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Motion In Straight Line

Question
CBSEENPH11017688

The gravitational potential due to some distribution of mass is given by straight V equals negative straight k over straight r to the power of 3 divided by 2 end exponent. Find the gravitational field due to that distribution of mass.

Solution

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Gravitational P.E is given by, straight V equals negative straight k over straight r to the power of 3 divided by 2 end exponent 
We know that field the negative gradient of gravitational potential energy. 
∴     straight E equals negative dV over dr equals fraction numerator negative straight d over denominator dr end fraction open parentheses negative straight k over straight r to the power of 3 divided by 2 end exponent close parentheses equals negative fraction numerator 3 straight k over denominator 2 straight r to the power of 5 divided by 2 end exponent end fraction 
This is the required expression for gravitational field due to the distribution of mass.