a) [tex]9.80 m/s^2[/tex]
The acceleration due to gravity at a certain location on Earth is given by
[tex]g=\frac{GM}{(R+h)^2}[/tex]
where
G is the gravitational constant
M is the Earth's mass
R is the Earth's radius
h is the altitude above the Earth's surface
At the top of Mt. Everest,
R = 6380 km = [tex]6.38\cdot 10^6 m[/tex]
[tex]h' = 8848 m[/tex]
[tex]g'=9.77 m/s^2[/tex]
With
[tex]g'=\frac{GM}{(R+h')^2}[/tex] (1)
At the Earth's surface,
R = 6380 km = [tex]6.38\cdot 10^6 m[/tex]
h = 0
g = ?
So
[tex]g=\frac{GM}{R^2}[/tex] (2)
By doing the ratio (2)/(1), we find an expression for g in terms of g':
[tex]\frac{g}{g'}=\frac{\frac{GM}{R^2}}{\frac{GM}{(R+h')^2}}=\frac{(R+h')^2}{R^2}=\frac{(6.38\cdot 10^6+8848)^2}{(6.38\cdot 10^6)^2}=1.003[/tex]
And therefore,
[tex]g=1.009g'=1.009(9.77)=9.80 m/s^2[/tex]
b) 519.3 N
The weight of an object near the Earth's surface is given by
[tex]W=mg[/tex]
where
m is the mass of the object
g is the acceleration of gravity at the object's location
In this problem,
m = 50 kg is the mass of the object
g' = 9.77 m/s^2 is the acceleration of gravity on top of Mt Everest
Susbtituting,
[tex]W=(50)(9.77)=519.3 N[/tex]
