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Suppose a 96.10 kg hiker has ascended to a height of 1841 m above sea level in the process of climbing Mt. Washington. By what percent has the hiker's weight changed from its value at sea level as a result of climbing to this elevation? Use g=9.807 m/s2 and R=6.371×106 m . Keep four significant figures in your weight calculation to find the percent difference, and enter your answer as a positive value.

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Answer:

0.029%

Explanation:

Weight on the surface is given by,

[tex]F_W = mg = (96.1Kg)(9.8m/s^2)=941.78N[/tex]

The acceleration due to gravity at given height is,

[tex]g_h=\frac{gR^2}{R+h^2}[/tex]

The weight in another height is equal to

[tex]F_h=mg_h[/tex]

[tex]F_h= m*\frac{gR^2}{R+h^2}[/tex]

[tex]F_h = \frac{F_WR^2}{(R+h)^2}[/tex]

[tex]F_h=\frac{(941.78)(6.371*10^6)}{(6.371*10^6+1841)}[/tex]

[tex]F_h = 941.50N[/tex]

The change of the Weight is,

[tex]\frac{F-F_h}{F}*100=\frac{941.78-941.50}{941.78}*100=0.029\%[/tex]

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