Question

compute the ratio of Earth's Orbital angular momentum and its rotational angular momentum about its Axis given that radius of the earth is 6.4×10^6m and the radius of the orbit of the earth = 1.5 × 10^11m.

Answer 1

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Hope you found below the solution to the asked query

Let M be the mass of the earth. Suppose that it revolves around the sun in a circular orbit of Radius r with an angular velocity ω.

Since the Earth completes its one revolution around the sun in one year, so

          ω = $\frac{2\pi} {T}$ = $\frac{2\pi}{365*24*60*60}$

              = $1.99*10^{-7}rad s^{-1}$

Earth's orbital angular momentum is=>

             $L_{orbit}$ = Mvr = M×rω×r = Mr^2ω

Here,   r = 1.5×10^11m

∴ L_{orbit} = M×(1.5×10^11)^2×1.99×10^{-7}

           = 4.48×10^{15} M

Now, Let's play with the question :)

Let ω' be the angular speed of rotation of  The Earth about its axis.

Since the Earth's complete one rotation about its axis in 24 hours,

                    ω' = $\frac{2\pi}{T'} = \frac{2\pi}{24*60*60}$

                = 7.27 × 10^{-5} rad s ^{-1}

Let R be radius of the Earth. If I is moment of inertia of the earth about the axis, then the Earth's rotational angular momentum,

                      L_{rot} = Iω' = $\frac{2}{5}MR^2$×ω'

Here, R = 6.4×10^6m

∴     L_{rot} = $\frac{2}{5}$

M×(6.4×10^6)^2×7.27×10^{-5} = 1.19×10^9 M

Hence,

$\frac{L_{orbit}}{L_{rot}}$ = $\frac{4.48*10^{15} M}{1.19*10^9M}$

                 = 3.76×10^6

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