Question

1. (constancy of "g") First, we ignore rotation. According to the law of gravitation the force per unit mass on an object above Earth's surface has amplitude GM/r2, wherer is the distance from the object to the center of Earth, G is the gravitational constant and M is the mass of the earth. For a point in the atmosphere at elevation z above the surface, we can write r = 2 + a where a is the radius of a (spherical) earth. Then the force per unit mass can be written GM GM 1 1 (1) = g". (1+z/a)2 r T2 a2 (1/a) (here g is a symbol used to denote the fraction а GM/a2 ) (a) Find g* (including its units), assuming the following values: G= 6.673 x 10-11m3kg_}sec2, M = 5.988 x 1024kg, and a = then, is the acceleration due to gravity at the surface (z = 0) Give your answer with just three decimal points: "nxxx" 6370 km. This, (b) How does this answer compare with the value 9.81 m/sec2 we've been using in class for g? (c) Find the value of the expression (1) at a typical tropopause level of z 12 km What, to three decimal places, is the difference between this value and the value at the surface found in part (a)? N CO

Answer 1

- we have a = 6,67810-11 M = 5.98 x 1024 *; 9 - 6370 kr. 6.673x10+11% 5.98 x1024 (6.378 10672 g* = 9.834 M/82 (b) we find that gx is slightly more than the value 3.81 mig? which we have been using forg, (C.) at z=12 km = 12x103 m. expression (1) = g* . 1_ (2 + 2 o putting a = 1,2x108m, a= 6,37 X106 I and gx=9.834 . we get 8(Z) = 9,834% 1 (1+ 1.28104 12 6.378106) → 9(7)= 9.797 m 192.) difference =9(2)- g* -0.0311

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