Question 46: Gravitation

Question

A planet of radius R = $${1 \over {10}} 	imes $$ (radius of Earth) has the same mass density as Earth. Scientists dig a well of
depth $${R \over 5}$$
on it and lower a wire of the same length and of linear mass density 10-3 kg m-1 into it. If the wire is not touching anywhere, the force applied at the top of the wire by a person holding it in place is (take the radius of Earth = 6 $$ 	imes $$ 106 m and the acceleration due to gravity of Earth is 10 ms -2)

Accepted Answer

Given R
e
= 10R = 6 $$	imes$$ 10
6
m and g
e
= 10 m/s
2
. Consider a wire element of length dr placed at a distance r from the centre O.
Let $$ho$$ 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., $$m = {4 \over 3}\pi {r^3}ho $$. 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
$$dF = {{Gm\lambda dr} \over {{r^2}}} = {4 \over 3}G\pi ho \lambda rdr$$.
Integrate dF from $$r = {4 \over 5}R$$ to r = R to get
$$F = {4 \over 3}G\pi ho \lambda \int_{{4 \over 5}R}^R {r\,dr = {4 \over 3}G\pi ho \lambda \left( {{{9{R^2}} \over {50}}} ight)} $$
$$ = {{{4 \over 3}\pi ho R_e^3G} \over {R_e^2}}\left( {{{9\lambda {R_e}} \over {5000}}} ight)$$
$$ = {g_e}{{9\lambda {R_e}} \over {5000}} = {{(10)(9)({{10}^{ - 3}})(6 	imes {{10}^6})} \over {5000}} = 108$$ N.

Answer

96 N

Answer

108 N

Answer

120 N

Answer

150 N

Gravitation

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