Problem 5-Gravity We now have an explanation for why the acceleration due to gravity near the surface of the Earth is 9.81 m/s2 a) If we naively treat the Earth as a perfect sphere with radius RE = 6371 km and mass ME = 5.972*1024 kg evenly distributed throughout its volume, what is the acceleration due to gravity standing on its surface? b) Mt Everest has a height of 8848 meters. What would the acceleration due to gravity standing on top of it be, maintaining our assumptions? c) The Marianas Trench reaches a depth of 11000 meters. What would the acceleration due to gravity be standing at its lowest point be, maintaining our assumptions? Solution (a) For standing on earth surface, we must have mg = GmM e /R e 2 => g = GM e /R e 2 So, g =6.67 * 10 -11 * 5.972 * 10 24 /(6371 * 10 3 ) 2 = 9.81 m/s 2 (b) Acceleration due to gravity at the top of mount everest is =g(1- 2h/R e ) = 9.81(1- 2*8848/6371*10 3 )m/s 2 = 9.78 m/s 2 (c) Acceleration due to gravity at the lowest point will be = g ( 1 - d/R e ) = 9.81 * ( 1 - 11000/6371 * 10 3 ) m/s 2 = 9.79 m/s 2 .