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Motion In Straight Line

Question
CBSEENPH11017649

Show that the gravitational potential due to the earth at a height h from the surface of the earth is
negative fraction numerator GM over denominator straight R plus straight h end fraction

Solution
Gravitational potential at any point is defined as the amount of work done in moving a unit mass from infinity to that point.
Mass, m = 1 kg

 
Let a unit mass be placed at P at a distance r from the centre of the earth.
The force acting on unit mass is, 
           straight F equals fraction numerator GM cross times 1 over denominator straight r squared end fraction equals GM over straight r squared 
If unit mass is moved by a distance dr towards O, then work done by gravitational force is,
dW equals straight F with rightwards arrow on top. dr with rightwards arrow on top equals straight F space dr 
    equals GM over straight r squared dr 
Total work done when the unit mass is moved from infinity to a distance R + h from the centre is, 
space space space straight W equals integral straight d space straight W equals integral subscript infinity superscript straight R plus straight h end superscript GM over straight r squared dr 

    equals open square brackets negative GM over straight r close square brackets subscript infinity superscript straight R plus straight h end superscript equals negative GM open square brackets fraction numerator 1 over denominator straight R plus straight h end fraction minus 1 over infinity close square brackets 

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This work done by definition is equal to gravitational potential. 
∴            straight V equals negative fraction numerator GM over denominator straight R plus straight h end fraction
Hence, the result.