The acceleration due to gravity at a height ( take g= 10 `ms^-2` on the earth surface ) above earth,s surface is 9 ms^-2 . It value at a point , at an equal distance below the surface of the earth is

Options

A. `9.5ms ^-2`
B. `8.5ms ^-2`
C. `7.5 ms ^-2`
D. `5.5 ms^-2`

Answer

A

Explanation

Acceleration above the earht surface ,
`g_1 = g((1-(2h))/(R_e))`
Acceleration below the earth surface ,
`g_2 = g((1-(d))/(R_e))`
From equation (1) , 9=10 `((1-(2h))/(R_e))` `(therefore g = 10 ms^-2)`
`9/10 = 1-(2h)/ (R_e)`
`(2h)/(R_e) = 1-(9)/(10)`
`therefore h= (R_e)/20 `
From equation (2) , `g_2 = g((1-(h))/(R_e))`
= 10 `((1-((R_e))/20 )/R_e))`
10`(1-(1)/(20))`
`10xx (19)/(20)`
`therefore g_2 = 9.5 ms^-2`