ch13-p025

# ch13-p025 - 25. (a) The magnitude of the force on a...

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M = (1.93 × 10 24 kg + 4.01 × 10 24 kg ) = 5.94 × 10 24 kg. The first term is the mass of the core and the second is the mass of the mantle. Thus, () ( ) 11 3 2 24 2 2 6 6.67 10 m /s kg 5.94 10 kg == 9 . 8 4 m / s . 6.345 10 m g a ×⋅ × × (c) A point 25 km below the surface is at the mantle-crust interface and is on the surface of a sphere with a radius of R = 6.345 × 10 6 m. Since the mass is now assumed to be uniformly distributed the mass within this sphere can be found by multiplying the mass per unit volume by the volume of the sphere: 33 (/) , ee M RRM = where M e is the total mass of Earth and R e is the radius of Earth. Thus, 3 6 24 24 6 = 5.98 10 kg = 5.91 10 kg. 6.37 10 m M §· × ×× ¨¸ × ©¹ The acceleration due to gravity is ( ) 11 3 2
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## This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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