The acceleration due to gravity, g, is constant at sea level on the Earth\'s sur

Question

The acceleration due to gravity, g, is constant at sea level on the Earth's surface. However, the acceleration decreases as an object moves away from the surface due to the increase in distance from the center of the Earth. Derive an expression for the acceleration due to gravity at a distance h above the surface of the Earth, gh. Express the equation in terms of the radius of the Earth RE, g and h. gh= A 74.35 kg hiker has ascended to a height of 1828 m in the process of climbing Mt. Washington. By what percent has the hiker's weight changed by climbing to this elevation? Use g = 9.807 m/s^2 and RE = 6.371 x 10^6 m. (Hint: Keep 4 significant figures in your weight calculation to find the percent difference.) % W change= Number %

Explanation / Answer

a)

let

mass of the earth is M,

radius of the earth is RE

mass of the object is m

then,

gravitational force on the object is,

F=G*M*m/(RE)^2 (if the object placed at the surface of the earth)

and

g=F/m

g=G*M/(RE)^2

now,

if the object is placed at a distance of h above the surface of earth

then,

gravitational force F=G*M*m/(RE+h)^2

and

gh=F/m

=G*M/(RE+h)^2

=[G*M/(RE)^2)]*(1+h/RE)^2

=g/(1+h/RE)^2

=g*(1+h/RE)^-2

gh=g*(1-2*h/RE)

b)

w=m*g

=74.35*9.8

=728.63 N

and

w'=m*gh

=74.35*(g*(1-2*h/RE))

=74.35*(9.807*(1-2*1828/6.371*10^6))

=728.63(0.9994)

=728.2118 N

change in weight=(728.63-728.2118)

=0.4181 N

% change in weight is=0.0049 N

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