PLEASE help me!! Part A) The acceleration due to gravity, g, is constant at sea

Question

PLEASE help me!! Part A) 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. PartB) A 96.10 kg hiker has ascended to a height of 1854 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/s2 and RE = 6.371 × 106 m. (Hint: Keep 4 significant figures in your weight calculation to find the percent difference.)

Explanation / Answer

  
we know
   F = mg ---------.1

       F = G Mm/R^2 --------------->2
  

Where F is the force, m is the mass of the body, g is the acceleration due to gravity,
M is the mass of the Earth, R is the radius of the Earth and G is the gravitational constant.

so    g = GM/R^2 when the body is on surface

Let the body be now placed at a height h above the Earth's surface. Let the acceleration due to gravity at that position be g|.

   g' = GM/(R+h)^2------------>3

ratio of g' and g is    g' = g(R/R+h)^2
               = g(1+h/R)^-2
   using binomial theorem

       g' = g(1- 2h/R) ------------>4

b)     mg' = mg - (2mgh/R)
       mg - mg' = (2mgh/R) -------------------->5

lossin weight at height h
                 hiker mass is 96.10 kg, h = 1854 m, g = 9.807 m/s2 ,
       R = 6.371*10^6 m
         
    mg- mg' = 2*96.10*9.807*1854 /(6.371*10^6) = 0.54851kg
         
percentage is 0.052

       mg = 942.4527, mg' = 941.9040 N

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