physics (Newton"s law of gravitation)

The top of Mt. Everest is 8850 m above sea level. Assume that sea level is at the average Earth radius of 6.38×106 m. What is the magnitude of the gravitational acceleration at the top of Mt. Everest? The mass of the Earth is 5.97×1024 kg.

Thank you.

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1. F = G m M / r^2
a = F/m
a = G M/r^2

a = 6.67*10^-11 (5.97*10^24) / (6,380,000+8,850)^2

a = (6.67*5.97/4.08)10^(-11+24-13)

a = 9.76 * 10^0
= 9.76 m/s^2

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posted by Damon

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