Re2 = 0
Which will give x = [(√5 - 1)/2]*Re.
Considering Re to be 6400km, will yield x = 3200{√5 - 1}
Hope this helps.
Best Regards,
Ashwin (IIT Madras).
Please explain in detail how did you derive the second equation, i.e., no. 2.
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