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.

g_h = g_s ( R/(R+H))²                                      where s is surface and h is height

g_d = g_s ( (R-d)R)                                       where d is depth

g_h = g_s

g_s ( R/(R+H))² = g_s ( (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 + R_ex - 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.