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A remote sensing satellite of earth revolves in a circular orbit at a height of 0.25 x 106 m above the surface of the earth. if earth 's radius is 6.38 x 106 m and g = 9.8 ms-2 then the orbital speed of the  satellite is 

  • 7.76 kms-1

  • 8.56 kms-1

  • 9.13 kms-1

  • 6.67 kms-1


A.

7.76 kms-1

Given, height of a satellite

h = 0.25 x 106

Earth's radius, Re = 6.38 x 106

For the satellite revolving around the  earth, orbital velocity of the satellite



straight v subscript straight o space equals space square root of GM subscript straight e over straight R subscript straight e end root space equals square root of space fraction numerator GM over denominator straight R subscript straight e open square brackets 1 plus space begin display style straight h over straight R subscript straight e end style close square brackets end fraction end root

straight v subscript straight o space equals space square root of fraction numerator gR over denominator 1 plus begin display style straight h over straight R subscript straight e end style end fraction end root

substitutes space the space values space of space straight g comma space straight R subscript straight e space and space straight h comma space we space get
straight v subscript straight o space equals space square root of 60 space straight x space 10 to the power of 6 space end root space straight m divided by straight s
straight v subscript straight o space equals space 7.76 space straight x space 10 cubed space straight m divided by straight s space equals space 7.76 space km divided by straight s 

Given, height of a satellite

h = 0.25 x 106

Earth's radius, Re = 6.38 x 106

For the satellite revolving around the  earth, orbital velocity of the satellite



straight v subscript straight o space equals space square root of GM subscript straight e over straight R subscript straight e end root space equals square root of space fraction numerator GM over denominator straight R subscript straight e open square brackets 1 plus space begin display style straight h over straight R subscript straight e end style close square brackets end fraction end root

straight v subscript straight o space equals space square root of fraction numerator gR over denominator 1 plus begin display style straight h over straight R subscript straight e end style end fraction end root

substitutes space the space values space of space straight g comma space straight R subscript straight e space and space straight h comma space we space get
straight v subscript straight o space equals space square root of 60 space straight x space 10 to the power of 6 space end root space straight m divided by straight s
straight v subscript straight o space equals space 7.76 space straight x space 10 cubed space straight m divided by straight s space equals space 7.76 space km divided by straight s 

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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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