the weight of a body at a height equal to be radius of the earth is 'N'. What will be it's weight at a height equal to three times the radius of the earth?
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Answered by
6
First:
mg¹= N
g¹= GM/(R+R)²= GM/4R²
Second:
New weight=mg2
g2=GM/(R+3R)²=GM/16R²
g¹=4g2
g2=g¹/4
Therefore new weight is N/4
mg¹= N
g¹= GM/(R+R)²= GM/4R²
Second:
New weight=mg2
g2=GM/(R+3R)²=GM/16R²
g¹=4g2
g2=g¹/4
Therefore new weight is N/4
Answered by
1
At height h=R
g'/g = (R/R')² = (R/2R)² = 1/4
g' = g/4
Weight = mg/4 = N
So , mg = 4N
At height h=3R
g'/g = (R/R')² = (R/4R)² = 1/16
g' = g/16
Weight = mg/16 = 4N/16 = N/4
g'/g = (R/R')² = (R/2R)² = 1/4
g' = g/4
Weight = mg/4 = N
So , mg = 4N
At height h=3R
g'/g = (R/R')² = (R/4R)² = 1/16
g' = g/16
Weight = mg/16 = 4N/16 = N/4
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