Suppose the charge of a proton and an electron differ slightly. One of them is –e, the other is (e + Δe). If the net of electrostatic force and gravitational force between two hydrogen atoms placed at a distance d (much greater than atomic size) apart is zero, then Δe is of the order of [Given mass of hydrogenmh = 1.67 × 10–27 kg] from Physics Gravitation Class 11 Manipur Board
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Suppose the charge of a proton and an electron differ slightly. One of them is –e, the other is (e + Δe). If the net of electrostatic force and gravitational force between two hydrogen atoms placed at a distance d (much greater than atomic size) apart is zero, then Δe is of the order of [Given mass of hydrogen
mh = 1.67 × 10–27 kg]

  • 10–20 C

  • 10–23 C

  • 10–37 C

  • 10–47 C


C.

10–37 C

Fe = Fg

fraction numerator 1 over denominator 4 πε subscript 0 end fraction fraction numerator increment straight e squared over denominator straight d squared end fraction space equals space Gm squared over straight d squared
9 space straight x space 10 to the power of 9 space left parenthesis increment straight e squared right parenthesis space equals space 6.67 space straight x space 10 to the power of negative 11 end exponent space straight x space 1.67
straight x space 10 to the power of negative 27 end exponent space space straight x space 1.67 space straight x space 10 to the power of negative 27 end exponent
increment straight e squared space equals space fraction numerator 6.67 space straight x space 1.67 space straight x space 1.67 over denominator 9 end fraction space straight x space 10 to the power of negative 74 end exponent
increment straight e almost equal to space 10 to the power of negative 37 end exponent

Fe = Fg

fraction numerator 1 over denominator 4 πε subscript 0 end fraction fraction numerator increment straight e squared over denominator straight d squared end fraction space equals space Gm squared over straight d squared
9 space straight x space 10 to the power of 9 space left parenthesis increment straight e squared right parenthesis space equals space 6.67 space straight x space 10 to the power of negative 11 end exponent space straight x space 1.67
straight x space 10 to the power of negative 27 end exponent space space straight x space 1.67 space straight x space 10 to the power of negative 27 end exponent
increment straight e squared space equals space fraction numerator 6.67 space straight x space 1.67 space straight x space 1.67 over denominator 9 end fraction space straight x space 10 to the power of negative 74 end exponent
increment straight e almost equal to space 10 to the power of negative 37 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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