If g is the acceleration due to gravity on the earth’s surface, the gain in the potential energy of object of mass m raised from the surface of the earth to a height equal to the radius R of the earth is from Physics Gravitation Class 11 Manipur Board
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If g is the acceleration due to gravity on the earth’s surface, the gain in the potential energy of object of mass m raised from the surface of the earth to a height equal to the radius R of the earth is

  • 2 mgR

  • mgR/2

  • mgR/4

  • mgR


B.

mgR/2

Gravitational potential energy of body on earth's surface

straight U space equals space minus space fraction numerator GM subscript straight e straight m over denominator straight R end fraction
At a height h from earth's surface, its value is

straight U subscript straight h space equals space minus space fraction numerator GM subscript straight e straight m over denominator left parenthesis straight R plus straight h right parenthesis end fraction
space equals space fraction numerator GM subscript straight e straight m over denominator 2 straight R end fraction

where Me = mass of earth
m = mass of body
R = radius of earth

therefore comma space Gain space in space potential space energy space equals space straight U subscript straight h minus straight U
equals space fraction numerator negative space GM subscript straight e straight m over denominator 2 straight R end fraction minus open parentheses fraction numerator GM subscript straight e straight m over denominator straight R end fraction close parentheses
space equals space fraction numerator GM subscript straight e straight m over denominator 2 straight R end fraction space plus space fraction numerator GM subscript straight e straight m over denominator straight R end fraction
space equals space fraction numerator GM subscript straight e straight m over denominator 2 straight R end fraction
space equals space fraction numerator gR squared straight m over denominator 2 straight R end fraction space space open parentheses straight g space equals space GM subscript straight e over straight R close parentheses
space equals space 1 half space mgR

Gravitational potential energy of body on earth's surface

straight U space equals space minus space fraction numerator GM subscript straight e straight m over denominator straight R end fraction
At a height h from earth's surface, its value is

straight U subscript straight h space equals space minus space fraction numerator GM subscript straight e straight m over denominator left parenthesis straight R plus straight h right parenthesis end fraction
space equals space fraction numerator GM subscript straight e straight m over denominator 2 straight R end fraction

where Me = mass of earth
m = mass of body
R = radius of earth

therefore comma space Gain space in space potential space energy space equals space straight U subscript straight h minus straight U
equals space fraction numerator negative space GM subscript straight e straight m over denominator 2 straight R end fraction minus open parentheses fraction numerator GM subscript straight e straight m over denominator straight R end fraction close parentheses
space equals space fraction numerator GM subscript straight e straight m over denominator 2 straight R end fraction space plus space fraction numerator GM subscript straight e straight m over denominator straight R end fraction
space equals space fraction numerator GM subscript straight e straight m over denominator 2 straight R end fraction
space equals space fraction numerator gR squared straight m over denominator 2 straight R end fraction space space open parentheses straight g space equals space GM subscript straight e over straight R close parentheses
space equals space 1 half space mgR

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The position co-ordinates of two particles of masses m1 and m2are (x1, y1, z1) and (x2, y2, z2) respectively. Find the coordinates of the centre of mass.

The position vectors of masses m1 and m2 are respectively,

space space space space space space space space space space space space r with rightwards harpoon with barb upwards on top subscript 1 equals space straight x subscript 1 straight i with overbrace on top plus straight y subscript 1 straight j with overbrace on top plus straight z subscript 1 straight k with overbrace on top

a n d space space space space space r with rightwards harpoon with barb upwards on top subscript 2 space equals space straight x subscript 2 straight i with overbrace on top plus straight y subscript 2 straight j with overbrace on top plus straight z subscript 2 straight k with overbrace on top

Let the position coordinates of the centre of mass be (X, Y, Z).

Therefore the position vector of centre of mass is,


R with rightwards harpoon with barb upwards on top equals straight X straight i with overbrace on top plus straight Y straight j with overbrace on top plus straight Z straight k with overbrace on top

Since,    R with rightwards harpoon with barb upwards on top equals fraction numerator straight m subscript 1 stack straight r subscript 1 with rightwards harpoon with barb upwards on top space plus straight m subscript 2 space straight r with rightwards harpoon with barb upwards on top subscript 2 over denominator straight m subscript 1 plus straight m subscript 2 end fraction space

therefore space space space straight X straight i with overbrace on top plus straight Y straight j with overbrace on top plus straight Z straight k with overbrace on top 

equals space space fraction numerator straight m subscript 1 left parenthesis straight x subscript 1 straight i with overbrace on top plus straight y subscript 1 straight j with overbrace on top plus straight z subscript 1 straight k with overbrace on top right parenthesis plus straight m subscript 2 left parenthesis straight x subscript 2 straight i with overbrace on top plus straight y subscript 2 straight j with overbrace on top plus straight z subscript 2 straight k with overbrace on top right parenthesis over denominator straight m subscript 1 plus straight m subscript 2 end fraction

space equals space space fraction numerator open parentheses straight m subscript 1 straight x subscript 1 plus straight m subscript 2 straight x subscript 2 close parentheses over denominator straight m subscript 1 plus straight m subscript 2 end fraction straight i with overbrace on top plus fraction numerator left parenthesis straight m subscript 1 straight y subscript 1 plus straight m subscript 2 straight y subscript 2 over denominator straight m subscript 1 plus straight m subscript 2 end fraction straight j with overbrace on topplus fraction numerator left parenthesis straight m subscript 1 straight z subscript 1 plus straight m subscript 2 straight z subscript 2 over denominator straight m subscript 1 plus straight m subscript 2 end fraction space straight k with overbrace on top

Comparing the coefficients of  straight i with overbrace on top comma space straight j with overbrace on top space and space straight k with overbrace on top, we get

straight X equals fraction numerator left parenthesis straight m subscript 1 straight x subscript 1 plus straight m subscript 2 straight x subscript 2 right parenthesis over denominator straight m subscript 1 space plus space straight m subscript 2 end fraction
straight Y space equals space straight X equals fraction numerator left parenthesis straight m subscript 1 straight y subscript 1 plus straight m subscript 2 straight y subscript 2 right parenthesis over denominator straight m subscript 1 space plus space straight m subscript 2 end fraction
straight z equals fraction numerator left parenthesis straight m subscript 1 straight z subscript 1 plus straight m subscript 2 straight z subscript 2 right parenthesis over denominator straight m subscript 1 space plus space straight m subscript 2 end fraction 


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