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Sir,The gravitational due to a uniform disc at a point in its axis is 2GMr/R²((1/r)-(1/√(r²+R²))).Andthe potential at a point in the axis is -2GM/R²(√(R²+r²)-r) where R-radius and r-distance in the axis. Also,We know that on differentiating gravitational potential we get the gravitational field at that point.But that is not the case for disc.On integrating the potential we are not getting the value of the field.Why is it so? Please help
Asked by das.atom1966 | 16 May, 2022, 05:34: PM
answered-by-expert Expert Answer
Your expression for gravitational force per unit mass due uniform disc is given as
 
begin mathsize 14px style F space equals space fraction numerator 2 G M over denominator R squared end fraction cross times r cross times open parentheses 1 over r space minus space fraction numerator 1 over denominator square root of R squared plus r squared end root end fraction close parentheses end style  .................................(1)
 
where G is universal gravitational constant, M is mass of disc , R is radius of disk and
r is ditance from centre of disc to the point where we are finding gravitational force
 
Above expression is simplified as
 
begin mathsize 14px style F space equals space fraction numerator 2 G M over denominator R squared end fraction cross times open parentheses 1 space minus space fraction numerator r over denominator square root of R squared plus r squared end root end fraction close parentheses end style ..................................(2)
 
Potential at poit r ,  begin mathsize 14px style V space left parenthesis r right parenthesis space equals space integral subscript 0 superscript r F left parenthesis r apostrophe right parenthesis d r apostrophe end style
begin mathsize 14px style V left parenthesis r right parenthesis space equals space fraction numerator 2 space G space M over denominator R squared end fraction integral subscript 0 superscript r open parentheses 1 space minus space fraction numerator r apostrophe over denominator square root of R squared plus r apostrophe squared end root end fraction close parentheses space d r apostrophe end style
 
By above integration we get
 
begin mathsize 14px style V left parenthesis r right parenthesis space equals space minus space fraction numerator 2 G M over denominator R squared end fraction open parentheses square root of R squared plus r squared end root space minus space r close parentheses end style ..................................(3)
If you want to get Force F(r) from potential , then F(r) = -dV/dr
 
Hence we get gravitational force by differentiating eqn.(3)
 
begin mathsize 14px style F left parenthesis r right parenthesis space equals space minus fraction numerator d V over denominator d r end fraction space equals space fraction numerator 2 G M over denominator R squared end fraction open parentheses 1 space minus space fraction numerator r over denominator square root of R squared plus r squared end root end fraction close parentheses end style
Answered by Thiyagarajan K | 16 May, 2022, 09:34: PM
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Sir,The gravitational due to a uniform disc at a point in its axis is 2GMr/R²((1/r)-(1/√(r²+R²))).Andthe potential at a point in the axis is -2GM/R²(√(R²+r²)-r) where R-radius and r-distance in the axis. Also,We know that on differentiating gravitational potential we get the gravitational field at that point.But that is not the case for disc.On integrating the potential we are not getting the value of the field.Why is it so? Please help
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