anonymous
  • anonymous
1 What is the gravitational field strength at a height h above the surface of the Earth? R is the radius of the earth gR2(R+h)2 gR(R+h) Rg(R−h) R2g(R−h)2 2 The instrument used for experimental determination of the value of the universal gravitational constant G is called Galilio’s telescope ballistic pendulum Newton’s balance Cavendish balance 3 At what altitude above the earth's surface would the acceleration due to gravity be 49ms−2? Assume the mean radius of the earth is 64106 metres and the acceleration due to gravity 98ms−2 on the surface of the earth 26106m 323106m 465106m 776106m
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
(a) Gravitational field strength at a height "h" from the surface from the earth is given by \[g' = GM/(R+h)^{2} = GM R ^{2}/R ^{2}(R+h)^{2}=gR ^{2}/(R+h)^{2}\] where is have used g=GM/R^2 ,the gravitational field on the earth's surface. (b) The experiment was performed by Cavendish. (c)For this use the above formula\[g' = gR ^{2}/(R+h)^{2}\] Put g' = g/2 and you can solve for altitude "h" easily since R is given.

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