a sensitive gravimeter at a mountain observatory finds that the free-fall accele
ID: 1372320 • Letter: A
Question
a sensitive gravimeter at a mountain observatory finds that the free-fall acceleration is 0.0065 m/s^2 less than that at sea level. (gsea level = 9.83 m/s^2) What is the observatory altitude? Assume Rearth = 6.37x10^6m h = ?I've looked up similar problems, but I don't understand how the answer were found. There were some numbers I didn't understand where they came from. Please help me ! a sensitive gravimeter at a mountain observatory finds that the free-fall acceleration is 0.0065 m/s^2 less than that at sea level. (gsea level = 9.83 m/s^2) What is the observatory altitude? Assume Rearth = 6.37x10^6m h = ?
I've looked up similar problems, but I don't understand how the answer were found. There were some numbers I didn't understand where they came from. Please help me ! a sensitive gravimeter at a mountain observatory finds that the free-fall acceleration is 0.0065 m/s^2 less than that at sea level. (gsea level = 9.83 m/s^2) What is the observatory altitude? Assume Rearth = 6.37x10^6m h = ?
I've looked up similar problems, but I don't understand how the answer were found. There were some numbers I didn't understand where they came from. Please help me !
Explanation / Answer
use accleration due to gravity g = GM/R^2
here g' = GM/(R+h)^2
use g' = g(1-2 h/r)
1-2h/R = 0.0065/9.81 = 6.62 e -4
2h/R = 1- 6.62 e -4
h = 0.999338 * 6.37 e 6/2
h = 3.18 e 6 m
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