Question

The tallest spot on Earth is Mt. Everest, which is 8857 m above sea level. If the radius of the Earth to sea level is 6369 km, how much does the magnitude of g change between sea level and the top of Mt. Everest?

Answer 1

`"Decrease in magnitude of g"~~0.0273m/s^2`

Explanation:

Let

`R->"Radius of the Earth to sea level"= 6369 km=6369000m`

`M->"the mass of the Earth"`

`h->"the height of the tallest spot of"`
`" Mt Everest from sea level"=8857m`

`g->"Acceleration due to gravity of the Earth"`
`" to sea level"= 9.8m/s^2`

`g'->"Acceleration due to gravity to tallest"`
`" "" spot on Earth"`

`G->"Gravitational constant"`

`m->"mass of a body"`

When the body of mass m is at sea level, we can write

`mg=G(mM)/R^2........(1)`

When the body of mass m is at the tallest spot on Everst, we can write

`mg'=G(mM)/(R+h)^2......(2)`

Dividing (2) by (1) we get

`(g')/g=(R/(R+h))^2=(1/(1+h/R))^2`

`=(1+h/R)^(-2)~~1-(2h)/R`
(Neglecting higher power terms of `h/R` as `h/R"<<"1`)

Now `g'=g(1-(2h)/R)`

So change (decrease) in magnitude of g

`Deltag=g-g'=(2hg)/R=(2xx8857xx9.8)/6369000~~0.0273m/s^2`

Answer 2

`approx -.027 m s^(-2)`

Explanation:

Newton's Law for Gravitation

`F = (GMm)/(r^2)`

And `g` is computed at the earth's surface `r_e` as follows:

`m g_e = (GMm)/(r_e^2)`

So `g_e = (GM)/(r_e^2)`

if we were to compute different `g`'s we would get

`g_(everest) - g_(sea) = GM ( 1/(r_(everest)^2) - 1/(r_(sea)^2))`

`GM= 3.986005 times 10^14 m^3 s^(-2)`

`approx 3.986005 times 10^14 * ( 1/(6369000 + 8857)^2) - 1/(6369000^2))`

`approx -.027 m s^(-2)`

Using differentials to double check:

`g_e = (GM)/(r_e^2)`

`implies ln (g_e) = ln( (GM)/(r_e^2)) = ln (GM) - 2 ln (r_e)`

`(dg_e)/(g_e)= - 2 (dr_e)/(r_e)`

`dg_e = - 2 (dr_e)/(r_e) g_e =-2 * 8857/6369000 * 9.81 = -0.027 ms^(-2)`