Question #276930

determine the force of gravitational attraction between the earth 5.98 x 10^24 kg and a 70 kg boy who is standing at sea level, a distance of 6.38 x 10^4 in from earth's center.

Expert's answer

F=(6.67⋅10−11)70(5.98⋅1024)(6.38⋅106)2=686 NF=(6.67\cdot10^{-11})\frac{70(5.98\cdot10^{24})}{(6.38\cdot10^{6})^2}=686\ N


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