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Physics Part I

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Physics

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Class 10 Class 12

At what height from the surface of earth the gravitation potential and the value of g are -5.4 x 107 J kg-2 and 6.0 m/s2 respectively? (Take the radius of the earth as 6400 km)

  • 1600 km

  • 1400 km

  • 2000 km

  • 2600 km


D.

2600 km

Gravitational potential at some height h from the surface of the earth is given by,

V = fraction numerator negative GM over denominator straight R plus straight h end fraction    ... (i)
Acceleration due to gravity at some height h from the earth surface can be given as,


straight g apostrophe space equals space fraction numerator GM over denominator left parenthesis straight R plus straight h right parenthesis squared end fraction  ... (ii)
From equation (i) and (ii), we get


space space space space space fraction numerator open vertical bar straight V close vertical bar over denominator straight g apostrophe end fraction equals fraction numerator G M over denominator R plus h end fraction space x space fraction numerator left parenthesis R plus h right parenthesis squared over denominator G M end fraction

rightwards double arrow space fraction numerator open vertical bar straight V close vertical bar over denominator straight g apostrophe end fraction space equals space R plus h    ... (iii)

because V = -54 x 107 J kg-2
g' = 6.0 m/s2

Radius of Earth, R = 6400 km


Putting the values in Eqn. (iii), we get

fraction numerator 5.4 space straight x space 10 to the power of 7 over denominator 6.0 end fraction = R+h
rightwards double arrow space 9 space straight x space 10 to the power of 6 space equals space straight R plus straight h
rightwards double arrowh = (9-6.4)x106 = 2.6 x 106 m

rightwards double arrow h = 2600 km

Gravitational potential at some height h from the surface of the earth is given by,

V = fraction numerator negative GM over denominator straight R plus straight h end fraction    ... (i)
Acceleration due to gravity at some height h from the earth surface can be given as,


straight g apostrophe space equals space fraction numerator GM over denominator left parenthesis straight R plus straight h right parenthesis squared end fraction  ... (ii)
From equation (i) and (ii), we get


space space space space space fraction numerator open vertical bar straight V close vertical bar over denominator straight g apostrophe end fraction equals fraction numerator G M over denominator R plus h end fraction space x space fraction numerator left parenthesis R plus h right parenthesis squared over denominator G M end fraction

rightwards double arrow space fraction numerator open vertical bar straight V close vertical bar over denominator straight g apostrophe end fraction space equals space R plus h    ... (iii)

because V = -54 x 107 J kg-2
g' = 6.0 m/s2

Radius of Earth, R = 6400 km


Putting the values in Eqn. (iii), we get

fraction numerator 5.4 space straight x space 10 to the power of 7 over denominator 6.0 end fraction = R+h
rightwards double arrow space 9 space straight x space 10 to the power of 6 space equals space straight R plus straight h
rightwards double arrowh = (9-6.4)x106 = 2.6 x 106 m

rightwards double arrow h = 2600 km
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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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