A sensitive gravimeter at a mountain observatory finds that the free-fall acceleration is 0.0090m/s2 less than that at sea level (gsealevel = 9.83 m/s2).
can somsone guide me with this problem pleaseee?
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From g' = g(r/R)^2 and e = g - g' = g(1 - (r/R)^2 we find e/g + (r/R)^2 = 1 so that r/R = sqrt(1 - e/g) and R = r/sqrt(1 - (e/g) = r/sqrt(1 - .009/9.83) = r/0.999542113 = 1.000458097 r. So the altitude is H = R - r = .000458 r where r is Earth's average radius. ANS. If you need H in meters look up the average Earth radius r.