saplinglearning.com A Lion-YouTube University of t Physics 1201 11/3/2015 11:00
ID: 1349071 • Letter: S
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saplinglearning.com A Lion-YouTube University of t Physics 1201 11/3/2015 11:00 PM A27.6/100 10/29/2015 10:40 AM Grade Print Calulator bed Periodic Table uestion 6 of 18 Map sapling 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, g Express the equation in terms of the radius of the Earth Re. g and h. A 70.00 kg hiker has ascended to a height of 1.880 x 10 m in the process of dlimbing Mt. Washington. By what dinbing lo then devaione 0mat percent has the hiker's weight changed by climbing to this elevation? Use g 9.807 m/s' and RE . 6.3 71 × 1O m. (Hint Keep 4 spificant figures in your weight cal lation to find the percent dierence.) Number % w change-10 Hin MacBook ProExplanation / Answer
g = GMe/Re^2 ...(1)
if it is h height above the earth surface
g' = GMe(Re+h)^2 .....(2)
(2) divide (1)
g' = g( Re/Re+h)^2
g' = g*Re^2/Re^2(1/(1+ h/Re)^2
g' = g(1+h/Re)^-2
by binomial expansion
g' = g(1-2h/Re)
part 2 )
mass at surface =
W = mg
W = 70 x 9.807 = 686.5 N
at 1.880 x 10^3m above the surface
g' = g(1-2h/Re)
g' = 9.801 m/s^2
W' = m*g'
W' = 686.1 N
% change = W'-W/W * 100
% change = 0.0583 %
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