Given R_e = 10R = 6 $$\times$$ 10⁶ m and g_e = 10 m/s². Consider a wire element of length dr placed at a distance r from the centre O.

Let $$\rho$$ be the common mass density of the planet and the earth and m be the mass of the planet inside the sphere of radius r i.e., $$m = {4 \over 3}\pi {r^3}\rho $$. The gravitational force on the wire element by the planet is equal to the force by a point mass of magnitude m placed at the centre O. Thus, force on the wire element is

$$dF = {{Gm\lambda dr} \over {{r^2}}} = {4 \over 3}G\pi \rho \lambda rdr$$.

Integrate dF from $$r = {4 \over 5}R$$ to r = R to get

$$F = {4 \over 3}G\pi \rho \lambda \int_{{4 \over 5}R}^R {r\,dr = {4 \over 3}G\pi \rho \lambda \left( {{{9{R^2}} \over {50}}} \right)} $$

$$ = {{{4 \over 3}\pi \rho R_e^3G} \over {R_e^2}}\left( {{{9\lambda {R_e}} \over {5000}}} \right)$$

$$ = {g_e}{{9\lambda {R_e}} \over {5000}} = {{(10)(9)({{10}^{ - 3}})(6 \times {{10}^6})} \over {5000}} = 108$$ N.