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?

2 Answers
Oct 11, 2016

Answer:

#"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#

Oct 11, 2016

Answer:

#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)#