physics (Newton"s law of gravitation)

Could anyone please help me with this problem? I would really appreciate it.

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.
Answer ________ N

Thank you.

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asked by Kathy
  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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