Question

Assume that a tunnel is dug along a chord of the earth, at a perpendicular distance $\frac{R}{2}$ from the earth's center where R is the radius of the earth. the wall of the tunnel is frictionless. Find the time period.

• A. $2\pi \sqrt{\frac{R}{M}}$
• B. $2\pi \sqrt{\frac{R^3}{GM}}$
• C. $\sqrt{\frac{2R^3}{GM}}$
• D. $2\pi \sqrt{\frac{M}{R}}$

Answer 1

The correct option is B $2\pi \sqrt{\frac{R^3}{GM}}$

The particle will be on the periphery of some circle (let's say of radius $X_1$)

In that case the force of gravity on the particle will be because of the sphere of radius $X_1$ only!

$F=\frac{G{M}_{1}m}{x_1^2}$

If the mass of inner sphere is $M_1$ then $M_1=\frac{\frac{4}{3}\pi x_1^3}{\frac{4}{3}\pi R^3}M$

$=\frac{x_1^{3}M}{R^3}$

$\therefore F=-\frac{G{x}_1^{3}Mm}{R^3{x}_1^2}=-\frac{GM{x}_{1}m}{R^3}$

Its component along O would be F $cos\theta$

$F_0=Fcos\theta =-\frac{GM{x}_{1}m}{R^3\left(\frac{x}{x_1}\right)}=-\frac{GM}{R^3}x\times m$

$a_0=-\frac{GMx}{R^3}$

comparing with $a=-{\omega }^{2}x$ [for a SHM]

${\omega }^2=\frac{GM}{R^3}\Rightarrow \omega =\sqrt{\frac{GM}{R^3}};T=2\pi \sqrt{\frac{R^3}{GM}}$