Question

Derive the concept of gravity.
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Answer 1

A knowledge of acceleration due to gravity enables us to determine the following three quantities, namely_

(i) Mass of the earth

(ii) Density of the earth

(iii) Mass of the sun

Now let's come to the point of derivation_

(i) Mass of the earth:

                                Mass 'M' of the earth can be calculated fro the equation (9) of universal gravitational constant (G).

Such that,

$M = \frac{gR^{2} }{G} \\\\Since,\\g = 9.8 ms^{-2} \\\\R= 6.38*10^{6} \\\\G= 6.67*10^{-11}$

∵ Metric system

$M= \frac{9.8*10^{6} }{6.67*10^{-11} } \\\\= 5.98 * 10^{24}\ kg$

So, the mass of the earth is 5.98*10²⁴ kg.

(ii) Density of the earth:

                                   If ' ρ' is the mean density of the earth then,

 $M = V*ρ\\= (4/3)πR³ρ$                            

Substituting for M in equation (9),

$g = G*(4/3)*πR³ρ\\⇒ g= (4/3)πGRP$

$ρ=3g/(4πGR)\\=(3*9.8)/ (4*3.142*6.667*10^-11 *6.38 * 10^6)\\= 5497 kg / m³$

So, the density of the earth is 5497 kg/m³

(iii) Mass of the sun:

                               If 'm' be the mass of the sun, gravitational attraction between the sun, and the earth is_

$F₈ = GMm/r²$

Where 'r' is the radius of orbit around the sun and M is the mass  of the earth. Centripital force 'Fc' required by the earth is_

$Fc= Mrω²$

Where, 'ω'  is the angular velocity of the earth.

In equlibrium,

$Fc =Fg\\Mrω²=GMm/r²$

$m= r³ω²/G\\=r³ω²/G\\=(r³/G)(2π/T)²\\=(4π²r³/GT²)$

Where 'T' is the time period of rotation of earth around the sun.

Here,

r = 150 million km

 = 150*10 ⁶ km

or, r = 1.5*10⁸ km

     = 1.5*10¹¹ km

G = 6.67*10⁻¹¹ M.K.S. units

   = 365*24*60*60s

$m ={4*(3.142)²*(1.5*10¹¹)³}/{6.67*10⁻¹¹*(365*24*3600)²}\\    = 2*10³⁰ kg$

So, the mass of the sun is 2*10³⁰ kg.