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Acceleration due to gravity




Acceleration due to gravity of earth

  • Earth attracts every object lying an its surface towards its centre with a force known as gravitational towards its centre with a force known as gravitational pull or gravity.
  • Whenever force acts on any body it produces acceleration and in case of gravitation this acceleration produced under effect of gravity is known as accelesation due to gravity (g)
  • Value of accelesation due to gravity is independent of mass of the body and its value near surface of earth is 9.8 ms-2
  • Expression for acceleration due to gravity
    Consider mass of earth to be as ME and its redius be RE Suppose a body of mass M (much smaller then that fo earth) is kept at the earth surface. Force eseerted by earth on the body of mass m is
    $F=\frac{GMM_{E}}{R_{E}^{2}}$
    The force for the body due to earth produces acceleration due to gravity
    (g) in the motion of the body. From Newton's Second law of motion
    $f=mg$
    from above equations
    $g=\frac{GM_{E}}{R_{E}^{2}}$
    which is acceleration due to gravity at earth's surface.


Acceleration due to gravity below and above the earth surface


(i)Above earth's surface

  • An object of mass m is placed at hight h above the earth's active in this object is
    $ F=\frac {-GMM_{E}}{(R_{E}+h)^{2}}$
    From this it can be concluded that value of g decreases as distance above surface of earth increases now,
    $g=\frac{GM_{E}}{R_{E}(1+\frac{h}{R_{E}})^{2}}$
    $g=\frac{g_{0}}{(1+\frac{h}{R_{E}})^{2}}$
    Where $g_{0}=\frac{GM_{E}}{R_{E}^{2}}$ value of acceleration due to gravity at earth surface
  • for h<<R
    $g=g_0(1+ \frac {h}{R_E})^{-2}$
    Expanding by Binomial theriom
    $g=g_0(1- \frac {2h}{R_E})$
  • Above equation tells us that for small hight h above surface of earth. value of g decreases by factor $(1-\frac {2h}{R_E})$

(ii)Below the earth's surface
  • If one goes inside the earth surface the value of g again decreases
  • $\rho$ = density of material of earth them
    $m=\frac {4}{3}(R_E)^3 \rho$
    From this acceleration due to gravity at earth's Surface is
    $g_{0}=\frac{G(4/3)R_{E}^{3} \rho}{R_{E}^{2}}$
    $g_0=\frac {4}{3}G R_E \rho$ ---(A)
  • Let g acceleration due to gravity at depth D below earth's surface
    Body at depth d will experience force only due to portion of reduce (RE-d) of earth's
    outer spherical shell of thickness d will not exeperience any force
    M is mass of the portion of earth with radius (RE-d) then
    $g=\frac{GM}{r^{2}}$
    $M=\frac {4}{3}(R_E-d)^3 \rho$
    $g=\frac{G(4/3)(R_{E}-d)^{3}P}{(R_{E}-d)^{2}}$
    $g= \frac {4}{3}G (R_E -d) \rho $ ---(B)
    Dividing epn (B) by (A)
    $ \frac {g}{g_0}= (1- \frac {d}{R_E})$
    or $g=g_0 (1- \frac {d}{R_E})$
  • from above equation is clear that acceeleration due to gravity also decreases with depth.


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