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Gravitation

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Physics Part I

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Physics

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Class 10 Class 12

A satellite of mass m revolves around the earth of radius R at a height x from its surface. If g is the acceleration due to gravity on the surface of the earth, the orbital speed of the satellite is

  • fraction numerator gR over denominator straight R minus straight x end fraction
  • gx

  • fraction numerator gR squared over denominator straight R space plus straight x end fraction
  • open parentheses fraction numerator gR squared over denominator straight R space plus straight x end fraction close parentheses to the power of 1 divided by 2 end exponent

D.

open parentheses fraction numerator gR squared over denominator straight R space plus straight x end fraction close parentheses to the power of 1 divided by 2 end exponent

The gravitational force exerted on satellite at a height x is

straight F subscript straight G space equals space fraction numerator GM subscript straight e straight m over denominator left parenthesis straight R plus straight x right parenthesis squared end fraction
where Me = mass of earth Since, gravitational force provides the necessary centripetal force, so,

fraction numerator GM subscript straight e straight m over denominator left parenthesis straight R plus straight x right parenthesis squared end fraction space equals space fraction numerator mv subscript 0 superscript 2 over denominator left parenthesis straight R plus straight x right parenthesis end fraction
rightwards double arrow space fraction numerator GM subscript straight e straight m over denominator left parenthesis straight R plus straight x right parenthesis end fraction space equals space mv subscript 0 superscript 2
rightwards double arrow space fraction numerator begin display style gR squared straight m end style over denominator left parenthesis straight R plus straight x right parenthesis end fraction space equals space mv subscript 0 superscript 2
rightwards double arrow space fraction numerator begin display style gR squared straight m end style over denominator left parenthesis straight R plus straight x right parenthesis end fraction space equals space mv subscript 0 superscript 2 space
open parentheses because space straight g space equals space fraction numerator begin display style GM subscript straight e end style over denominator straight R end fraction close parentheses
straight v subscript straight o space equals space square root of open square brackets fraction numerator gR squared over denominator left parenthesis straight R plus straight x right parenthesis end fraction close square brackets end root
space equals space open square brackets fraction numerator gR squared over denominator left parenthesis straight R plus straight x right parenthesis end fraction close square brackets to the power of 1 divided by 2 end exponent

The gravitational force exerted on satellite at a height x is

straight F subscript straight G space equals space fraction numerator GM subscript straight e straight m over denominator left parenthesis straight R plus straight x right parenthesis squared end fraction
where Me = mass of earth Since, gravitational force provides the necessary centripetal force, so,

fraction numerator GM subscript straight e straight m over denominator left parenthesis straight R plus straight x right parenthesis squared end fraction space equals space fraction numerator mv subscript 0 superscript 2 over denominator left parenthesis straight R plus straight x right parenthesis end fraction
rightwards double arrow space fraction numerator GM subscript straight e straight m over denominator left parenthesis straight R plus straight x right parenthesis end fraction space equals space mv subscript 0 superscript 2
rightwards double arrow space fraction numerator begin display style gR squared straight m end style over denominator left parenthesis straight R plus straight x right parenthesis end fraction space equals space mv subscript 0 superscript 2
rightwards double arrow space fraction numerator begin display style gR squared straight m end style over denominator left parenthesis straight R plus straight x right parenthesis end fraction space equals space mv subscript 0 superscript 2 space
open parentheses because space straight g space equals space fraction numerator begin display style GM subscript straight e end style over denominator straight R end fraction close parentheses
straight v subscript straight o space equals space square root of open square brackets fraction numerator gR squared over denominator left parenthesis straight R plus straight x right parenthesis end fraction close square brackets end root
space equals space open square brackets fraction numerator gR squared over denominator left parenthesis straight R plus straight x right parenthesis end fraction close square brackets to the power of 1 divided by 2 end exponent

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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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What is Heliocentric theory?

According to the Heliocentric theory, the sun is at the centre and various planets revolve around the sun at their axis. 
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What is the celestial sphere?

At night, if we see the planets and the stars in the sky, all appear to lie in the hemisphere (rest of the hemisphere we are unable to see because of being on the other side of the earth). This sphere is called the celestial sphere.
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What is Geocentric theory?

According to the geocentric theory, all the astronomical bodies like the moon, the sun and stars revolve around the earth, and the earth is at the centre of the universe. 
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Is Geodesic always a straight line?

No, Geodesic is a straight line if and only if,  the two points lie on the flat surface. If the two points lie on the curved surface then it is a curved line.
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